Optics - AP Physics 2

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Question

The speed of light in a vacuum, , is calculated to be $2.99 \cdot 10^8 \frac{m}{s}$ . The speed of light in a diamond is calculated to be $1.25 \cdot 10^8 \frac{m}{s}$ . What is the refractive index of diamond?

Answer

The definition of refractive index of a medium is the speed of light in a vacuum divided by the speed of light in the medium:

$n = \frac{c}{v}$

We have values for and , so we can plug in our numbers into the equation.

$n&=\frac{2.99\cdot 10^8\ \frac{m}{s}}{1.25\cdot 10^8\ \frac{m}{s}}$

Because we're dividing two values with the same units, our answer is unitless.

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Question

What is the speed at which light travels through the water?

$c=3 *10^8:\frac{m}{s}$ and $n_{water}=1.33$

Answer

Write the formula relating the speed of light, velocity of light in specific medium, and index of refraction of the medium.

$n_{material}=\frac{c}{v}$

Rewrite the equation so that we are solving for the velocity of light in water. Substitute given values.

$v=\frac{c}{n_{water}}=\frac{3 * 10^8 \frac{m}{s}}{1.33}=2.26 * 10^8\frac{m}{s}$

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Question

A beam of light travelling through air enters a transparent material at an angle of incidence at $34^{\textup o}$ , with the refracted beam's angle being $26^{\textup o}$ . What is the index of refraction of the material? Assume the index of refraction of air is 1.

Answer

To find the index of refraction given two angles and another index of refraction, we use Snell's law.

$\begin{align} n_1\sin\theta_1&=n_2\sin\theta_2 \ \frac{n_1\sin\theta_1}{\sin\theta_2}&= n_2 \end{align}$

We can plug in the numbers we have to find the answer.

$\frac{n_1\sin\theta_1}{\sin\theta_2}&= n_2$

$\frac{(1)(\sin(34^o))}{\sin(26^o)}&=n_2$

Simplifying, we get:

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Question

A beam of light travelling through air enters a medium with an index of refraction of and goes through it at an angle of $41^{\text o}$ . What is the angle of incidence?

Assume the index of refraction of air is 1.

Answer

To find the angle of incidence, we use Snell's law.

$n_1\sin\theta_1&=n_2\sin\theta_2$

$\sin\theta_1&=\frac{n_2\sin\theta_2}{n_1}$

We're given both indices of refraction, and the second angle, so we can plug in our numbers.

$\sin\theta_1=\frac{n_2\sin\theta_2}{n_1}$

$\sin \theta_1=\frac{(1.51)(\sin(41^o))}{1}$

$\theta_1 = 82.16^{\text o}$

Therefore, the angle of incidence is $82.16^{\textup o}$ .

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Question

An argon/krypton laser puts out green light at . It takes to travel through of honey. What is the index of refraction of the honey?

Answer

First off we can find the velocity of the laser beam since we know the distance of honey it travels through and the transit time.

$v=\frac{d}{t}$

However, this doesn't need to be explicitly solved for. The velocity of light inside a material is inversely proportional to the index of refraction of that material where

$v_{honey}=\frac{c}{n_{honey}}$

where is the speed of light in vacuum. Setting these equal, we can solve for the index of refraction of the honey.

$\frac{d}{t}=\frac{c}{n_{honey}}$

$n_{honey}=\frac{c t}{d}=1.47$

We expect that the index of refraction will be greater than that of air and a velocity less than .

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Question

The speed of light through a particular material is measured to be $v=2.5 * 10^{8} \frac{m}{s}$ , calculate the index of refraction of the material.

Answer

The speed of light passing through a medium is given by:

$v_{light,n}=\frac{c}{n}$

Rearrange to solve for the index of refraction and plug in numbers.

$n=\frac{c}{v_{light,n}}=\frac{3.0\times 10^{8}\frac{m}{s}}{2.5\times 10^{8}\frac{m}{s}}=1.2$

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Question

You are passing a ray of light through clear alcohol to determine properties. You shine the light ray exactly to the surface of alcohol.

Determine the index of refraction of alcohol if the light ray bends to to the normal. Assume index of refraction of air is .

Answer

We can use our knowledge about the indices of refraction to come up with our equation:

, where is the index of refraction of air and is the index of refraction for our alcohol.

Since

$\frac{sin(45^o)}{sin(22^o)}=n_2=1.9$

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Question

You are passing a ray of light through clear alcohol to determine properties. You shine the light ray exactly to the surface of alcohol.

Determine the index of refraction required in the alcohol to have total internal reflection?

Answer

To have total internal reflection, our equation will become:

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Question

Suppose that a ray of light traveling through air strikes a new medium. Upon doing so, the light bends away from the normal. Which of the following could this new medium be?

Answer

For this question, we're told that light is passing from air into another medium. In doing so, the light is refracted such that it bends away from the normal. We're asked to identify a possible medium.

The most important thing to understand about refraction is that when light passes into a new medium at an angle with respect to the normal, that light will be refracted, either away from or toward the normal. This is because light will travel through different media at different speeds. The faster light travels, the more it will bend away from the normal.

Generally, the angle at which light is bent can be predicted by Snell's law. In doing so, this equation takes use of the refractive index, a value unique to each medium. The expression for refractive index is as follows.

$n=\frac{c}{v}$

Where refers to the refractive index, refers to the speed of light in a vacuum, and refers to the speed of light in a given medium.

Since the speed of light is fastest when in a vacuum, the refractive index can never be less than . Only when the light is in a vacuum is the refractive index equal to . In any other medium, the refractive index will be greater than , even if it is slight.

Light will always refract away from the normal when it passes into a medium with a lower refractive index (indicating the light is traveling faster). Starting from air, the only way this can happen is for the new medium to have an index of refraction that is less than air. Of the answer choices shown, the only one that fits that criteria is the vacuum.

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Question

Two small blue LEDs, both with a wavelength of are located away from you. You have a telescope with a diameter objective. How far apart are the LEDs in if you can barely resolve them as two objects?

Answer

A telescope with an objective lens with a diameter with a wavelength of light $\lambda$ can barely resolve two objects with an angular separation of $\theta_{1}$ using

$\theta_{1}=\frac{1.22 \lambda}{D}$

The separation distance between the two objects, we will call this , and the distance to the objects from the telescope objective $\left ( L\right )$ can be related to the tangent of that angle,

$\tan \left ( \theta_{1}\right )=\frac{d}{L}$

You are far away from the lights and this angle is very small so the small angle approximation can be used,

$\theta_{1}\approx\tan\left ( \theta_{1}\right )=\frac{d}{L}$

Setting these expressions for $\theta_{1}$ equal and solving for and converting from to ,

$d=\frac{1.22\lambda L}{D}=0.22m\left ( \frac{100cm}{1m} \right )=22cm$

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Question

Suppose that an object is located in front of a thin convex lens whose radius of curvature is equal to . Will the image be oriented up or down? Will it be a real image, or a virtual one?

Answer

In this question, we're presented with a scenario in which an object is placed a certain distance in front of a thin convex lens. After being given the radius of curvature for this lens, we're asked to determine certain properties of the image; whether it will be real or virtual and whether it will be up-right or upside down.

To begin with, let's use the following equation:

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

Where is equal to the distance of the object from the lens

is equal to the distance of the image from the lens, and

is equal to the focal length of the lens

In order to calculate the focal length, we'll need to use the radius of curvature. The focal length is simply half this value:

$f=\frac{R}{2}=\frac{100: cm}{2}=50:cm$

Now that we have the focal length, we can calculate the image distance by first isolating the term:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

Next, we can plug in values to obtain the image distance:

$\frac{1}{q}=\frac{1}{0.5: m}-\frac{1}{0.2: m}=-3:m^{-1}$

$q=-\frac{1}{3}:m$

The value that we have obtained for the image distance is negative. This means that the image formed in this scenario will be a virtual image.

Moving forward, we can use the image distance and object distance in order to determine the orientation of the image (the magnification):

$M=-\frac{q}{p}=-\frac{-\frac{1}{3}: m}{0.2: m}=1.667$

Since the value we obtained is positive, we can conclude that the image will be formed up-right.

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Question

A tall man stands in front of a converging lens of focal length . What is the height of his image?

Answer

In order to solve this problem, we first must find where the image is located, and then we can use that to find the image's height.

The formula needed is:

$\frac{1}{f}=\frac{1}{d_{o}}+\frac{1}{d_{i}}$

-focal length

$d_{o}$ -distance man is from lens

$d_{i}$ -distance image is from lens (positive is on the other side of the lens with respect to the man)

In this case we have

$\frac{1}{.25}=\frac{1}{.5}+\frac{1}{d_{i}}$

We get

$d_{i}=.5m$

Now we can use the formula

$-\frac{d_{i}}{d_{o}}=\frac{h_{i}}{h_{o}}$

$h_{i}$ -image height

$h_{o}$ -object height

to find the height of the image.

$-\frac{.5}{.5}=\frac{h_{i}}{1}$

This means the image height is , or inverted.

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Question

A ray of light passes through the focal point and then strikes a thin lens. After refraction, this ray:

Answer

One of the rules of drawing ray diagrams is that if the ray goes through the focal point on one side and hits a thin lens, then the ray will emerge parallel to the principal axis.

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Question

How will the image of an object appear if it is located outside the focal length of a converging lens?

Answer

This question is asking us to determine the properties of the image formed by a converging lens when the object is located at a distance that is beyond the focal point of the lens.

One way to go about solving this problem is to draw a visual diagram showing ray vectors, as shown below.

In the above diagram, we have an object located on a flat surface, along with a converging lens to the right of it and the focal points on either side of the lens. After drawing these in, we need to draw our light rays as coming from the top of the object, and we'll need to use a total of three lines.

The first line is drawn horizontally from the top of the object to the middle of the lens. From there, this line is drawn passing through the focal point on the right.

The second line drawn is on the bottom in the picture above. This line is drawn from the top of the object and through the left focal point until it reaches the middle of the lens. From there, the ray is drawn from the middle of the mirror and straight horizontally.

Finally, the third line is drawn from the top of the object and through the very center of the lens.

At the point where all three of these lines intersect is where we will find the top of the image. As can be seen above, the top of the image will be facing downwards, and is therefore inverted.

To determine if the image is real or virtual, we have to ask ourselves: Are the light rays going where they're supposed to go, or did we have to extrapolate the rays to find our image? In this case, the rays went where they are supposed to go, which is through the lens and to the opposite side. Therefore, the image formed is real.

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Question

An object tall is from a mirror. If the image distance is from the mirror, what is the image height?

Answer

The Magnification Equation is as follows:

$M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

The negative on the last part is very important. If you don't include it (and it's easy to forget), you will get the wrong answer.

We'll use the parts without for our problem. We want , so we just need to multiply both sides by .

$h_i=-\frac{d_i}{d_o}\cdot h_o$

Now, we can plug in our numbers.

$h_i&=-\frac{d_i}{d_o} \cdot h_o$

$h_1=-\frac{-3\cdot 10^{-2}}{9\cdot 10^{-2}}\cdot (11\cdot 10^{-2})$

Therefore, the image height is .

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Question

You have a concave mirror and a candle. The candle is tall, and the image is tall. What is the the magnification of the mirror?

Answer

The magnification equation is

$M=\frac{h_i}{h_o}$

where is the height of the image, and is the height of the object. We can plug these given values into the equation.

$M&=\frac{h_i}{h_o}$

$M=\frac{21.9}{14.7}$

The magnification is a scalar value, so it's unitless. We can verify this by examining what goes into the equation. Both heights have the same units, which cancel, so that leaves us with no units.

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Question

The image produced by a concave mirror is at , and the magnification is . What is the object distance?

Answer

The equation for magnification is

$M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

We're given the image distance and the magnification, so we'd use the second equality.

$M&=-\frac{d_i}{d_o}$

$M\cdot d_o&=-d_i$

$d_o&=-\frac{d_i}{M}$

Now, we can plug in our numbers.

$d_o&=-\frac{d_i}{M}$

$d_o= -\frac{-16.2}{1.79}$

The negative in front of the equation is very important. If you forget it, the answer will be incorrect.

The distance from the mirror to the object is . Note that the object distance is always positive.

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Question

An object of height object is placed in front of a convex mirror that has a radius of curvature of . Determine the magnification of the image.

Answer

Using the relationship:

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{r}$

Where:

is the object distance from the mirror

is the image distance from the mirror

is the focal length of the mirror

is the radius of curvature of the mirror

Plugging in values:

$\frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{67}$

Solving for :

$i=\frac{1}{-\frac{2}{67}+\frac{1}{10}}$

Using the equation for magnification:

$M=\frac{h_i}{h_p}=-\frac{i}{p}$

Where:

is magnification

$M=-\frac{14.25}{-10}$

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Question

You are passing a ray of light through a thin tank of alcohol to determine properties. Assume that the light ray is not effected by the tank's exterior. You find that if a piece of white paper is put underneath the alcohol tank, there is an image on the metal.

Assuming that the image is the size of and is actually a picture of a tree of size . What can we say about the magnification of this alcohol solution?

Answer

To determine magnification, we simply divide the object length from the image length.

$M=-\frac{1.510^{-2}m}{10m}=-1.510^{-3}$

The negative sign is used to designate that the image is real.

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Question

Suppose that an object is placed 15cm in front of a concave mirror that has a radius of curvature equal to 20cm. Will the resulting image be upright or inverted? Will it be a real or virtual image?

Answer

To solve this problem, it is essential to use the mirror equation:

$\frac{1}{d_{o}}+\frac{1}{d_{i}}=\frac{1}{f}$

Where:

$d_{o}$ is the distance the object is from the mirror

$d_{i}$ is the distance the image is from the mirror

is the focal length of the mirror

To calculate the focal length, we'll have to use the radius of curvature:

$f=\frac{r}{2}=\frac{20cm}{2}=10cm$

Plugging this value into the top equation and rearranging, we obtain:

$\frac{1}{d_{i}}=\frac{1}{f}-\frac{1}{d_{o}}=\frac{1}{10cm}-\frac{1}{15cm}=\frac{1}{30cm}$

Therefore, the image distance is:

$d_{i}=30cm$

Since the image distance calculated above is a positive value, this means that the image forms on the same side of the mirror that the reflected light traveled. Thus, the image is real.

To determine whether the image is inverted or upright, we'll need to use the magnification equation:

$M=\frac{h_{i}}{h_{o}}=-\frac{d_{i}}{d_{o}}$

$M=-\frac{30cm}{15cm}=-2$

The magnification obtained above is a negative number. As a result, the image formed from this scenario will be inverted.

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