A double convex lens has focal length 25 cm. The radius of curvature of one of the surfaces is double of the other. Find the radii, if the refractive index of the material of the lens is 1.5.
Solution
Given,
Double convex lens of focal length, f = 25 cm
Refractive index of the material, μ = 1.5
As per the question, the radius of curvature of one surface is twice that of the other.
i.e. R1= R and R2= −2R
(according to sign conventions)
Using the lens maker formula
\[\frac{1}{f} = (\mu - 1)\left[ \frac{1}{R_1} - \frac{1}{R_2} \right]\]' we have:
\[\frac{1}{25} = \left( 1 . 5 - 1 \right)\left[ \frac{1}{R} - \left( \frac{1}{- 2R} \right) \right]\]
\[\Rightarrow \frac{1}{25} = 0 . 5\left[ \frac{3R}{2} \right]\]
\[ \Rightarrow \frac{1}{25} = \left[ \frac{3R}{4} \right]\]
\[ \Rightarrow R = 18 . 75 \text{ cm }\]
R1 = 18.75 cm and R2 = 37.5 cm
Hence, the required radii of the double convex lens are 18.75 cm and 37.5 cm.
