Snell's law problem about a hemispherical glass ball

[issacnewton]

Homework Statement

I have posted the snapshot of the problem.

Homework Equations

Snell's law equations

The Attempt at a Solution

Now the problem says that laser travels through air vertically upward to reach the outer surface of the glass half-cylinder. If that is the case, then from Snell's law, the angle at which the laser is entering the bottom should be zero with the vertical. But the diagram clearly shows that laser enters at some non zero angle to the vertical. So, is there any other way that laser will be going vertically upward to reach outer surface of the glass half-cylinder ? I am confused here.

Thanks

 

[haruspex]

What are the possible paths of the light ray having entered the tank?

 

[issacnewton]

Laser will first go inside the lower glass and then enter water and then hit the right glass wall. And it will again enter the right glass wall. The only way the laser will turn back in is the total internal reflection at the interface of glass and air outside the right glass wall. And from there it will again enter water I think and from water it will enter the air inside the glass tank and from there it will enter upper glass surface of the tank and then it will emerge vertically from the tank. This is the only possibility I can think of. What do you think haruspex ?

 

[haruspex]

The only way the laser will turn back in is the total internal reflection at the interface of glass and air outside the right glass wall

What if it doesn't quite do that?

 

[issacnewton]

Jyoti, I did not say that at point A, the angle of incidence is zero. Assuming that its non zero, one of the possibilities that laser can emerge vertically up is the way I have described in post # 3

 

[issacnewton]

What if it doesn't quite do that?

In that case, the angle of incidence at point A would have to be zero, but the problem clearly says that the angle of incidence at A is not zero.

 

[haruspex]

In that case, the angle of incidence at point A would have to be zero, but the problem clearly says that the angle of incidence at A is not zero.

You may be right.. don't have time to check right now, but I did not think that was the case.
Try writing out the algebra for that possibility and for the one you mention in post #3.

Back in 4 hours or so.

 

[issacnewton]

I think I got the path as described as in post # 3 but with one change. The laser beam is incident on the glass air interface on the outer glass surface on the right side of the tank with the critical angle. This is possible since laser will be going from denser media to less denser media. So, after this, the laser will travel vertically upward parallel to the right glass surface. This is the only way the laser will strike the glass cylinder vertically. With this consideration and after lot of algebra, I am getting ##(A_x, A_y) = (81.3,-1.28)\;cm## and ##(B_x, B_y) = (0, 128.3)\;cm## Can people confirm this ?

 

[haruspex]

The laser beam is incident on the glass air interface on the outer glass surface on the right side of the tank with the critical angle.

That's what I was hinting at in post #4.

 

[issacnewton]

Thanks haruspex... problem involves lot of trig algebra