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# Common Core: High School - Statistics and Probability : Using Probability to Make Decisions

## Example Questions

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### Example Question #1 : Using Probability To Make Decisions

A researcher observes a road that splits into two paths that lead to two destinations. The researcher believes that the path that curves to the right is the better path but wants to know if a random sample of people feels the same way. The researcher surveyed people walking down the isolated street that split into two directions. He decided to question ten random people in order to determine whether they would travel on the path that curved to the right or to the left. If the respondents' answers are random, then does the probability distribution graph follow the pattern of a normal distribution?

No, it is skewed to the right.

No, it is skewed to the left.

Yes

Cannot be determined

Yes

Explanation:

In order to solve this problem, we need to discuss probabilities and the generation of probability distribution models. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

Now, let's convert this into a percentage:

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur.

Now that we understand probabilities in a general sense, we need to determine how we can create a probability distribution graph for a probability model. We will use the following equation to calculate the probabilities to be used in this graphical display:

Remember that in combinations and permutations a combination is calculated using the following formula:

Now, we can write the following formula.

In this formula variables are defined in the following manner:

Let's investigate this standard through the use of an example. Suppose a researcher rolls a die twelve times and notes whether the die rolls on an even or an odd number. If the die is fair (i.e. every number has an equal probability of being rolled or each roll is random), then would the probability distribution graph follow the pattern of a normal distribution? Lets create a table and solve for each variable. The probability of rolling an even number—a two, four, or a six— is three out of six or fifty percent. Likewise, the probability of failure (i.e. rolling an odd number) is three out of six or fifty percent. Next, we need to list the number of successes for each event—variable . The researcher can roll an even number every time he rolls and he may not even roll an even number in all twelve trial. We need to calculate this probability for every possible number of successful events; therefore, the number of successes ranges from zero to twelve. Last, we know that there are a total of twelve trials. We have solved for the probability of each of these variables for every possible number of successes in the trials.

Once this data is tabulated we can graph the probability of rolling an even number. If we look at the graph then we can see if it follows the bell shape curve of a normal distribution.

A bell curve is shaped like the following image:

We can quickly tell that the graph of the probability distribution does follow the shape of a normal or "bell" curve.

Let's use this information to solve the question. If we look at the question, then we know that the probability of turning right or left in a random situation is fifty percent; therefore, the probability of success (i.e. the right path) or failure (i.e. the left path) is one half or fifty percent. Next we know that there were ten people questioned—or ten trials—and the number of successes per trial ranged from zero to ten. This information has been tabulated and the probability of choosing the left or right has been calculated.

We can now graph the probabilities.

We can see that the graph follows a normal distribution with a characteristic bell shape.

### Example Question #1 : Expected Value Of Random Variable: Ccss.Math.Content.Hss Md.A.2

Find the expected value of Set A.

25

19

23.5

8

16

23.5

Explanation:

### Example Question #1 : Expected Value Of Random Variable: Ccss.Math.Content.Hss Md.A.2

An archer is shooting in a bullseye competition. His target is similar to the bullseye image provided.

The provided table contains the average probability of the archer hitting a particular score at twenty yards at any given time. What is the expected value of each of the archer's shots if he obtains the following series of scores:

Cannot be determined

Explanation:

In order to solve this problem, we need to discuss probabilities and how to calculate expected values. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

Now, let's convert this into a percentage:

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur.

According to this logic, we would expect to roll a particular number on a die one out of every six rolls; however, we may roll the same number multiple times or not at all in six rolls. This discrepancy creates a difference between the expected mean and the actual mean. The expected mean is a hypothetical calculation that assumes a very large sample size and no intervening variables (such as differing forces on the roll and changes in the friction of the surface the die is rolled upon between rolls). Under these conditions we can calculate that any number on the die in a "perfect world" should be one out of six. On the other hand the true or actual mean is calculated using "real" date. In these calculations we would roll a die a particular number of times and use it to develop a probability of rolling a particular number. It is important to note that, theoretically, actual means will eventually equal expected means over a large—or near infinite—amount of trials. In other words, over many many rolls we would eventually find that each number on the die has a one in six chance of being rolled.

Let's discuss how expected means are determined. The expected mean is calculated using the following formula:

In this equation the variables are identified as the following:

Let's use this information to solve the problem. In order to solve the archer problem, we need to use the expected mean formula. We will substitute each of the archer's scores with its respective probability and solve.

It is important to note that we cannot solve this equation using the following formula:

This is because each shot has a different probability. The archer does not consistently score each value because the values occupy different positions on the target; therefore, there is a different probability for each shot not one out of six as the incorrect formula assumes.

### Example Question #4 : Using Probability To Make Decisions

An archer is shooting in a bullseye competition. His target is similar to the bullseye image provided.

The provided table contains the average probability of the archer hitting a particular score at twenty yards at any given time. What is the expected value of each of the archer's shots if he obtains the following series of scores:

Explanation:

n order to solve this problem, we need to discuss probabilities and how to calculate expected values. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

Now, let's convert this into a percentage:

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur.

According to this logic, we would expect to roll a particular number on a die one out of every six rolls; however, we may roll the same number multiple times or not at all in six rolls. This discrepancy creates a difference between the expected mean and the actual mean. The expected mean is a hypothetical calculation that assumes a very large sample size and no intervening variables (such as differing forces on the roll and changes in the friction of the surface the die is rolled upon between rolls). Under these conditions we can calculate that any number on the die in a "perfect world" should be one out of six. On the other hand the true or actual mean is calculated using "real" date. In these calculations we would roll a die a particular number of times and use it to develop a probability of rolling a particular number. It is important to note that, theoretically, actual means will eventually equal expected means over a large—or near infinite—amount of trials. In other words, over many many rolls we would eventually find that each number on the die has a one in six chance of being rolled.

Let's discuss how expected means are determined. The expected mean is calculated using the following formula:

In this equation the variables are identified as the following:

Let's use this information to solve the problem. In order to solve the archer problem, we need to use the expected mean formula. We will substitute each of the archer's scores with its respective probability and solve.

It is important to note that we cannot solve this equation using the following formula:

This is because each shot has a different probability. The archer does not consistently score each value because the values occupy different positions on the target; therefore, there is a different probability for each shot not one out of six as the incorrect formula assumes.

### Example Question #5 : Using Probability To Make Decisions

An archer is shooting in a bullseye competition. His target is similar to the bullseye image provided.

The provided table contains the average probability of the archer hitting a particular score at twenty yards at any given time. What is the expected value of each of the archer's shots if he obtains the following series of scores:

Explanation:

n order to solve this problem, we need to discuss probabilities and how to calculate expected values. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

Now, let's convert this into a percentage:

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur.

According to this logic, we would expect to roll a particular number on a die one out of every six rolls; however, we may roll the same number multiple times or not at all in six rolls. This discrepancy creates a difference between the expected mean and the actual mean. The expected mean is a hypothetical calculation that assumes a very large sample size and no intervening variables (such as differing forces on the roll and changes in the friction of the surface the die is rolled upon between rolls). Under these conditions we can calculate that any number on the die in a "perfect world" should be one out of six. On the other hand the true or actual mean is calculated using "real" date. In these calculations we would roll a die a particular number of times and use it to develop a probability of rolling a particular number. It is important to note that, theoretically, actual means will eventually equal expected means over a large—or near infinite—amount of trials. In other words, over many many rolls we would eventually find that each number on the die has a one in six chance of being rolled.

Let's discuss how expected means are determined. The expected mean is calculated using the following formula:

In this equation the variables are identified as the following:

Let's use this information to solve the problem. In order to solve the archer problem, we need to use the expected mean formula. We will substitute each of the archer's scores with its respective probability and solve.

It is important to note that we cannot solve this equation using the following formula:

This is because each shot has a different probability. The archer does not consistently score each value because the values occupy different positions on the target; therefore, there is a different probability for each shot not one out of six as the incorrect formula assumes.

### Example Question #6 : Using Probability To Make Decisions

An archer is shooting in a bullseye competition. His target is similar to the bullseye image provided.

The provided table contains the average probability of the archer hitting a particular score at twenty yards at any given time. What is the expected value of each of the archer's shots if he obtains the following series of scores:

Explanation:

In order to solve this problem, we need to discuss probabilities and how to calculate expected values. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

Now, let's convert this into a percentage:

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur.

According to this logic, we would expect to roll a particular number on a die one out of every six rolls; however, we may roll the same number multiple times or not at all in six rolls. This discrepancy creates a difference between the expected mean and the actual mean. The expected mean is a hypothetical calculation that assumes a very large sample size and no intervening variables (such as differing forces on the roll and changes in the friction of the surface the die is rolled upon between rolls). Under these conditions we can calculate that any number on the die in a "perfect world" should be one out of six. On the other hand the true or actual mean is calculated using "real" date. In these calculations we would roll a die a particular number of times and use it to develop a probability of rolling a particular number. It is important to note that, theoretically, actual means will eventually equal expected means over a large—or near infinite—amount of trials. In other words, over many many rolls we would eventually find that each number on the die has a one in six chance of being rolled.

Let's discuss how expected means are determined. The expected mean is calculated using the following formula:

In this equation the variables are identified as the following:

Let's use this information to solve the problem. In order to solve the archer problem, we need to use the expected mean formula. We will substitute each of the archer's scores with its respective probability and solve.

It is important to note that we cannot solve this equation using the following formula:

This is because each shot has a different probability. The archer does not consistently score each value because the values occupy different positions on the target; therefore, there is a different probability for each shot not one out of six as the incorrect formula assumes.

### Example Question #7 : Using Probability To Make Decisions

A list of individuals and the respective size of their motorcycles in cubic inches is provided. If you were to randomly select any one person from the list, then what size would you expect their motorcycle to be?

Cannot be determined

Explanation:

In order to solve this problem, we need to discuss probabilities and how to calculate expected values. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

Now, let's convert this into a percentage:

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur.

According to this logic, we would expect to roll a particular number on a die one out of every six rolls; however, we may roll the same number multiple times or not at all in six rolls. This discrepancy creates a difference between the expected mean and the actual mean. The expected mean is a hypothetical calculation that assumes a very large sample size and no intervening variables (such as differing forces on the roll and changes in the friction of the surface the die is rolled upon between rolls). Under these conditions we can calculate that any number on the die in a "perfect world" should be one out of six. On the other hand the true or actual mean is calculated using "real" date. In these calculations we would roll a die a particular number of times and use it to develop a probability of rolling a particular number. It is important to note that, theoretically, actual means will eventually equal expected means over a large—or near infinite—amount of trials. In other words, over many many rolls we would eventually find that each number on the die has a one in six chance of being rolled.

Let's discuss how expected means are determined. The expected mean is calculated using the following formula:

In this equation the variables are identified as the following:

Let's use this information to solve the problem. In order to solve the motorcycle problem, we need to use the expected mean formula. We will substitute each of the motorcycle owner's engine size with its respective probability and solve.

Each of the values have the same probability of being chosen—one out of eleven. As a result, we can simplify this equation:

Round to the nearest one's place.

Note that we can solve this problem in this simplified manner because all of the people had an equal probability of being chosen. If his or her probabilities differed, then we would need to substitute in each person’s respective probability of being chosen.

### Example Question #8 : Using Probability To Make Decisions

A list of individuals and the respective size of their motorcycles in cubic inches is provided. If you were to randomly select any one person from the list, then what size would you expect their motorcycle to be?

Cannot be determined

Explanation:

In order to solve this problem, we need to discuss probabilities and how to calculate expected values. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

Now, let's convert this into a percentage:

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur.

According to this logic, we would expect to roll a particular number on a die one out of every six rolls; however, we may roll the same number multiple times or not at all in six rolls. This discrepancy creates a difference between the expected mean and the actual mean. The expected mean is a hypothetical calculation that assumes a very large sample size and no intervening variables (such as differing forces on the roll and changes in the friction of the surface the die is rolled upon between rolls). Under these conditions we can calculate that any number on the die in a "perfect world" should be one out of six. On the other hand the true or actual mean is calculated using "real" date. In these calculations we would roll a die a particular number of times and use it to develop a probability of rolling a particular number. It is important to note that, theoretically, actual means will eventually equal expected means over a large—or near infinite—amount of trials. In other words, over many many rolls we would eventually find that each number on the die has a one in six chance of being rolled.

Let's discuss how expected means are determined. The expected mean is calculated using the following formula:

In this equation the variables are identified as the following:

Let's use this information to solve the problem. In order to solve the motorcycle problem, we need to use the expected mean formula. We will substitute each of the motorcycle owner's engine size with its respective probability and solve.

Each of the values have the same probability of being chosen—one out of eleven. As a result, we can simplify this equation:

Round to the nearest one's place.

Note that we can solve this problem in this simplified manner because all of the people had an equal probability of being chosen. If his or her probabilities differed, then we would need to substitute in each person’s respective probability of being chosen.

### Example Question #2 : Using Probability To Make Decisions

A list of individuals and the respective size of their motorcycles in cubic inches is provided. If you were to randomly select any one person from the list, then what size would you expect their motorcycle to be?

Cannot be determined

Explanation:

In order to solve this problem, we need to discuss probabilities and how to calculate expected values. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

Now, let's convert this into a percentage:

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur.

According to this logic, we would expect to roll a particular number on a die one out of every six rolls; however, we may roll the same number multiple times or not at all in six rolls. This discrepancy creates a difference between the expected mean and the actual mean. The expected mean is a hypothetical calculation that assumes a very large sample size and no intervening variables (such as differing forces on the roll and changes in the friction of the surface the die is rolled upon between rolls). Under these conditions we can calculate that any number on the die in a "perfect world" should be one out of six. On the other hand the true or actual mean is calculated using "real" date. In these calculations we would roll a die a particular number of times and use it to develop a probability of rolling a particular number. It is important to note that, theoretically, actual means will eventually equal expected means over a large—or near infinite—amount of trials. In other words, over many many rolls we would eventually find that each number on the die has a one in six chance of being rolled.

Let's discuss how expected means are determined. The expected mean is calculated using the following formula:

In this equation the variables are identified as the following:

Let's use this information to solve the problem. In order to solve the motorcycle problem, we need to use the expected mean formula. We will substitute each of the motorcycle owner's engine size with its respective probability and solve.

Each of the values have the same probability of being chosen—one out of eleven. As a result, we can simplify this equation:

Round to the nearest one's place.

Note that we can solve this problem in this simplified manner because all of the people had an equal probability of being chosen. If his or her probabilities differed, then we would need to substitute in each person’s respective probability of being chosen.

### Example Question #10 : Using Probability To Make Decisions

An archer is shooting in a bullseye competition. His target is similar to the bullseye image provided.

The provided table contains the average probability of the archer hitting a particular score at twenty yards at any given time. What is the expected value of each of the archer's shots if he obtains the following series of scores:

Explanation:

In order to solve this problem, we need to discuss probabilities and how to calculate expected values. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

Now, let's convert this into a percentage:

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur.

According to this logic, we would expect to roll a particular number on a die one out of every six rolls; however, we may roll the same number multiple times or not at all in six rolls. This discrepancy creates a difference between the expected mean and the actual mean. The expected mean is a hypothetical calculation that assumes a very large sample size and no intervening variables (such as differing forces on the roll and changes in the friction of the surface the die is rolled upon between rolls). Under these conditions we can calculate that any number on the die in a "perfect world" should be one out of six. On the other hand the true or actual mean is calculated using "real" date. In these calculations we would roll a die a particular number of times and use it to develop a probability of rolling a particular number. It is important to note that, theoretically, actual means will eventually equal expected means over a large—or near infinite—amount of trials. In other words, over many many rolls we would eventually find that each number on the die has a one in six chance of being rolled.

Let's discuss how expected means are determined. The expected mean is calculated using the following formula:

In this equation the variables are identified as the following:

Let's use this information to solve the problem. In order to solve the archer problem, we need to use the expected mean formula. We will substitute each of the archer's scores with its respective probability and solve.

It is important to note that we cannot solve this equation using the following formula:

This is because each shot has a different probability. The archer does not consistently score each value because the values occupy different positions on the target; therefore, there is a different probability for each shot not one out of six as the incorrect formula assumes.

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# Rotating regular polygon

Let $R_1$ be regular $n$-sided polygon on the plane (square, pentagon, hexagon, etc).
Now from this position we start to rotate this polygon about its center of gravity obtaining figure $R_2$.

• How to calculate the angle of rotation $\alpha$ for the case where common area of $R_1$ and $R_2$ i.e. area ($R_1 \cap R_2)$ will be minimal? (intuition tells what possible solution could be but how to prove it?)
• Does some simple method exist for solution of this problem in general case? (preferably with the use of rotations matrices)
• The procedure for $n$-odd and $n$-even could be the same or we should to differentiate between these two cases?

• Could it be proven that the shape obtained for situation of minimal area is also a regular polygon ( $2n$ sided) as we see in the below picture of pentagon made by Joseph?

The minimum is achieved at $\alpha = \frac{\pi}{n}$ and at minimum, $R_1 \cap R_2$ is a regular $2n$-gon.

Choose a coordinate system so that $R_1$ is centered at origin and one of its vertices lies on $x$-axis.
Let $\rho(\theta)$ be the function which allow us to parametrize $\partial R_1$ in following manner:

$$\mathbb{R} \ni \theta \quad\mapsto\quad (x,y) = (\sqrt{2\rho(\theta)}\cos\theta,\sqrt{2\rho(\theta)}\sin\theta) \in \partial R_1$$

In terms of $\rho(\theta)$, we have

$$f(\alpha) \stackrel{def}{=} \verb/Area/(R_1 \cap R_2) = \int_0^{2\pi} \min(\rho(\theta),\rho(\theta-\alpha)) d\theta$$

Since $R_1$ is a regular $n$-gon and one of its vertices lies on $x$-axis, $\rho(\theta)$ is even and periodic with period $\frac{2\pi}{n}$. In fact, it strictly decreases on $[0,\frac{\pi}{n}]$ and strictly increases on $[\frac{\pi}{n},\frac{2\pi}{n}]$.

As a result of these, $f(\alpha)$ is even and periodic with same period. To determine the minimum of $f(\alpha)$, we only need to study the case where $\alpha \in \left[0,\frac{\pi}{n}\right]$.

For $\alpha \in \left[0,\frac{\pi}{n}\right]$ and $\theta \in \left[0,\frac{2\pi}{n}\right]$, the curve $\rho(\theta)$ and $\rho(\theta - \alpha)$ intersect at $\frac{\alpha}{2}$ and $\frac{\alpha}{2} + \frac{\pi}{n}$.
This leads to \begin{align}f(\alpha) &= n\left[ \int_{\frac{\alpha}{2}}^{\frac{\alpha}{2}+\frac{\pi}{n}} \rho(\theta) d\theta + \left( \int_0^{\frac{\alpha}{2}} + \int_{\frac{\alpha}{2}+\frac{\pi}{n}}^{\frac{2\pi}{n}} \right)\rho(\theta-\alpha)d\theta \right] = 2n\int_{\frac{\alpha}{2}}^{\frac{\alpha}{2}+\frac{\pi}{n}} \rho(\theta) d\theta\\ \implies \frac{df(\alpha)}{d\alpha} &= n\left(\rho\left(\frac{\alpha}{2}+\frac{\pi}{n}\right) - \rho\left(\frac{\alpha}{2}\right)\right) \end{align} At the minimum, we have $$\frac{df(\alpha)}{d\alpha} = 0 \implies \rho\left(\frac{\alpha}{2}\right) = \rho\left(\frac{\alpha}{2} + \frac{\pi}{n}\right) = \rho\left(\frac{\alpha}{2} - \frac{\pi}{n}\right) = \rho\left(\frac{\pi}{n} - \frac{\alpha}{2}\right)$$ But $\frac{\pi}{n} - \frac{\alpha}{2}$ also belongs to $[0,\frac{\pi}{n}]$ and $\rho(\theta)$ is strictly decreasing there, this means

$$\frac{\alpha}{2} = \frac{\pi}{n} - \frac{\alpha}{2}\quad\implies\quad \alpha = \frac{\pi}{n}$$

Please note that this argument doesn't use the explicit form of regular $n$-gon. It uses

• $n$-fold rotation symmetry about center,
• $2$-fold reflection symmetry about a ray through a vertex,
• $\rho(\theta)$ is strictly decreasing on suitable intervals of $\theta$.

This means the same argument should work for other shapes with similar properties. e.g. those obtain from filling the "interior" of a regular star polygon.

• what a fine solution .... it should be a classic.. and I like its general interpretation very much.. – Widawensen Nov 28 '16 at 10:53

"Does some simple method exist for solution of this problem in general case?"

Presumably by the "general case" you mean $R_1$ is an arbitrary convex polygon? Or maybe an arbitrary simple polygon, perhaps nonconvex? I don't think this will have a simple answer. Below I computed that the minimum intersection area for the blue quadrilateral is achieved with $R_2$ rotated about $94.5^\circ$ degrees. In general you might have to resort to numerical optimization.

And, Yes, the minimum for a regular pentagon is achieved at a rotation of $36^\circ$, half the $72^\circ$ angle subtended by each edge from the centroid. Etc.

• In, the case of pentagon, intuition gives us answer as you have presented , I expected it, but how to prove it provided regularity of the shape - it's not clear for me. If the shapes are overlapping it is trivial that the common area is maximal but why exactly in the half of regular angle for pentagon is minimal it is not known to me, only intuition tells that since there is a symmetry it has to be true. But how to use this symmetry in the proof ? And thank you Joseph for drawings, they made question much more available.. – Widawensen Nov 27 '16 at 22:23
• I mean regular polygon but considered by some method independently from the number of sides i.e. $n$ is just the parameter used in the method but we don't concentrate whether it is square or hexagon. I understand that for irregular shapes it would be hard to find a general method, probably for complicated shapes only experimental method would be appropriate.. but maybe if we had some kind of a simple description of the shape we could infer somehow also about common area of the original figure and rotated one. – Widawensen Nov 27 '16 at 22:26

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# What is $\mathrm{Var}(E(X|I))$?

I randomly pick one of two coins (with equal weight) from a bag. Flipping coin 1 has a probability of $p_1$ of yielding heads. Flipping coin 2 has a probability of $p_2$ of yielding heads. Let $I$ be the indicator variable for whether coin 1 was picked (so $E(I) = \frac{1}{2}$). I flip the picked coin $n$ times. Let $X$ be the total number of heads out of $n$ that result.

What is $\mathrm{Var}(E(X\mid I))$?

Calculating $E(X\mid I)$ is not too bad I think: $E(X\mid I)$ is $np_1$ since $E(X\mid I)$ is distributed $\mathrm{Binomial}(n, p_1)$. However, how do I calculate the variance of $E(X\mid I)$ (not of $X\mid I$, but of $E(X\mid I)$)?

-
One thing that confused me when I started learning probability should be stated: While the expression $E(X)$ stands for a "final" value, the expression $E(X|I)$ is again a random variable. –  Yoni Rozenshein Dec 12 '12 at 9:07


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Math Help - Finding solutions for trig equations

1. Finding solutions for trig equations

Im sure this is a trivial problem for many, but not having done this for a while im a bit rusty and cant figure it out

I have the answers to the following question, but i do not know how to calculate them from the first:

Find all the solutions to the following equation

Sin( ((5x)/2) + 15°) = 0.34

Where 0°≤x≥360°

My attempt

sin-1(0.34)=19.88

19.88-15 = 4.88

(4.88*2)/5 = 1.95

How are the other answer worked out from here?

Any input appreciated

Thanks

2. Re: Finding solutions for trig equations

Remember that there are two quadrants where sin(x) = 0.34 (both first and second quadrant), so x may be either 19.88 degrees or 180 -19.88 = 160.12 degrees. And furher since the final answer must be in the range 0 < x < 360, that means that 5x/2+15 will be in the range 15 to 915 degrees. Which means in addition to 5x/2= 19.88 you also have 5x/2=19.88 + 360 and 5x/2+15 = 19.88+ 2x360 degrees. Same with 5x/2+15=160.12, 5x/2+15= 160.12+360, etc.

One other comment - no fault of yours given the answers in the form that that you were given, but it drives me crazy to see answers taken out to 10 decimal places of accuracy when the data you were given is only two decimal places. The proper answer for the first one is x=1.95, not x= 1.950749628.

3. Re: Finding solutions for trig equations

Thanks, that sorted the problem. Do you know how i would rearrange cos(56deg -3x) = -0.305. iv tried to work this out as inverse of cos 107.758 then use (107.758-56)/-3 which gives me the wrong answers, i also tried to get the other cos angle as 107.758-90 = 17.758 then staying in the negative quadrant of cos 270 + 17.758 = 287.758 to which iv got the wrong answers

Thanks

4. Re: Finding solutions for trig equations

Originally Posted by MMCS
Do you know how i would rearrange cos(56deg -3x) = -0.305. iv tried to work this out as inverse of cos 107.758 then use (107.758-56)/-3 which gives me the wrong answers,
I think it gives the correct answer of x = -17.25 degrees. What have you been told is the answer?

Originally Posted by MMCS
i also tried to get the other cos angle as 107.758-90 = 17.758
Remember for that cos(x) = cos(-x) = 360 - cos(x). So try -107.8 degrees, which is the same as +252.2 degrees.

5. Re: Finding solutions for trig equations

Inspect roots before finding their numerical values (Radian):


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How to create a \underset{\text{label}}{...} from the GUI?

There seems to be a problem for display a "underbrace" item.
The attached screenshot decoration_import_problem.png show the issue. To reproduce:
Create a file in SWP5.5 like this: decoration.tex
Start SWP 6.0.26 and choose File:Import TeX from the menu to import decoration.tex
The result is like this: decoration.sci
It is displayed wrong in "normal" edit view but the PDF is fine.

There is also a problem creating typeset objects like this in SWP6.0.26:
There is no menu entry for creating a "label".
In SWP 5.5 you can choose from the menu Inset:label and choose below|above.
In SWP 6 i did not find a way other than "import existing tex file" or "put raw TeX in typeset field". This should be added again.

AttachmentSize
decoration.tex1.89 KB
decoration.sci81.36 KB
decoration_import_problem.png89.79 KB

Labels are now part of the

Labels are now part of the Decorations dialog. They are the rightmost entries in the list of Decorations above and Decorations below.

Using them in conjunction with the underbrace, though, can be problematic. We will try to make improvements in this portion of the interface.


text:
?

Free Version
Easy

# The Periodic Table: Size of Atoms

CHEM-IGQZLN

Which of the following sets of elements is arranged in order of INCREASING metallic radii?

A

$Cs<{ Ba }<{ La }<{ Hf }<{ Ta }$

B

$Al<{ Si }<{ P }<{ S }<{ Cl }$

C

$Be<{ Mg }<{ Ca }<{ Sr }<{ Ba }$

D

$Rb<{ K }<{ Na }<{ Li }<{ H }$


text: Refractive Indices of Water And Turpentine Oil Physics Astronomy Project Topics, Physics Science Fair Projects, Pyhsical Science, Astrology, Planets Solar Experiments for Kids and also Organics Physics Science ideas for CBSE, ICSE, GCSE, Middleschool, Elementary School for 5th, 6th, 7th, 8th, 9th and High School Students for the year 2016. The Refractive Index - n - of a medium is a measure for how much the velocity of light is reduced inside the medium. Similar to our Optical Illusion science experiment, light experiments also rely on what our eyes see. Snell’s law: It is an equation connecting the angles of incidence and refraction with the refractive indices of the two mediums. Solution: Relation between Angle of Incidence and Angle of Refraction Experiment. The material consists of a two-dimensional array of repeated unit cells of copper strips and split ring resonators on interlocking strips of standard circuit board material. Refractive Index Database The table below contains links to refractive index data for common materials. You can accept all cookies by clicking on the button or define your cookie settings using the link "Customize your cookie settings". Refraction is an effect that occurs between transparent materials of different densities. Since it is characteristic for each material, it can be used for identification purposes or for the prediction of other properties. We’re continuing with our simple science experiments this week by taking a look at light refraction. These essential oils are often obtained by distillation. Tags:physics, optics, light, lenses, refraction, index of refraction, water, oil -> The refractive indices of water and olive oil are 1.33 and 1.47, respectively. These measurements can therefore also provide insight into the quality of essential oils, as any change in their optimal composition will affect the refractive index. Typical immersion oils have a refractive index of 1.51 and a dispersion similar to that of glass coverslips. and the liquids under investigation were benzene, ethanol, water . Wesson oil has a higher index of refraction than water (n = 1.33), but a lower mass density, so it floats on water. An experimental study and analysis of fiber optic cable based refractive index sensor has been carried out to measure the change in signal for different solutions of known refractive indices. Copyright © 2002 Elsevier Science Ltd. All rights reserved. I’ve found information online for things like vegetable oils. Typical immersion oils have a refractive index of 1.51 and a dispersion similar to that of glass coverslips. The refractive indices (n) of eight standard oils from Physikalisch Technische Bundesanstalt, Germany were determined with an accuracy of ±1×10−4 by using Abbe Refractometer. It has been established that the density and refractive index are strongly correlated [1-5]. Determination of this index through experiment is capital-intensive, time consuming, and also toil. Temperature of the sample can be varied by using the slider. The refractive index can then be used to determine the mixing ratio of two components. It is not the province of a Compiler to accept or reject data according to their aspect of verisimilitude or otherwise. Do you have a refractive index list of common household liquids that can be used for immersions? Data security: Varying degrees of security can be set, which include electronic signature and audit trail (not Abbemat 200). But many of these are expensive and/or poisonous. If we keep the mirror behind a lens and put an object at the focus point of the lens above it, the image of the object will form at the same focus point where the object is. v = c / n (1) where . Today they are standard tools in many laboratories in the fields of engineering, biology, optical analysis, forensics, instrumentation, cancer research and particle identification such as in the asbestos industry. Fats and oils are macronutrients that have a very varied composition. Experiments were conducted to determine what by-products would contribute to this increased refractive index by investigating the effects of individual groups on the refractive index change. Refractive Index A refractometer measures the extent to which light is bent (i.e. parameter for the experimental determination of viscosity and other transport properties. A refractive index is a numerical value that indicates the way in which light will travel through a particular substance in relation to how it travels through a vacuum. for the shopping cart), others help us make our online content better and easier for you over the long term via analysis, external media, and marketing services. refraction, respectively Fig 2. In all cases, the refractive index decreased monotonically with increasing wavelength. To redo the experiment, click the ‘Reset’ button. The refractive indices of the studied oils and the uncertainty in their values are calculated at λ=0.589. We collect and combine data on our visitors and their behavior on our website. Cargille refractive index (RI) liquids evolved from the Mineralogical industry where they were used in identification. This might be one of the most simple to set up science experiments we’ve ever tried. For many years, cedar wood oil was routinely used for immersion (and is still commercially available). This technique allows us the determination of the refractive index of a given liquid (or a transparent sample) accurately to high precision, with an accuracy limited only by the precision in the determinationoftherotationangle. The refractive index for cooking oil is 1.50. 4. Different types of glass have different indices of refraction. The Necessary cookies enable the basic functioning of the website. Master-RI is refractive-index model of the Master-Series. Refractive index (RI) provides valuable information about various reservoir engineering calculations, making it a key parameter for characterizing crude oils. Choose the liquid sample. The prediction that I made was correct in some aspects as I said that the refractive index of cooking oil would be higher than that of both ethanol and water, which it was. Density and refractive index of light crude oil (API = 46.2), heavy crude oil (API = 16.4) and hydrocarbon liquids, Propanol and Butanol were measured as a function of temperature. Refractive Indices of Water And Turpentine Oil Physics Astronomy Project Topics, Physics Science Fair Projects, Pyhsical Science, Astrology, Planets Solar Experiments for Kids and also Organics Physics Science ideas for CBSE, ICSE, GCSE, Middleschool, Elementary School for 5th, 6th, 7th, 8th, 9th and High School Students for the year 2016. The knowledge of the refractive index is crucial in all optical applications of transparent polymers. Manage your online orders and purchase settings, Get full access to download all documents, Pharmaceutical instrument qualification service, Peltier temperature control: The refractive index is dependent on the temperature of the sample.The, Measurement accuracy: Depending on the chosen. Accelerated aging experiments were conducted in a laboratory and increases in the refractive indices of these artificially aged oils were observed. They do this by tracking which websites visitors go to. You can make Pyrex glass disappear by immersing it in glycerin or mineral oil. (a) Refractive Index in terms of Speed of Light: The experimental setup is based on a modiﬁed version of the Michelson interferometer employing a rotation stage. The measurements were performed at temperature 20°C in the spectral range 0.4– 0.7 μm. Easily determine the quality of your essential oil Essential oils such as lemon or orange oil are also referred to as volatile oils. Although this oil has a refractive index of 1.516, it has a tendency to harden and can … For example, oil refineries depend on many oils and fats whose origin is different and unknown. Use this sample to know how much liquid you need to obtain about 4 mL of distillate during this part of the experiment. The data of the samples were analyzed using the Lorentz−Lorenz theory and simple mixing rules. Again a mild trend of increase with density is evident, a typical value of 1.448 having been noted in Chapter 4 for kerosenes and values of around 1.42 having been noted in Chapter 3 for gasoline. Essential oils such as lemon or orange oil are also referred to as volatile oils. FAQ: How important are refractive index and absorption for nanoparticles? 5. Water wave: Refraction is observed in water waves. The velocity of light in a medium can be expressed as . EXPERIMENT # 3: Essential Oils (EO) the other peaks can be calculated using the pentane peak. Moreover, it helps to differentiate between different essential oils. March 30, 2020 By Emma Vanstone Leave a Comment. Here you can find an overview of all used cookies, get detailed information, and decide which cookie types to accept. Encyclopædia Britannica, Inc. Higher the refractive index, the higher the optical density and slower is the speed of light. This produces the wavelengths of light required with high accuracy, and determines refractive indices within a value range of 1.3200 to 1.7000. The refractive indices measured with mirrors that had longer focal lengths were more accurate than mirrors with short focal lengths. The index of refraction depends not only on density but also on the chemical composition of a material. Continue collecting 4-mL samples until you have collected about 30 mL of distillate. Measurement range is RI 1.435 to 1.520 Master-RI can be used for measuring the refractive index of vegetable oil, petroleum, oil slicks, etc Hence, it is essential to search for an efficient and accurate estimation of crude oil RI. Determination of this index through experiment is capital-intensive, time consuming, and also toil. I’ve found information online for things like vegetable oils. In general, the index is indicated as a decimal value, but it is based on the speed of light in a vacuum divided by the speed of light traveling through a particular medium, such as glass or water. It is defined as: n = c/v. For the industry, it is important to know the origin and concentration of its main elements. Thus, (2) Where n vac is the absolute refractive index of M 2. I did found a change in the refractive index with increasing temperature. The present study was designed to correlate the refractive index (RI) obtained using SPR with … Thus n vac can be obtained by measuring the two angles θ 1 and θ 2 Abbe's refractometer is used to measure the refractive index of the given organic liquid. The refractive index for cooking oil is 1.50. They are concentrated hydrophobic liquids which obtain essential aroma compounds from plants. Good results have been obtained with an oil that has a refractive index of n = 1.518, which is close to the refractive index of glass. (b) What is the speed of light in a medium with a refractive index of 2.4? For a given color, the refractive index of a medium depends on the density of the medium. This may seem a small difference – until you see the effect it has with your own eyes. Cite Aim: To find the relationship between the angle of incidence and the angle of refraction. The Refractive Index - n - of a medium is a measure for how much the velocity of light is reduced inside the medium. The refractive index of a material is defined as the ratio between the speed of light in a vacuum(c) and the speed of light in that material (c m). v = velocity of light in the medium (m/s, ft/s) c = speed of light in air or vacuum - approximately 3 10 8 m/s (or 299792458 m/s) n = refractive index. The refractive index (n) is an important optical property of polymers and is widely used in material science. refracted) when it moves from air into a sample and is typically used to determine the index of refraction(“refractive index” or “n”) of a liquid sample.The refractive index is a unitless number, between 1.3000 and 1.7000 for most compounds, and is Diagram of a light ray being refracted. Each material in the database has refractive index listed as a function of wavelength over a range typically required for thin-film thickness measurement. On that occasion, make your students apply what they know about lenses and learn a useful trick to find the focal length. Refractive Index of Some Oxide Glasses Roy M. Waxler and G. W. Cleek Institute for Materials Research, National Bureau of Standards, Washington, D.C. 20234. The method we used in the experiment to measure the refractive index of sugar solution and salt solution consists of intense monochromatic source of light, LASER. In this work, we measure the refractive index of crude oils and asphaltene in toluene solutions using a fiber optic refractometer designed to work with high-viscosity and high optical density samples. This information is used to improve the website. Mie theory can fully describe these phenomena. The acid value (AV) of edible vegetable oils provides important information about the quality and the degree of purity of oil during processing and storage. Most objectives in the magnification range between 60x and 100x (and higher) are designed for use with immersion oil. But many of these are expensive and/or poisonous. The experimental data were fitted to the simple Cauchy dispersion formula and the results were found to be consistent within the limits of experimental error. The determination of the refractive index or butyro reading is a simple, accurate and fast method. If you accept cookies from external media, accessing these contents no longer requires giving manual consent. By measuring the refractive index of the original oil… Some of them are necessary (e.g. The refractive indices (n) of eight standard oils from Physikalisch Technische Bundesanstalt, Germany were determined with an accuracy of ±1×10 −4 by using Abbe Refractometer. 6. Easily determine the quality of your essential oil . 1. Predefined methods: The Abbemat series from Anton Paar has the predefined methods "butyro fat" (40 °C) and "butyro oils" (20 °C) which directly show the butyro readings and are in … where, c is the speed of light in vacuum and v is the phase velocity of light in the medium. Content from video platforms and social media platforms is blocked by default. In general, the optical properties of the scattering material have a tremendous influence on the observed scattering behavior. A dramatic increase in numerical aperture is observed when the objective is designed to operate with an immersion medium such as oil, glycerin, or water between the front lens and the specimen cover glass. OELCHECK measures refractive indices using the RX40 Refractive Index Cell Module from Mettler Toledo. You can calculate the refractive index of the liquid using the equation, You can verify your results by clicking on the ‘Show result’ button. Refractive Index A refractometer measures the extent to which light is bent (i.e. Refractive Index of Essential Oils The refractive index of essential oils is a quickly measured, easily generated quality parameter. Essential oils such as lemon or orange oil are also referred to as volatile oils. Therefore Mie theory is the best choice. Different mediums can bend light in different ways due to how much they are able to slow it down, referred to as its index of refraction. The refractive indices (n) of these oils have been measured as a function of the temperature t(20°C up to 50°C) at λ=0.589μm and were found to have linear temperature dependencies. refractive index. We present experimental scattering data at microwave frequencies on a structured metamaterial that exhibits a frequency band where the effective index of refraction (n) is negative. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Refractive index of standard oils as a function of wavelength and temperature. Determine the refractive index of each collected sample, as well as determining the refractive index for pure ethanol and pure water. Read the index of refraction using the telescope scale. Refractive Index of Polymers. The refractive index of thermodynamically ideal liquid mixtures is demonstrated to be given by the volume‐fraction mixing rule of the pure‐component squared refractive indices (Newton formula). Question: Your article on gemstones and refraction liquids has a refractive index list for fluids frequently used in gemstone testing. The refractive index determines how much light is bent, or refracted, when entering a material. In any type of solution, the density of the solution increase with increase in concentration of solute. It can thus provide insight as to their quality. Refractive Index Database The table below contains links to refractive index data for common materials. I found a great light refraction experiment video on YouTube and decided to try it with Tigger. I therefore predict that the refractive index of cooking oil will be approximately 1.40. The refractive index of essential oils is a quickly measured, easily generated quality parameter. When light moves from one medium to another, it changes its direction, i.e., it is refracted. Snell’s law is used to find the angle of refraction or the refractive index of a material (e.g., a glass slab). The velocity of light in a medium can be expressed as Repeat the experiment with different liquids and study the variation of refractive inde… Their varying composition allows for them to be characterized by refractive index measurement. https://doi.org/10.1016/S0030-3992(01)00101-3. Another important property of crude oil is the refractive index which can give information about the intermolecular interaction of the system. refracted) when it moves from air into a sample and is typically used to determine the index of refraction(“refractive index” or “n”) of a liquid sample.The refractive index is a unitless number, between 1.3000 and 1.7000 for most compounds, and is 7. The refractive index of the imaging medium is critical in determining the working numerical aperture of a microscope objective. Light rays passing through the specimen encounter a homogeneous medium between the coverslip and immersion oil and are not refracted as they enter the lens, but only as they leave its upper surface. Refractive Index Formula: The value of this constant, $$\frac{\sin \,i}{\sin \,r}$$ for a light ray passing through a vacuum into a given medium is called the refractive index, n of the medium. Where, n is the refractive index; c is the velocity of light in a vacuum ( 3 × 10 8 m/s); v is the velocity of light in a substance; The vacuum has a refractive index of 1. The experiment described in this project is an effective and simple method of measuring the refractive index of any liquid (transparent) using a convex lens and plane mirror. The refractive index of other materials can be calculated from the above equation. Wesson vegetable oil has nearly the same index of refraction (n) as Pyrex glass (n = 1.474). Refractive surgery: A non-essential eye surgery used to improve the refractive state of the eye and decrease or eliminate dependency on glasses or contact lenses. I therefore predict that the refractive index of cooking oil will be approximately 1.40. Moreover, it helps to differentiate between different essential oils. By continuing you agree to the use of cookies. This video will run you through on how the refractive index of a glass block can be obtained through the use of a series of carefully placed optical pins The refractive index of the prism was ~1.75. The refractive indices of the low-sulfur diesel which feature in the final row of the table at the end of the previous section are 1.4595 and 1.4745. We use cookies to help provide and enhance our service and tailor content and ads. The relative amounts of carvone and limonene in each fraction and the original oil may be calculated by using the area under the appropriate peaks. In this experiment, the independent variable is the angle of incidence, and the dependent variable is the angle of refraction. Full disclosure: I didn’t know that refraction was what this experiment demonstrated. Do you have a refractive index list of common household liquids that can be used for immersions? Refractive indices of coolants and oils. The Lorentz–Lorenz (L–L) formula has been tested and it was found to be valid with a maximum deviation of 0.4% and was used to calculate the molecular polarizability θ. It is a bit tricky to explain, but impressive even if you can’t quite get your head around it! Water, for instance has an index of refraction of 1.33. Refractive index (RI) provides valuable information about various reservoir engineering calculations, making it a key parameter for characterizing crude oils. The refractive index of essential oils can easily be measured with an Abbemat refractometer. Turn on the light source. Marketing cookies are used by third parties or publishers to show you personalized advertising. We use cookies on our website. Only use oil which is recommended by the objective manufacturer. The measurements were performed at temperature 20°C in the spectral range 0.4–0.7μm. In an experiment, the values of refractive indices of glass were found to be 1.54, 1.53, 1.44, 1.54,1.56 and 1.45 in successive measurements. Turn the refractometer scale knob to get the clear interface between the illuminated and dark regions. Pyrex and vegetable oil both have refractive indexes of 1.47 and air is at about 1.00. The refractive index of thermodynamically ideal liquid mixtures is demonstrated to be given by the volume‐fraction mixing rule of the pure‐component squared refractive indices (Newton formula). The determination of the refractive index is a simple, accurate and fast method. All reflections on the path from the object to the objective are eliminated in this way. Refractive index is also equal to the velocity of light c of a given wavelength in empty space divided by its velocity v in a substance, or n = c/v. Abstract: Refractive Index of Gold, Silver, Titanium and Other NanoParticles. Each material in the database has refractive index listed as a function of wavelength over a range typically required for thin-film thickness measurement. Refractive Index of Edible Oils and Fats, why and how to measure it? concept of refractive index (From the 2nd to the 19th century) Jean-Luc Godet Laboratoire de photonique d'Angers (LPhiA-EA 4464), Universitéd'Angers Centre François Viète (EA 1161), Universitéde Nantes. Light Refraction Experiment. To switch the views of the equipment, click the left and right arrows on the bottom of the simulator. Calculate the refractive index of the medium. Copyright © 2020 Elsevier B.V. or its licensors or contributors. Therefore, light rays passing through the specimen encounter a homogeneous medium between the coverslip and immersion oil and are not refracted as they enter the lens, but only as they leave its upper surface. Essential oils consist of many different compounds. Fortunately for us, a great deal of laboratory glassware and home kitchen glassware is made from Pyrex glass. The MCP polarimeters from Anton Paar are ideal for determining the optical rotation of optically active essential oils. International regulations such as the official AOAC method 920.141 "Refractive Index of Lemon and Orange Oils", the ISO 280 "Essential oils - Determination of Refractive Index" as well as international pharmacopoeias require the measurement of essential oils with a refractometer. 3. The prediction that I made was correct in some aspects as I said that the refractive index of cooking oil would be higher than that of both ethanol and water, which it was. Many materials have a well-characterized refractive index, but these indexes often depend … 2. The radius of curvature of the lens is R cm. A high precision density meter was used for density measurements and the refractive index was measured using an Abbe Refractometer. For such complex substances as oils, fats and waxes - many of them inevitably mixed with other bodies from the very nature of their commercial production - seemingly discrepant data cannot be rejected off-hand. Refractive Index Of Different Liquids Using Hollow Prism Physics Kids Projects, Physics Science Fair Project, Pyhsical Science, Astrology, Planets Solar Experiments for Kids and also Organics Physics Science ideas for CBSE, ICSE, GCSE, Middleschool, Elementary School for 5th, 6th, 7th, 8th, 9th and High School Students. It is defined as: This relationship is also known as Snell’s law. Question: Your article on gemstones and refraction liquids has a refractive index list for fluids frequently used in gemstone testing. The Refractive Index. Refraction & Light Experiments: Easy Science for Kids. The experiment produced a refractive index close to that of water, excluding the flexible mirror. (August IS, 1973) The change in refractive index with temperature has been determined for some oxide glasses from about -200 to 700 °C. Water is used first to calibrate the instrument. Anton Paar specialists are close to you to provide service, support, and training. Tracking which websites visitors go to oil has nearly the same index of refraction video... A refractive index of M 2 eliminated in this experiment demonstrated look at refraction. Support, and also toil reservoir engineering calculations, making it a key parameter for characterizing oils. Sample to know the origin and concentration of its main elements the medium for fluids frequently used in testing... Capital-Intensive, time consuming, and also toil, cedar wood oil routinely... Concentration of its main elements to that of water, for instance has an of! The MCP polarimeters from Anton Paar specialists are close to that of glass.! Data on our website: Relation between angle of incidence and the liquids investigation. Are close to that of glass have different indices of refraction using the telescope scale can calculated. Temperature of the refractive index which can give information about various reservoir engineering calculations, making it a parameter. The magnification range between 60x and 100x ( and is widely used in identification around. Nearly the same index of the refractive index of Edible oils and fats, why and to... The chemical composition of a Compiler to accept or reject data according to their.! Tremendous influence on the button or define your cookie settings '' the studied oils and the liquids investigation! ) liquids evolved from the above equation n ) is an important optical property of and! Material in the medium employing a rotation stage Relation between angle of refraction of. Measurements and the dependent variable is the angle of refraction with Tigger rotation of optically active oils... Measurements and the uncertainty in their values are calculated at λ=0.589 increasing.... In vacuum and v is the angle of incidence, and the uncertainty in their values are calculated λ=0.589... Vegetable oils, Pyrex disappears, but impressive even if you accept cookies from media...: this relationship is also known as Snell ’ s law can easily be measured with mirrors had... This produces the wavelengths of light in the spectral range 0.4– 0.7 μm (. Question: your article on gemstones and refraction liquids has a refractive index refraction..., which include electronic signature and audit trail ( not Abbemat 200 ) to explain but... Security: varying degrees of security can refractive index of oil experiment used for immersions home kitchen glassware is made from glass. Is defined as: this relationship is also known as Snell ’ s law higher ) designed! Curvature of the simulator third parties or publishers to show you personalized advertising cooking oil will be approximately 1.40 on. Switch the views of the samples were analyzed using the Lorentz−Lorenz theory and simple mixing rules accurate mirrors! Experiment produced a refractive index of a medium depends on the button or your... Not only on density but also on the chemical composition of a medium with a refractive of... Becomes equal to c in equation ( 1 ) where n vac is the speed of light vacuum... Other transport properties with mirrors that had longer focal lengths were more accurate than mirrors with short focal lengths and... Found information online for things like vegetable oils n ) is an effect that occurs transparent. Refraction liquids has a refractive index a refractometer measures the extent to which light is bent, or,! Other materials can be used to determine the quality of your essential oil oils! The aim of this index through experiment is capital-intensive, time consuming, and training absorption for?! 1.474 ), water easily determine the quality of your essential oil essential oils such as lemon orange... Index can then be used for immersion ( and is widely used identification. And higher ) are designed for use with immersion oil refraction is an effect that occurs between materials... Aim of this index through experiment is capital-intensive, time consuming, and also toil oil is! 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text:
# Normal Space is Tychonoff Space

## Theorem

Let $\struct {S, \tau}$ be a normal space.

Then $\struct {S, \tau}$ is also a Tychonoff (completely regular) space.

## Proof

Let $T = \struct {S, \tau}$ be a normal space.

From the definition of normal space:

$\struct {S, \tau}$ is a $T_4$ space
$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.

Let $F$ be any closed set in $T$.

Let $y \in \relcomp S F$, that is, $y \in S$ such that $y \notin F$.

As $T$ is a $T_1$ (Fréchet) space it follows from Equivalence of Definitions of $T_1$ Space that $\set y$ is closed.

As $T = \struct {S, \tau}$ is a $T_4$ space, we have that for any two disjoint closed sets $A, B \subseteq S$ there exists an Urysohn function for $A$ and $B$.

But $F$ and $\set y$ are disjoint closed sets.

So there exists an Urysohn function for $F$ and $\set y$.

This is the definition of a $T_{3 \frac 1 2}$ space.

Next we note that as $T$ is a $T_1$ (Fréchet) space, from $T_1$ Space is $T_0$ Space it follows that $T$ is a $T_0$ (Kolmogorov) space.

So $T$ is both a $T_{3 \frac 1 2}$ space and $T_0$ (Kolmogorov) space.

So, by definition, $T$ is a Tychonoff space.

$\blacksquare$


text:
Previous issue ·  Next issue ·  Most recent issue · All issues

# Journal of Operator Theory

Volume 67, Issue 2, Spring 2012  pp. 301-316.

Properties of $L_2$-solutions of refinement equations

Authors Victor D. Didenko
Author institution: Faculty of Science, Universiti Brunei Darussalam, Bandar Seri Begawan, BE 1410, Brunei

Summary:  This paper is devoted to the study of $L_2$-solutions of the operator equations $f-B_M \mathfrak{F} a \mathfrak{F}^{-1} f= g$, where $a$ is the operator of multiplication by a matrix $a\in L_\infty^{m \times m}(\sR^s)$, $m,s\in\sN$, $\mathfrak{F}$ denotes the Fourier transform, and $B_M$ is the dilation operator $B_Mf(x):=f(Mx)$, $x\in \sR^s$, generated by a non-singular matrix $M\in \sR^{m\times m}$. This class of equations contains discrete and continuous refinement equations widely used in wavelet analysis, signal processing, computer graphics and other fields of mathematics and in applications. It is shown that the set of nontrivial solutions of the homogeneous equation is either empty or contains a subset isomorphic to a space $L_\infty(\mathbb{V}_M)$, where $\mathbb{V}_M$ is a Lebesgue measurable set with positive Lebesgue measure. It follows that the operator $I-B_M \mathfrak{F} a \mathfrak{F}^{-1}$ is Fredholm if and only if it is invertible. Moreover, if the dilation $M$ satisfies some mild conditions, then $\ker \,(I-B_M \mathfrak{F} a \mathfrak{F}^{-1})\subset \overline{\im(I-B_M \mathfrak{F} a \mathfrak{F}^{-1})}$.

Contents    Full-Text PDF


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# An easy algebra problem

Algebra Level 2

Find the positive real value of $$x$$ which satisfies:

$x^2 + \left( \frac{ x}{x-1} \right) ^2 = \frac{5}{4}.$

×


text:
# Let X1, X2, X3, and X4 be independent random variables. Assume that X2, X3, and X4 each are Poisson.

Let X1, X2, X3, and X4 be independent random variables. Assume that X2, X3, and X4 each are Poisson distributed with mean 5 and suppose that Y = X1 + X2 + X3 + X4 P01(25). What is the distribution of X,? What is the distribution of W X1 + X2'

Let X1 and X2 be a random sample of size n = 2 from a continuous distribution with pdf of the formf(x) = 2x if O < i="" and="" zero="" otherwise.="" find="" the="" marginal="" pdfs="" of="" the="" smallest="" and="" largest="" order="" statistics,="" y1="" and="" y2.="" find="" the="" joint="" pdf="" of="" y1="" and="" y2.="" find="" the="" pdf="" of="" the="" sample="" range="" r="Y2" -="" y1.="">


text:
# Beam of particles focused by magnetic field

1. Apr 15, 2009

### phyzmatix

1. The problem statement, all variables and given/known data

A beam of particles of charge q and velocity v is emitted from a point source, roughly parallel with a magnetic field B, but with a small angular dispersion. Show that the effect of the field is to focus the beam to a point at a distance

$$z=\frac{2\pi mv}{qB}$$

from the source

2. The attempt at a solution

All that I know for sure is that

$$\vec{F}=q \vec{v} \wedge \vec{B}$$

I also think that, due to the angular dispersion, I'll have to somehow find an equation that will relate to all particles in the beam. We know that if v is initially perpendicular to B then the particle describes a circle with radius r = mv/qB. But what happens when the velocity is NOT perpendicular to B (nor parallel, in which case the particle will experience no force)?

My understanding is that with "focus" they mean that particles will move in a spiral around the z-axis (set in the direction of B) from where they were emitted to where they are focused, i.e. if distance r between the particle and the z-axis is initially r, then the final distance at point z will be 0. Is this correct?

So far, however, all attempts at a solution have been unsuccessful.

Could someone please give me a hint?

Thanks!
phyz

Last edited: Apr 15, 2009
2. Apr 16, 2009

### Redbelly98

Staff Emeritus
For problems like these, it's usually best to consider two parts of the motion separately:

• Motion along B
Velocity = vz = constant

and

• Motion perpendicular to B
The trajectory is circular
Centripetal force & acceleration play a role

Hope that helps.

3. Apr 18, 2009

### phyzmatix

OK, I've got some time today so I'll try and work it again from this angle. Thanks Red!

4. Apr 28, 2009

### phyzmatix

Hi Red!

Well, I've been working on this problem in between the half a million other things I need to do and I believe I've made some progress. From the theory, we can easily determine that the force component along B (z-axis) will be zero, and the component perpendicular to B (z-axis) will result in the particle describing a circle of radius

$$F=qvB$$
$$ma_r=qvB$$
$$m \frac{v^2}{r}=qvB$$
$$mv^2=rqvB$$
$$r=\frac{mv}{qB}$$

Now, it seems from the question (pls see OP) that this ultimately means that the distance z we're looking for is equal to 2 pi r. I don't believe it is coincidental that this distance z is therefore equal to the circumference of the circle described by the particle's motion, but now I can't seem to make the connection between the two.

Is there some theorem/reasoning I can use to prove that z = s (circumference of the circle) or even that the linear velocity component of the particle in the xy plane will be equal to the velocity component in the z direction (which will lead to the same result)?

5. Apr 28, 2009

### Redbelly98

Staff Emeritus
You're on the right track for the most part. However, note that the vB term should actually be a cross product,

|v×B| = vBsinθ

where θ is the angle that an individual particle makes with respect to the z-axis. vsinθ is simply the v-component perpendicular to B. I.e., when ignoring the z-component of motion, the speed is vsinθ

Hint for next step: how long does it take a particle to complete one loop of the circle?

6. May 12, 2009

### RaStevey

Hi, I seem to be struggling with the same problem...

(P.S. phyz, you're not perhaps from south africa and studying at unisa...)

I have been searching all over the internet and in all the relevant textbooks i have, but can't seem to find a solution...

I also got something similar to what phyz got, but i don't know where to go from there, i don't even know if i understand the problem correctly, because surely i must have found some hints as to what to do by now...

Hope you can help Red, please!!!

7. May 13, 2009

### Redbelly98

Staff Emeritus
Welcome to Physics Forums

If you show how far you have gotten, we'll take it from there.

8. May 17, 2009

### RaStevey

Ok, what I was trying to do is to derive the equation for the position of the particle, which i found to be:

$$\bar{r}$$=$$\left\{v_{x0}\frac{m}{qB}sin\left(\frac{q}{m}Bt\right)-v_{y0}\left[\frac{m}{qB}cos\left(\frac{q}{m}Bt\right)-\frac{m}{qB}\right]\right\}$$$$\hat{i}$$ + $$\left\{v_{x0}\left[\frac{m}{qB}cos\left(\frac{q}{m}Bt\right)-\frac{m}{qB}\right]+v_{y0}\frac{m}{qB}sin\left(\frac{q}{m}Bt\right)\right\}$$$$\hat{j}$$ + $$\left(v_{z0}t\right)$$$$\hat{k}$$

Where $$v_{x0}$$, $$v_{y0}$$ and $$v_{z0}$$ are the seperate initial components (at time=0) of the velocity in the x, y and z directions.

Now this seems to be a valid result, since I plotted the position equation with the help of Matlab and it gave the expected result, a helix which starts at x=0, y=0, z=0, since this was accepted as the initial starting position of the particle.

Now to determine the position of focus, I was planning to work out at what time the x and y components of the position returns back to (0; 0), since then the particle will be in line with the origin of the beam (along the z axis) and thus focused, and then substitute this time into the equation for the z component of the position to determine how far this point is from the origin along the z-axis.

Now I tried to do the above process, but couldn't get a reasonable result for the time and the z component of the position that looks anything like the result the expect and if I do get a result for the z component, it is dependent on only the z component of the initial velocity and not the whole initial velocity as they expect you to get, which can be seen from the fact that the z component of the position is equal to the z component of the initial velocity times time.

I'm sure there must be a much shorter and easier way of doing this, which I'm just not thinking about, so please tell me what you think and if I'm just missing something.

Thanks

9. May 17, 2009

### Redbelly98

Staff Emeritus
You're almost there!

Your expressions for the x and y components involve periodic functions. One period corresponds to the time it takes the argument of the sin or cos functions to change by 2π.

Does that help?

10. May 18, 2009

### RaStevey

Ok, so you basically have to work out what the time will be when the arguments for the sine and cosine functions are equal to $$2\pi$$ and then substitute this value for time into the equation for the z component of the position, which will give you the right answer that they expect you to prove, except the answer is in terms of $$v_{z0}$$ instead of the complete $$v_{0}$$ (the size of the initial velocity), which the expect, but then i thought, they say that the beam is "roughly parallel" to the magnetic field with only "a small angular dispersion", which implies that $$v_{z0}>>v_{y0}$$ and $$v_{z0}>>v_{x0}$$ and thus $$v_{0} \approx v_{z0}$$, which will give you the correct answer as they expect it, right?

11. May 18, 2009

### Redbelly98

Staff Emeritus
Yes.

12. May 18, 2009

### RaStevey

Well, I guess then it's solved Thanks for guiding my thoughts a bit.

13. May 19, 2009

### phyzmatix

Hello hello!

Hope you are both doing well and sorry for the slow response time but the last two weeks were spent on exams

I think I finally have a complete answer here after combining both of your inputs into something that works for my way of reasoning. It's pretty straightforward so I'd appreciate it if you can check it for me one last time. Here goes:

The component in the xy-plane (perpendicular to the z-axis and with theta = pi/2) will result in the particle describing a circle with radius r and angular velocity shown to be

$$F=qvB\sin\theta$$
$$ma_r=qvB\sin(\frac{\pi}{2})$$
$$m \frac{v^2}{r}=qvB$$
$$\frac{v}{r}=\omega =\frac{qB}{m}$$

The time taken by the particle to complete one loop of the circle is therefore

$$t=\frac{2\pi}{\omega}$$

$$t=\frac{2\pi m}{qB}$$

Applying the reasoning from your last couple of posts and substitution then gives us our final answer. Having got there, I still have some questions though. RaStevey, I know this is probably going to be a nightmare of LaTex for you, but could you show me how you got to:

Please. I think I might be lacking some maths for this one...Finally, why does this approach yield the correct answer?

How do we know that the beam is focused after one period and not two or three or whatever? What I'm trying to understand is why the time taken for the field to focus the beam is the same as the time taken for a particle to complete one revolution of the z-axis.

Thank you both for your help! Would never have got this far on my own

14. May 19, 2009

### Redbelly98

Staff Emeritus
The beam will be in focus after any integer multiple of a period.

To be in focus just means that x=y=0 for all particles in the beam.

15. May 20, 2009

### phyzmatix

Oh, ok cool. Thanks Red! You're a legend!

16. May 24, 2009

### RaStevey

$$\bar{F}=q\bar{v}\wedge\bar{B}$$
$$m\frac{d\bar{v}}{dt}=q\bar{v}\wedge\bar{B}$$
$$\frac{d\bar{v}}{dt}=-\frac{q}{m}\bar{B}\wedge\bar{v}$$

Now if you break this up into its x, y and z components you will get:

$$\frac{dv_{x}}{dt}=\frac{q}{m}Bv_{y}$$
$$\frac{dv_{y}}{dt}=-\frac{q}{m}Bv_{x}$$
$$\frac{dv_{z}}{dt}=0$$

Now if you solve the differential equations for $$v_{x}$$ and $$v_{y}$$ you will get the velocity to be:

$$\bar{v}=\left[v_{x0}cos\left(\frac{q}{m}Bt\right)+v_{y0}sin\left(\frac{q}{m}Bt\right)\right]\hat{i} + \left[-v_{x0}sin\left(\frac{q}{m}Bt\right)+v_{y0}cos\left(\frac{q}{m}Bt\right)\right]\hat{j} + \left(v_{z0}\right)\hat{k}$$

The displacement can now just be determined by integrating the velocity

17. May 24, 2009

### phyzmatix

Legend! Thanks!


text:
Question

# Demand for a good will tend to be relatively more price elastic if it exhibits which of the following characteristics?

A
It represents a small part of the consumer's income.
B
The good has many substitutes available.
C
It is a necessity (as opposed to a luxury).
D
There is little time for the consumer to adjust to the price change.

Solution

## The correct option is A The good has many substitutes available.Demand for a good will tend to be more elastic if the good has a many substitutes, because many substitutes have more elasticity.Hence, the correct option is B.Maths

Suggest Corrections

0

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text:
# How to move object on local space not object space in unity

I am watching this (game dev, space parenting & rotation) Sebastian Lague video and found that there are three spaces actually (not two) and these are world space, local space and object space.

1. World space: The static space (0,0,0)
2. Object space: related to the object space
3. Local Space: related to the parent of the object

I am amazed that i didn't find the distinction between these two spaces(local and object) on official unity forms but actually it exists. My question is that why there is no Space.Local? I found that there are Space. Self and Space.World. Space.Self is refer to object space. I can move my object to object space using this

    void Update () {
transform.Translate(new Vector3(0,0,1) * Time.deltaTime * 2,Space.Self);
}

And i can move my object to world space using this

void Update () {
transform.Translate(new Vector3(0,0,1) * Time.deltaTime * 2,Space.World);
}

but there is no support for local space that i could move the object to local space (means move the object related to its parent object). There is a fair distinction between Local and Object space but unity didn't consider it i guess or i am wrong.

I think that there is a bit of misunderstanding with these 3 spaces.

Basically you should focus on two spaces: World and Local. Object space has meaning when you take about vertices positions. Example, if an object has a local position of (1, 0, 0), if means its center is one unit away from the parent center on the x axis. but so what would an object space position of (1, 0, 0) mean? That the objects center is one unit away from itself? Only the object's vertices are given in object space (meaning, where is this vertex compared to the center of the object). Vertex shaders use this for example. But in your everyday use of Unity you don't use it.

So why is there a Space.Self?

Well, according to the doc, what it does is:

Applies transformation relative to the local coordinate system.

So Space.World is world coordinate, and Space.Self is local coordinate.

In your example, you move the object in local space. Moving it in object space doesn't mean anything.

Finally note that you can also move in local space by editing transform.localPosition instead of transform.position, But this would be the same as a translation using Space.Self or Space.World respectively.

Hope this clears things up a bit

transform.Translate does something a little funny. The Space you select picks the directions that it moves the object, but doesn't scale the translation. The actual translation vector is always applied in worldspace, and you're just selecting whether to rotate the direction to match the object's local axes, or not.

Since an object might not have a parent at all to fetch a direction from, I'm guessing they skipped the extra conditional logic for this case because it doesn't come up in practice much that you want to move an object with respect to its parent while ignoring its parent's scale. More often, if we're doing parent-space positioning, we want to use something like:

transform.localPosition += translation;

which allows the parent's scale to affect the resulting position.

But, if you really want transform.Translate style behaviour and syntax for this case, we can add it as an extension method:

public enum Spaces {
World,
Parent,
Self
}

public static class SpaceExtension {

public static void Translate(this Transform transform, Vector3 translation, Spaces ralativeTo) {
switch(relativeTo) {
case Spaces.World:
transform.position += translation;
break;
case Spaces.Self:
transform.position += transform.rotation * translation;
break;
case Spaces.Parent:
// Fall back on world space if there's no parent.
Quaternion rotation = Quaternion.identity;
if(transform.parent != null)
rotation = transform.parent.rotation;
transform.position += rotation * translation;
break;
}
}
}

This gives you a new overload of transform.Translate to use if you like, with the alternative Spaces enumeration to pass the parent space option.

I haven't tested it extensively to verify whether it handles non-uniform scales the same way the original transform.Translate does, though I'd caution that if you're using non-uniform scale in a parent hierarchy you're likely going to run into bigger troubles. ;)