Question

The left face of a biconvex lens has a radius of curvature of magnitude 11.7 cm, and the right face has a radius of curvature of magnitude 18.7 cm. The index of refraction of the glass is 1.52.
(a) Calculate the focal length of the lens.(b) Calculate the focal length if the radii of curvature of the two faces are interchanged.

Answer 1

Answer:

Explanation:

R1 = + 11.7 cm

R2 = - 18.7 cm

n = 1.52

(a) Use lens maker formula

$\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )$

$\frac{1}{f}=\left ( 1.52-1 \right )\left ( \frac{1}{11.7}}+\frac{1}{18.7} \right )$

$\frac{1}{f}=\frac{0.52\times 30.4}{218.79}$

f = 13.84 cm

(b)

R1 = + 18.7 cm

R2 = - 11.7 cm

n = 1.52

(a) Use lens maker formula

$\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )$

$\frac{1}{f}=\left ( 1.52-1 \right )\left ( \frac{1}{18.7}}+\frac{1}{11.7} \right )$

$\frac{1}{f}=\frac{0.52\times 30.4}{218.79}$

f = 13.84 cm