Finding refractive index for point image at back of sphere

[afcwestwarrior]

A beam of parallel light rays from a laser is incident on a solid transparent sphere of index of refraction n. (a) If a point image is produced at the back of the sphere, what is the index of refraction of the sphere.

Here's the equation you use to find the index of refraction of the sphere.

(n1/p ) + (n2/ i) = (n2 - n1 ) / (r)

P (object distance or laser distance) = infinity (makes sense)

i (image distance) = 2r

r = radius of curvature

n1 (object's side refraction index) = 1.000 (air)

n2 (refractive index of the sphere ) = ?




(1/ infinity) + (n2 / 2r) = (n2 - 1) / r


somehow 1 / infinity goes away

so it becomes
(n2 / 2r) = (n2 - 1) / r

now it becomes n2 = 2 (n2-1)

r's cancel out


now it becomes n2 = 2n2 - 2


The answer is n2 = 2

How did they just get 2 when I have n2 = 2n2 - 2

 

[afcwestwarrior]

Nevermind I figured it out. It becomes

n2/n2 = ( 2n2/n2) - (2/n2)

then it becomes

1 = 2 - 2/n2

then it becomes

1 + 2/n2 =2

then it becomes

2/n2 = 2 -1

then it becomes

2/n2 = 1

then 2 = n2