NEET Physics: Power of a Lens Practice Questions
11 free multiple-choice questions on Power of a Lens (Optics) for NEET Physics, each with the correct answer and a step-by-step explanation. Attempt a question, then reveal the solution to check your reasoning.
Practice thousands more NEET questions in the app →Q1. What is the SI unit of the power of a lens?
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Correct answer: D — Diopter
The power of a lens is defined as the reciprocal of the focal length, and its SI unit is the Diopter (D).
Q2. If the power of a lens is doubled, what happens to its focal length?
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Correct answer: B — It is halved.
Since power P = 1/f, if the power is doubled (P' = 2P), the new focal length f' must be f' = 1/P' = 1/(2P) = f/2. Therefore, the focal length is halved.
Q3. The power of a lens is defined as the reciprocal of its focal length. If a lens has a focal length of 25 cm, what is its power?
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Correct answer: A — 4D
Power (P) = 1/f. If f = 25 cm, then P = 1/0.25 m = 4 D.
Q4. A convex lens has a positive power. What does a positive power indicate about the lens?
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Correct answer: C — It is a converging lens
A positive power corresponds to a converging lens, which is characterized by a positive focal length.
Q5. The relationship between the power of a lens (P) and its focal length (f) is given by the formula P = 1/f (when f is measured in meters). If the focal length of a lens is 10 cm, what is its power in Diopters?
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Correct answer: B — 100 D
Focal length f must be in meters for the power calculation in Diopters. 10 cm = 0.1 m. Power P = 1/f = 1 / 0.1 = 10 D.
Q6. A lens has a power of -2.5 Diopters. What does this indicate about the type of lens?
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Correct answer: A — It is a concave lens.
A negative power indicates that the lens is diverging, which corresponds to a concave lens. This power value is related to the focal length of the lens.
Q7. A lens has a power of 2 D. If the focal length is measured in centimeters, what is the focal length?
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Correct answer: C — 50 cm
Power P = 1/f. If P = 2 D, then f = 1/2 = 0.5 m. Converting to centimeters: f = 0.5 × 100 = 50 cm.
Q8. Two thin convex lenses, each having a power of +3.0 D, are placed in contact. What is the power of the combination?
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Correct answer: A — A 6.0 D
When two thin lenses are placed in contact, the combined power (P₍eq)) is the algebraic sum of the individual powers. P₍eq) = P₁ + P₂ = 3.0 D + 3.0 D = 6.0 D.
Q9. A lens has a power of -1.75 D. What is its focal length in meters?
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Correct answer: D — -0.56 m
Power P = 1/f. If P = -1.75 D, then f = 1/(-1.75) = -0.5714 m. The focal length is -0.56 m (approximately). (Note: The actual relationship used is f = 1/P).
Q10. Two thin lenses, one with power +3.0 D and the other with power -1.5 D, are placed in contact. What is the power of the combination?
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Correct answer: B — +1.5 D
The total power of a combination of lenses in contact is the algebraic sum of their individual powers. Pₜₒₜₐₗ = P₁ + P₂ = (+3.0 D) + (-1.5 D) = +1.5 D.
Q11. If an object is placed at 30 cm from a converging lens, and the image formed is real, what must be the power of the lens?
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Correct answer: D — 1/20 D
For a real image to form when the object is placed outside the focal length (for a converging lens), the power must be positive. If we assume the focal length f=30 cm (for P=1/30), the object at 30 cm results in an image at infinity (or use the lens formula: 1/v - 1/u = 1/f. To find the power directly, we need the image position. If we assume the image is at 30 cm (which would be the focal point for P=1/30), this is the most direct application. Let's rephrase for clearer application based on P calculation. If P = 1/30 D, then f = 30 cm. An object at 30 cm placed at the focal point forms an image at infinity, which is real.
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