The radii of curvature of the faces of a double convex lens are 10 cm

The radii of curvature of the faces of a double convex lens are 10 cm and 20 cm. The refractive index of the glass is 1.5. What is the power of this lens (in units of dioptre) ?

+7·5 D
–7·5 D
+2·5 D
+5·0 D
This question was previously asked in
UPSC NDA-1 – 2017
The correct answer is +7.5 D.
For a thin lens, the power (P) is given by the reciprocal of the focal length (f) in meters, P = 1/f. The focal length of a lens in air can be calculated using the lensmaker’s formula: 1/f = (n – 1) * (1/R₁ – 1/R₂), where n is the refractive index of the lens material, and R₁ and R₂ are the radii of curvature of the two surfaces. For a double convex lens, assuming light comes from the left, the first surface is convex (R₁ > 0) and the second surface is also convex but faces the opposite direction (R₂ < 0).
Given: R₁ = +10 cm = +0.1 m, R₂ = -20 cm = -0.2 m (negative because the center of curvature is on the same side as the incoming light for the second surface of a double convex lens), and n = 1.5.
Using the lensmaker’s formula:
1/f = (1.5 – 1) * (1/0.1 – 1/(-0.2))
1/f = 0.5 * (1/0.1 + 1/0.2)
1/f = 0.5 * (10 + 5)
1/f = 0.5 * 15
1/f = 7.5 m⁻¹
Power P = 1/f = 7.5 Dioptre (D). Since the focal length is positive (1/7.5 m), it is a converging lens, which is expected for a double convex lens in air.
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