Question

A contact lens is made of plastic with an index of refraction of 1.30. The lens has an outer radius of curvature of 2.00 cm and an inner radius of curvature of 2.50 cm. What is the focal length of the lens?
a. 33 cm
b. None of them
c. 20 cm
d. -20 cm

Answer 1

Using analogy, from the lens-maker equation that states :

$\sf{ \frac{1}{f} = [n - 1] \frac{1}{R_1} - \frac{1}{R_2} }$

Solving for the focal length, we substitute initial values $\sf{R_1 = 2\;cm}$ and.$\sf{R_2 = 2.5 \;cm }$ to get :

$\small \sf{ \frac{1}{f} = [n - 1] \frac{1}{R_1} - \frac{1}{R_2} =[ 1.5 - 1] \: \frac{1}{2 \: cm} - \frac{1}{2.5 \: cm} = \frac{1}{0.05( \frac{1}{cm} )} }$

Solution is :

${ \boxed{ \sf{ \green {\underline{\overline{f = 20 \: cm}}}}}}$