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?

Respuesta :

Answer:

The value  [tex]f = 21.49 \ cm[/tex]

Explanation:

From the question we are told that

    The index of refraction is  [tex]n = 1.60[/tex]

     The outer radius of curvature is  [tex]R = 2.08 \ cm = 0.0208 \ m[/tex]

     The  inner radius is  [tex]r = 2.48 \ cm = 0.0248 \ m[/tex]

Generally from  Maker's equation we have that

       [tex]\frac{1}{f} = [n - 1 ][\frac{1}{(R)} - \frac{1}{r} ][/tex]

=>    [tex]\frac{1}{f} = [1.60 - 1 ][\frac{1}{( 0.0208 )} - \frac{1}{0.0248} ][/tex]

=>    [tex]\frac{1}{f} = 4.65[/tex]

=>    [tex]f = 0.2149 \ m[/tex]

=>    [tex]f = 21.49 \ cm[/tex]