Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A biconvex lens has radii of curvature, 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is:

infinity

+2 D

+20 D

+5 D

Refractive index, $\mu = 1.5$ .
Radii of curvature for a biconvex lens: $R_1 = +20 \text{ cm} = +0.2 \text{ m}$ and $R_2 = -20 \text{ cm} = -0.2 \text{ m}$ (using Cartesian sign convention).

Lens Maker's Formula: The power ( $P$ ) of a lens is given by: $P = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Calculation: Substituting the values: $P = (1.5 - 1) \left( \frac{1}{0.2} - \frac{1}{-0.2} \right)$ $P = 0.5 \left( \frac{1}{0.2} + \frac{1}{0.2} \right)$ $P = 0.5 \left( \frac{2}{0.2} \right)$ $P = 0.5 \times 10$ $P = +5 \text{ D}$
Conclusion: The power of the lens is +5 Diopters.

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