Question

Focal length of a equiconvex lens, made of a material of refractive index 1.5 is 20 cm. Calculate the radius of curvatine of its surfaces.

Answer 1

Answer:

20cm.

Explanation:

From the question we get that the refractive index of the lense is 1.5 and the focal length is given as 20cm. So, by using the lens maker formulae we can calculate the radius of curvature of the lense. Since, the given lense is equiconvex lense, hence the focal length of both the lenses will be equal and the radius of curvature will also be equal. So, R1=R2=R=f.

So, 1/f=(μ-1)(1/R1 + 1/R2), where μ is the refractive index of the lense. So, on substituting the values we will get that 1/f=(1.5-1)(1/R + 1/R) which will be 1/20=0.5*2/R which on solving we will get R=20cm.