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Lenses and optical instruments
6 MDCAT Physics questions with answers and explanations.
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The total magnification of a compound microscope is:
- 1 The sum of the magnifications of the two lenses
- 2 The product of the magnifications of the objective and the eyepiece
- 3 The difference of the two
- 4 Equal to that of the objective only
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Correct answer: The product of the magnifications of the objective and the eyepiece
The objective forms a magnified image, which the eyepiece magnifies again, so the magnifications multiply.
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An object is 30 cm in front of a convex lens of focal length 15 cm. The image is:
- 1 Virtual and smaller
- 2 Real, inverted and the same size
- 3 Real and magnified
- 4 Virtual and magnified
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Correct answer: Real, inverted and the same size
An object at twice the focal length gives a real, inverted image of the same size at twice the focal length.
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Short-sightedness (myopia) is corrected by using a:
- 1 Convex lens
- 2 Concave lens
- 3 Cylindrical lens only
- 4 Plane mirror
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Correct answer: Concave lens
A concave lens spreads the rays so that the image falls on the retina instead of in front of it.
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What is the power of a lens of focal length -25 cm?
- 1 +4 D
- 2 -4 D
- 3 -0.25 D
- 4 +0.25 D
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Correct answer: -4 D
Power = 1 / f in metres = 1 / (-0.25) = -4 dioptres.
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A coin at the bottom of a pool appears:
- 1 Deeper than it is
- 2 Shallower than it is
- 3 At its true depth
- 4 Larger and deeper
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Correct answer: Shallower than it is
Light bends as it leaves the water, so the coin seems to be raised.
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What is the speed of light in a medium of refractive index 2? (c = 3 x 10^8 m/s)
- 1 6 x 10^8 m/s
- 2 1.5 x 10^8 m/s
- 3 3 x 10^8 m/s
- 4 1 x 10^8 m/s
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Correct answer: 1.5 x 10^8 m/s
v = c / n = 3 x 10^8 / 2 = 1.5 x 10^8 m/s.