Question

The cornea behaves as a thin lens of focal length approximately 1.80 cm , although this varies a bit. The material of which it is made has an index of refraction of 1.38, and its front surface is convex, with a radius of curvature of 5.00 mm . (Note: The results obtained here are not strictly accurate, because, on one side, the cornea has a fluid with a refractive index different from that of air.)

(Part A) If this focal length is in air, what is the radius of curvature of the back side of the cornea?

Answer 1

`R_2 = 1.86 cm`

Step-by-step explanation:

By lens makers formula we know that focal length is given as

`(1)/(f) = ((\mu_2)/(\mu_1) - 1)((1)/(R_1) - (1)/(R_2))`

now we know that

`\mu_1 = 1`

`\mu_2 = 1.38`

`R_1 = 0.5 cm`

f = 1.80 cm

now from above expression

`(1)/(1.80) = (1.38 - 1)((1)/(0.5) - (1)/(R_2))`

`R_2 = 1.86 cm`