Question

The radii of curvature of a convex lens are 20 cm and 30 cm. The refractive index of the substance of lens is 1.5. If lens is kept in water(1.33) then find focal length.​

Answer 1

Answer:

From lens maker's formula, f1=(μ−1)(R11−R21)

We have,

 Refractive index, μ=1.5 and R1=20 cm and R2=30 cm

f1=(1.5−1)(R11−R21)

f1=(1.5−1)(−201−−301)

f1=(1.5−1)20×30−30+20

f=−120cm

I hope it helps you to understand.