Question

Equiconcave lens of crown glass has to be made. How much radii of the surfaces of the lens should be kept so that its power would be -2.5 D? The refractive index of crown glass is 1.65.

Hint:

For an equiconcave lens, the radii of curvature are equal in magnitude but opposite in sign.

Answer 1

The power \(P\) of a lens is given by the formula: \[ P = \frac{1}{f} \] Where \(f\) is the focal length of the lens. The lens formula for a lens in air is given by: \[ \frac{1}{f} = (n - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Where: - \(n = 1.65\) (refractive index of crown glass), - \(R_1\) and \(R_2\) are the radii of curvature of the two surfaces of the lens. For an equiconcave lens, both radii of curvature are equal in magnitude but opposite in sign. Thus, we have: \[ \frac{1}{f} = (1.65 - 1)\left( \frac{1}{R_1} - \frac{1}{-R_1} \right) = 0.65 \times \frac{2}{R_1} \] Now, using the formula for power: \[ P = \frac{1}{f} = -2.5 \, \text{D} \] Thus, \[ -2.5 = 0.65 \times \frac{2}{R_1} \] Solving for \(R_1\): \[ R_1 = \frac{0.65 \times 2}{2.5} = 0.52 \, \text{m} \] Thus, the radius of curvature of each surface of the lens should be \(0.52 \, \text{m}\).