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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

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6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Types Of Image

For a nearsighted person, the image of a distant object is formed __________.

Possible Answers:

in front of the retina, and corrected using a converging lens

behind the retina, and corrected using a diverging lens

in front of the retina, and corrected using a diverging lens

behind the retina, and corrected using a converging lens

Correct answer:

in front of the retina, and corrected using a diverging lens

Explanation:

In nearsightedness, the person cannot see far objects due to increased refraction. This causes the image to be formed in front of the retina. This condition is corrected using a diverging lens to compensate for the "over refraction" by the deformed cornea.

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Example Question #1 : Mirrors And Lenses

In a microscope has a $\small 20cm$ tube length and objective lens with a focal length of $\small 5cm$ . The viewer's eye is $\small 25cm$ from the objective lens, and they desire a magnification of $\small -20$ . What must the focal length of the eyepiece lens be to achieve this magnification?

Possible Answers:

4cm

-5cm

-1.25cm

5cm

1.25cm

Correct answer:

5cm

Explanation:

The total magnification of a compound microscope is the product of the objective lens magnification and the eyepiece magnification.

M = m_oM_e

Objective magnification and eyepiece magnification are given by the following equations:

$m_o = -\frac{L}{f_o}$

$M_e=\frac{d_{eye}}{f_e}$

We are given the tube length and focal length of the objective lens, allowing us to solve for its magnification.

$m_o=-\frac{L}{f_o}=-\frac{20cm}{5cm}=-4$

We also know the distance of the viewer's eye. Use this value in the eyepiece magnification equation.

$M_e=\frac{d_{eye}}{f_e}=\frac{25cm}{f_e}$

Finally, combine the eyepiece magnification and objective lens magnification into the original equation for total magnification.

M=m_oM_e

$M=(-4)(\frac{25cm}{f_e})$

Use the given value for total magnification to solve for the focal length of the eyepiece lens.

$-20=(-4)(\frac{25cm}{f_e})$

$5=\frac{25cm}{f_e}$

f_e=5cm

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Example Question #11 : Mirrors And Lenses

Jimmy is farsighted and uses a convex lens to correct his vision. Wendy is nearsighted and uses a concave lens to correct her vision. They both wear glasses. During a camping trip, they notice they do not have any matches, and decide to use their glasses to start the fire. Whose glasses could be used to start the fire?

Possible Answers:

Wendy

Jimmy

Neither glasses would work

Both glasses would work equally well

Correct answer:

Jimmy

Explanation:

This question deals with an application of optics. In this case we have a farsighted person and a near sighted person. The farsighted person would use a convex lens, which is a converging lens. This would allow all of the rays of light to converge on a single point, allowing them to heat the object up and start a fire. Wendy’s glasses are diverging lenses, which would cause the rays to separate.

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Example Question #121 : Ap Physics 2

A monochromatic light ray passes from air (n = 1.00) into glass (n = 1.50) at an angle of 20^o with respect to the normal. What is the approximate angle of refraction?

Possible Answers:

24^o

27^o

17^o

31^o

13^o

Correct answer:

13^o

Explanation:

To compare angles of incidence and refraction, use Snell's law.

$n_{1}sin(\theta _{1})=n_{2}sin(\theta _{2})$

$1.00sin(20)=1.50sin(\theta _{2})$

$.228=sin(\theta _{2})$

$\theta _{2}=sin^{-1}(.228)=13.2^o$

Notice that as the light enters a more dense medium, it bends towards the normal.

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Example Question #1 : Polarization

Unpolarized light passes through three polarizing filters. The angle between the first and second filters is 45^o. The angle between the second and third filters is also 45^o. What is the relationship between the intensity of light emerging from the third filter, $I_{3}$ , and the incident intensity, $I_{o}$ ?

Possible Answers:

$I_{3}=\frac{1}{6}I_{o}$

$I_{3}=\frac{1}{4}I_{o}$

$I_{3}=\frac{1}{2}I_{o}$

$I_{3}=\frac{1}{8}I_{o}$

No light passes through the third filter

Correct answer:

$I_{3}=\frac{1}{8}I_{o}$

Explanation:

Going through the first filter, the intensity of initially unpolarized light decreases by 1/2, so $I_{1}=\frac{1}{2}I_{o}$ . For each filter after the first, the intensity decreases by an additional factor of $cos^{2}(\theta )$ , where $\theta$ is the angle between adjacent polarizers. So, after the second filter, we have can see that $I_{2}=\frac{1}{2}I_{o}*cos^{2}(45)$ . Similarly, after the third filter, we have $I_{3}=\frac{1}{2}I_{o}*cos^{2}(45)*cos^{2}(45) = \frac{1}{8}I_{o}$ .

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Example Question #122 : Ap Physics 2

A man stands ten meters away from a converging mirror with a focal length of two meters. What is true of the image he sees?

Possible Answers:

His image is virtual, upright, and magnified

His image is real, inverted, and magnified

His image is virtual, upright, and minimized

His image is real, upright, and magnified

His image is real, inverted, and minimized

Correct answer:

His image is real, inverted, and minimized

Explanation:

The first thing to consider when answering this question is the fact that real images are always inverted and virtual images are always upright. Once you have determined one or the other, two answer choices can be eliminated.

The first equation that is necessary for this question is $\frac{1}{f} = \frac{1}{d_o}+\frac{1}{d_i}$ .

$\frac{1}{2}=\frac{1}{10}+\frac{1}{d_i}$

$\frac{1}{d_i}=\frac{2}{5}\rightarrow d_i=2.5$

From this we can determine that d_i is equal to 2.5 . Since 2.5 is a positive number we know the image is real, and thus inverted.

The second equation to consider is for magnification: $m = \frac{-d_i}{d_o}$ .

If the absolute value of is greater than one, the image is magnified, and if the value is less than one, it is minimized.

$m=-\frac{2.5}{10}<1$

We would expect the image to be minimized.

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Example Question #1 : Thin Lens Equation

How far from a converging lens must an object be placed to produce an image that is NOT real and inverted? Given the answer as d_o in terms of the focal length, .

Possible Answers:

d_o >2f

f<d_o<2f

d_o=2f

d_o=f

Correct answer:

d_o=f

Explanation:

When an object is placed a distance from a converging lens or mirror that is equal to the focal length, no image is produced. To test this out, stand in front of a single concave mirror and continue to back up until you no longer see an image. Once you've reached this point, you will be standing one focal length away from the mirror.

When the object is less than one focal length away from the converging lens/mirror, the image will be virtual and upright.

If you don't have these trends committed to memory, you can derive them from the equation $\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$ .

When d_i is a negative integer the image is virtual, and when it is a positive integer the image is real.

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Example Question #1 : Thin Lens Equation

A certain farsighted person cannot focus on objects closer to his eyes than $\small 140cm$ . What focal length eyeglass lenses are needed in order to focus on a newspaper held at $\small 35cm$ from the person's eyes, if the glasses are worn $\small 2cm$ from his eyes?

Possible Answers:

2cm

27cm

47cm

43cm

28cm

Correct answer:

43cm

Explanation:

First we need the object and image distances away from the eyeglass lenses. Here, the newspaper is the object and the $\small 140cm$ focal point is where the image needs to be located. Find the distance from the object to the lens, and the distance of the image to the lens, by subtracting out the distance from the lens to the eye.

$d_{ol}=d_{oe}-d_{le}=35cm-2cm=33cm$

$d_{il}=d_{ie}-d_{le}=140cm-2cm=138cm$

Now apply the thin lens equation to determine focal length.

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

Recall that if the image is on the same side of the lens as the object, then image distance is negative.

$\frac{1}{f}=\frac{1}{-138}+\frac{1}{33}\approx0.023$

$f = \frac{1}{0.023}\approx 43cm$

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Example Question #1 : Mirrors And Lenses

A person approaches a plane mirror at 5m/s. How fast do they approach the mirror image?

Possible Answers:

5m/s

2.5m/s

10m/s

7.5m/s

Correct answer:

10m/s

Explanation:

The image distance for a plane mirror is always equal to the object distance because the magnification is 1.

$m=\frac{d_i}{d_o}=1$

If the object and image are the same distance from the mirror and magnification is 1, then as the object approaches the mirror at a certain speed, the image is approaching the plane mirror at the same speed, therefore you approach the image more quickly than you approach the mirror, since you travel 5m/s toward the mirror and the image travels 5m/s toward the mirror.

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Example Question #4 : Gre Subject Test: Physics

A virtual image is formed $\small 8cm$ from a convex mirror with a focal length of $\small 10cm$ . How far from the mirror is the object that created this image?

Possible Answers:

4.4cm

40cm

2.5cm

22.5cm

2cm

Correct answer:

40cm

Explanation:

Use the equation:

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

Focal length is negative for convex mirrors, and image distance is negative for virtual images. We are given these values in the question, allowing us to calculate the object distance.

$\frac{1}{-10}=\frac{1}{-8}+\frac{1}{d_o}$

$\frac{1}{d_o}=\frac{1}{40}$

$d_{o}= 40cm$

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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #1 : Types Of Image

Example Question #1 : Mirrors And Lenses

Example Question #11 : Mirrors And Lenses

Example Question #121 : Ap Physics 2

Example Question #1 : Polarization

Example Question #122 : Ap Physics 2

Example Question #1 : Thin Lens Equation

Example Question #1 : Thin Lens Equation

Example Question #1 : Mirrors And Lenses

Example Question #4 : Gre Subject Test: Physics

All AP Physics 2 Resources

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