MCAT Optics Practice Questions: Mirrors, Lenses and Vision

Answer: A. 30 cm behind the lens, real, inverted and twice the object's height

1/d_i = 1/10 − 1/15 = 1/30, so d_i = +30 cm, a real image on the far side of the lens. The magnification is −30/15 = −2, so the image is inverted and twice as tall. An object between f and 2f always gives a real, inverted, enlarged image beyond 2f, which is how a projector works.

Answer: D. Virtual, upright, twice as large, 20 cm behind the mirror

1/d_i = 1/20 − 1/10 = −1/20, so d_i = −20 cm: a virtual image behind the mirror. The magnification is −(−20)/10 = +2, an upright image twice as large. An object inside the focal point of a concave mirror always gives an enlarged virtual image, which is why such mirrors magnify the face. An image does form, so C is wrong.

Answer: D. It can be projected onto a screen placed in front of the mirror

A convex mirror has a negative focal length, so for any real object the image distance is negative: the image is virtual, upright and reduced, located behind the mirror between the surface and the focal point. Virtual images form where reflected rays only appear to come from, so no light actually converges there and they cannot be projected onto a screen.

Answer: B. −2.0 diopters

Power is the reciprocal of focal length in meters: P = 1/(−0.50 m) = −2.0 diopters. The negative sign marks a diverging lens. Using centimeters instead of meters gives the −0.020 choice, and simply copying the focal length gives −50. A positive power would describe a converging lens.

Answer: C. 33 cm

Powers of thin lenses in contact add: +5.0 + (−2.0) = +3.0 diopters. The focal length is 1/3.0 m, about 0.33 m or 33 cm, and the combination is converging. Adding the magnitudes as if both lenses were converging gives 7.0 diopters and about 14 cm, and 3.0 cm confuses diopters with centimeters.

Answer: A. A diverging lens of power −0.50 diopters

The lens must take an object at infinity and form a virtual image at her far point, d_i = −2.0 m. Then 1/f = 1/∞ + 1/(−2.0) = −0.50 diopters, a diverging lens. Myopic eyes focus too strongly, so the correction spreads light out. The −2.0 diopter choice wrongly uses the far point distance itself as the power.

Answer: A. +3.0 diopters

The lens must form a virtual image of the page (d_o = 0.25 m) at his near point (d_i = −1.0 m). Then P = 1/0.25 + 1/(−1.0) = 4.0 − 1.0 = +3.0 diopters, a converging lens, as expected for hyperopia. The +4.0 choice ignores the near point, and +5.0 adds the terms instead of subtracting.

Answer: C. +1

Lens 1: 1/d_i = 1/10 − 1/20, so d_i = 20 cm and m₁ = −1. That image lies 30 − 20 = 10 cm in front of lens 2, so 1/d_i = 1/5.0 − 1/10 gives d_i = 10 cm and m₂ = −1. The total is m₁m₂ = +1: the final image is real, upright and the same size as the object.

Answer: B. 12 cm

Every row satisfies the mirror equation with the same f: 1/36 + 1/18 = 1/12, 1/24 + 1/24 = 1/12, and 1/18 + 1/36 = 1/12, so f = 12 cm. The middle row is the fastest check: equal object and image distances mean both sit at the center of curvature, 2f = 24 cm. Reading 24 cm as f is the common slip.

Answer: D. The ciliary muscle contracts and the lens becomes rounder, raising its power

A near object sends diverging light into the eye, so the eye needs more converging power to focus it on the retina. Contracting the ciliary muscle relaxes tension on the suspensory ligaments, letting the elastic lens round up, which shortens its focal length. Choice A describes focusing on a distant object, and the cornea's shape does not change for accommodation.