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

Answer:

v = 6.315 cm

Explanation:

given,

R₁ = 4 cm = 0.04 m

R₂ = 15 cm = 0.15 m

n =1.5

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

$\dfrac{1}{v}-\dfrac{1}{-1}=(1.5-1)(\dfrac{1}{0.04}+\dfrac{1}{0.15})$

$\dfrac{1}{v}+1 = 0.5 \times 31.66$

$\dfrac{1}{v} = 15.833$

v = 0.06315 m

v = 6.315 cm  

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