Question

The radii of curvature of a biconvex lens are 4 cm and 15 cm. The lens is in air, its index of refraction is 1.5. An object is at 1 m before the front surface of the lens. Calculate the distance of the image from the back surface of the lens.

Answer 1

v = 6.315 cm

Step-by-step explanation:

given,

R₁ = 4 cm = 0.04 m

R₂ = 15 cm = 0.15 m

n =1.5

`(1)/(f)=(1)/(v)-(1)/(u)=(n-1)((1)/(R_1)-(1)/(R_2))`

`(1)/(v)-(1)/(-1)=(1.5-1)((1)/(0.04)+(1)/(0.15))`

`(1)/(v)+1 = 0.5 * 31.66`

`(1)/(v) = 15.833`

v = 0.06315 m

v = 6.315 cm

hence, the distance of the image from the back surface is v = 6.315 cm