What Is the Angle of Incidence for Light to Emerge Parallel from a Sphere?

[JonNash]

[Mentor's note: this thread does not use the normal homework forum template because it was originally posted in a non-homework forum, then moved here.]

A light beam is incident on a sphere with refractive index n=√3 at an angle i from air and emerges parallel to the horizontal axis passing through the center of the sphere. Find i?

Here I made a line from the center of the sphere

to the exit point of the ray after which it becomes parallel to the axis. I attached a rough diagram I made in paint for it.

Now the triangle formed inside the circle is isosceles and hence two sides and corresponding angles are equal so we have two sides as r and I named the last side as d and the angle at the center is unknown. Here I only have one numerical value (n=√3) and I have to find the angle of incidence. I know there is some geometry involved so I used the cosine rule to solve for the refracted angle but even combined with snells law and other properties of triangles (S and Δ) I just don't see how to proceed, can you guys please help me out?

 

[user3]

There you go, this image should help.

note that θ=Θ-θ because it's and isosceles triangle. so Θ=2θ (θ is the angle of refraction from air to the sphere)

so √3 sin(2θ-θ) = sin(2θ)

expand the sin on the RHS you get: √3 sin(θ) = 2sin(θ)cos(θ)

solve that you get θ = 30°

now it's easy to solve this equation sin(i)=√3 sin(θ) to get i.

 

[JonNash]

note that θ=Θ-θ because it's and isosceles triangle. so Θ=2θ (θ is the angle of refraction from air to the sphere)

so √3 sin(2θ-θ) = sin(2θ)

expand the sin on the RHS you get: √3 sin(θ) = 2sin(θ)cos(θ)

solve that you get θ = 30°

now it's easy to solve this equation sin(i)=√3 sin(θ) to get i.

i=60⁰
Thanks a gazillion. Just extending one line makes a hell of a difference, well, I guess that's geometry for you. Anyhow, thanks again. World is simpler again till I find another brainteaser from Euclid.