Question

A contact lens is made of plastic with an index of refraction of 1.60. The lens has an outer radius of curvature of 2.08 cm and an inner radius of curvature of 2.48 cm. What is the focal length of the lens?

Answer 1

The value
`f = 21.49 \ cm`

Step-by-step explanation:

From the question we are told that

The index of refraction is
`n = 1.60`

The outer radius of curvature is
`R = 2.08 \ cm = 0.0208 \ m`

The inner radius is
`r = 2.48 \ cm = 0.0248 \ m`

Generally from Maker's equation we have that

`(1)/(f) = [n - 1 ][(1)/((R)) - (1)/(r) ]`

=>
`(1)/(f) = [1.60 - 1 ][(1)/(( 0.0208 )) - (1)/(0.0248) ]`

=>
`(1)/(f) = 4.65`

=>
`f = 0.2149 \ m`

=>
`f = 21.49 \ cm`