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Wave Optics

Question

Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

Solution

Given,
Refractive index, μ = 1.55
Radius of curvature, R₁ = R and R₂ = – R
Focal length, f = 20 cm.

Using the lensmaker formula,

\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})

we have,

\frac{1}{20} = (1.55 - 1) (\frac{1}{R} + \frac{1}{R}) = \frac{1.10}{R}

Hence,

R = 20 \times 1.1 = 22 cm, which is the required radius of curvature .

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