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AP Physics 2 : Optics

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Example Question #1 : Optics

Suppose a student who was farsighted wears glasses that allows him to read at a distance of 20cm from his eyes to the book. His near-point distance is 63cm . If his glasses are 1.5cm from his eyes, what is the refractive power of his glasses lenses?

Possible Answers:

3.846D

6.346D

1.613D

4.536D

2.342D

Correct answer:

3.846D

Explanation:

Write the thin-lens equation.

$\frac{1}{f}= \frac{1}{d_o}+\frac{1}{d_i}$

d_o represents the object distance from the lens, subtracting the distance between the student's eyes and the lens.

d_o=20cm-1.5cm=18.5cm

The image distance is:

d_i=1.5cm-63cm=-61.5cm

Substitute the givens to the equation.

$\frac{1}{f}= \frac{1}{18.5}+\frac{1}{-61.5}= 0.037794 cm^{-1}$

$f=\frac{1}{0.037794cm}=26.4592cm = 0.26 m$

Refractive power is the inverse of focal length in .

$\frac{1}{0.26m}= 3.846D$

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Example Question #1 : Optics

A flower is placed 20cm away from a converging lens. An upside down image of the flower is measured as 6cm tall and at a distance of 30cm from the lens. Find the focal length of the lens and the real height of the flower.

Possible Answers:

f=20cm

$h_{o}=12cm$

f=0.08cm

$h_{o}=6.7cm$

f=12cm

$h_{o}=4cm$

f=30.5cm

$h_{o}=4.5cm$

Correct answer:

f=12cm

$h_{o}=4cm$

Explanation:

Let the object distance be $d_{0}$ and the image distance be $d_{i}$ . The object distance is the distance the flower is from the lens. The image distance is the distance the image is located from the lens. The thin lens equation is

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

$f=\left ( \frac{1}{d_{o}}+\frac{1}{d_{i}}\right )^{-1}=12cm$

Alternatively you can use

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

To find the height, the equation for the magnification of an object is given by

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

it is also given as

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

The negative sign indicates an image that is upside down. Setting these equal gives

$\frac{d_{i}}{d_{o}}=\frac{h_{i}}{h_{o}} \rightarrow h_{o}=\frac{d_{o}}{d_{i}}h_{i}=4cm$

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Example Question #1 : Optics

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.

If the top of the image is 10cm from the center of the tank, and the top of the object is located 3mm from the center of the tank, what is the focal length of this lens?

Possible Answers:

f= 2.9

f=340

f=5.5

f= .0029

Correct answer:

f= .0029

Explanation:

Using object distance, image distance, and focal length formula:

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

Where is object distance, is image distance, and is focal length.

Plugging everything in:

$\frac{1}{.003}+\frac{1}{.1}=\frac{1}{f}$

Solving for ,

f= .0029

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Example Question #1 : Optics

An object 11cm tall is 9cm from a mirror. If the image distance is -3cm from the mirror, what is the image height?

Possible Answers:

33.0cm

-33.0cm

-3.67cm

3.67cm

11.0cm

Correct answer:

3.67cm

Explanation:

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 h_i , so we just need to multiply both sides by h_o .

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

Now, we can plug in our numbers.

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

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

h_1= 0.0367

Therefore, the image height is 3.67cm .

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Example Question #2 : Magnification

You have a concave mirror and a candle. The candle is 14.7cm tall, and the image is 21.9cm tall. What is the the magnification of the mirror?

Possible Answers:

1.49

0.75

0.66

1.68

1.79

Correct answer:

1.49

Explanation:

The magnification equation is

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

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

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

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

M= 1.49

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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Example Question #1 : Optics

The image produced by a concave mirror is at -16.2m , and the magnification is 1.79 . What is the object distance?

Possible Answers:

29.00m

-11.56m

-9.05m

9.05m

-29.00m

Correct answer:

9.05m

Explanation:

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.

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

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

d_o= 9.05

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 9.05m . Note that the object distance is always positive.

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Example Question #4 : Magnification

An object of height 2.9 cm object is placed 10 cm in front of a convex mirror that has a radius of curvature of 67.0 cm . Determine the magnification of the image.

Possible Answers:

None of these

6.11

1.88

1.425

Correct answer:

1.425

Explanation:

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 :

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

i=14.25cm

Using the equation for magnification:

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

Where:

is magnification

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

M=1.425

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Example Question #5 : Magnification

Assuming that the image is the size of 1.5cm and is actually a picture of a tree of size 10m . What can we say about the magnification of this alcohol solution?

Possible Answers:

$6.7*10^{2}$

$1.5*10^{-3}$

$-1.5*10^{-3}$

$-6.7*10^{2}$

Correct answer:

$-1.5*10^{-3}$

Explanation:

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

$M=-\frac{1.5*10^{-2}m}{10m}=-1.5*10^{-3}$

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

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Example Question #1 : Mirrors

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?

Possible Answers:

Upright and real

Inverted and virtual

Upright and virtual

Inverted and real

There is no way to determine whether the image is real, virtual, inverted, or upright based on the information given

Correct answer:

Inverted and real

Explanation:

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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Example Question #1 : Optics

You have an object 17.0cm away from a concave mirror with a focal length of 7.40cm. What is the image distance?

Possible Answers:

There is not enough information to determine the image distance

17.0cm

15.8cm

19.3cm

13.1cm

Correct answer:

13.1cm

Explanation:

Because we're dealing with a mirror, appropriately we would use the Mirror Equation:

$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

where is the focal length, d_o is the object distance, and d_i is the image distance. Because we're trying to solve for d_i , we need to rearrange the equation to solve for it.

$\frac{1}{f}&=\frac{1}{d_o}+\frac{1}{d_i}$

$\frac{1}{f}-\frac{1}{d_o}&=\frac{1}{d_i}$

$(\frac{1}{f}-\frac{1}{d_o})^{-1}&=d_i$

Now, we can just plug in our values for and d_o :

$(\frac{1}{f}-\frac{1}{d_o})^{-1}&=d_i$

$(\frac{1}{7.40\cdot 10^{-2}}-\frac{1}{17.0\cdot 10^{-2}})^{-1}&=d_1$

$(7.631)^{-1}&= d_1$

0.131= d_1

Therefore, the image distance is 13.1cm.

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AP Physics 2 : Optics

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Optics

Example Question #1 : Optics

Example Question #1 : Optics

Example Question #1 : Optics

Example Question #2 : Magnification

Example Question #1 : Optics

Example Question #4 : Magnification

Example Question #5 : Magnification

Example Question #1 : Mirrors

Example Question #1 : Optics

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