- #1
afcwestwarrior
- 457
- 0
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
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