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: - Physics MCQ NEET 2022 2026 | ExamTest.live

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1 mark

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

Solution & Step-by-step Explanation

Using Lens Maker's Formula: \frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) Given μ = 1.5, R_{1} = + 20 cm, R_{2} = - 20 cm (for biconvex). \frac{1}{f} = (1.5 - 1) (\frac{1}{20} - (- \frac{1}{20})) \frac{1}{f} = 0.5 (\frac{2}{20}) = 0.5 \times 0.1 = 0.05 cm^{- 1} f = \frac{1}{0.05} = 20 cm = 0.2 m .Power P = \frac{1}{f ( m )} = \frac{1}{0.2} = + 5 D .

Try it yourself before checking the explanation above.

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

+ 2 D

+ 20 D

+ 5 D

Infinity

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