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AP Physics 2- 13.4 Images Formed by Lenses- FRQs- New Syllabus

Images Formed by Lenses AP  Physics 2 FRQ

Unit 13: Geometric and Physical Optics

Weightage : 15–18%

AP Physics 2 Exam Style Questions – All Topics

Question

Some students are asked to determine the focal length of a convex lens. They have the equipment shown above, which includes a waterproof light box with a plate on one side, a lens, and a screen. The box has a bright light inside, and the plate on the side has shapes cut out of it through which the light shines to create a bright object. This particular plate has a cutout that is a vertical arrow and a horizontal bar with a circle at one end. In the view shown above, the circle is near the right edge of the plate.
With the screen and light box on opposite sides of the lens, the box is aligned so that the plate is \(20\,\text{cm}\) from the center of the lens, and an image of the arrow and bar is formed on the screen. The students find that the image is clear on the screen when the screen is \(30\,\text{cm}\) from the lens.
(a) On the figure below, sketch how the image on the screen appears to the students.
(b)
i. Calculate the focal length of the lens.
ii. Calculate the magnitude of the magnification of the image.
(c)
i. In the side view below, the arrow represents the bright object created by the plate. Draw a ray diagram on the figure below that is consistent with your calculations in parts (b)(i) and (b)(ii). Show at least two rays, as well as the location and orientation of the image.
ii. Explain how your diagram is consistent with your calculated focal length and magnification in parts (b)(i) and (b)(ii).
(d) The entire apparatus is now submerged in water, whose index of refraction is greater than that of air but less than that of the lens.
i. The figures below show cross sections of the top portion of the convex lens in air and the convex lens in water. An incident ray is shown in both cases. On each figure, draw the ray as it passes through the lens and back into the air or water.
ii. Describe how the focal length of the lens and the position and size of the image formed by the lens when it is in the water compare to when the lens is in air. Explain how the rays drawn in the figures in part (d)(i) support your answer.

Most-appropriate topic codes (AP Physics 2):

• Topic \(13.3\) — Refraction (Part \( \mathrm{(d)} \))
• Topic \(13.4\) — Images Formed by Lenses (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \), Part \( \mathrm{(d)} \))
▶️ Answer/Explanation

(a)
The image formed on the screen is a real image, so it is inverted. The arrow should point downward, and the bar-circle shape should be reversed left-to-right compared with the object.

\(\boxed{\text{The image is inverted: arrow down and left-right reversed.}}\)

(b)(i)
Use the thin-lens equation:

\(\dfrac{1}{f}=\dfrac{1}{d_o}+\dfrac{1}{d_i}\)

Substitute \(d_o=20\,\text{cm}\) and \(d_i=30\,\text{cm}\):

\(\dfrac{1}{f}=\dfrac{1}{20\,\text{cm}}+\dfrac{1}{30\,\text{cm}}\)

\(\dfrac{1}{f}=\dfrac{3}{60}+\dfrac{2}{60}=\dfrac{5}{60}\)

\(f=12\,\text{cm}\)

\(\boxed{f=12\,\text{cm}}\)

(b)(ii)
The magnitude of magnification is

\(|M|=\dfrac{d_i}{d_o}\)

\(|M|=\dfrac{30\,\text{cm}}{20\,\text{cm}}=1.5\)

\(\boxed{|M|=1.5}\)

(c)(i)
A correct ray diagram should show the object \(20\,\text{cm}\) from the lens and the image \(30\,\text{cm}\) on the opposite side. The image should be inverted and about \(1.5\) times as tall as the object.

(c)(ii)
The diagram is consistent with \(f=12\,\text{cm}\) because a ray parallel to the principal axis refracts through the focal point \(12\,\text{cm}\) from the lens.

The image is \(30\,\text{cm}\) from the lens, while the object is \(20\,\text{cm}\) from the lens, so \(|M|=\dfrac{30}{20}=1.5\). Therefore, the image should be inverted and \(1.5\) times larger than the object.

(d)(i)
In air, the lens bends the ray more strongly because the difference between the index of refraction of the lens and the surrounding medium is larger.

In water, the ray still bends toward the normal when entering the lens and away from the normal when leaving the lens, but it bends less than it does in air.

(d)(ii)
When the lens is in water, the focal length is greater than when the lens is in air.

This is because the index difference between the lens and water is smaller than the index difference between the lens and air. Therefore, rays bend less in water and converge farther from the lens.

Since the rays converge farther away, the image distance increases. For the same object distance, the image is also larger because \(|M|=\dfrac{d_i}{d_o}\), and \(d_i\) is larger.

\(\boxed{\text{In water, the focal length is larger, the image forms farther away, and the image is larger.}}\)

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