Refraction by a transparent spherical surface

[bcjochim07]

Homework Statement

Parallel light rays enter a transparent sphere along a line passing through the center of the sphere. The rays come to a focus on the far surface of the sphere. What is the sphere's index of refraction?

Homework Equations

The Attempt at a Solution

so I used the equation that relates object distances and image distances for spherical surfaces: n1/s + n2/s' = (n2-n1)/R

I am calling the radius of the sphere R

(1.00)/s + n2/2R = (n2-1.00)/R and I think for this case s$$\rightarrow$$$$\infty$$
so n2/2R = (n2-1)/R
n2 = 2(n2-1)
n2 = 2n2 -2
-n2=-2
so n2= 2.00 Is this correct?

 

[Kurdt]

Looks fine to me.

 

[baba89]

Your approach and calculation are correct. The sphere's index of refraction is indeed 2. This means that the speed of light is twice as slow inside the sphere compared to outside. This phenomenon is known as refraction, where light bends as it passes through a medium with a different index of refraction. It is a fundamental concept in optics and has many practical applications, such as in lenses and prisms. Keep up the good work!