How can I find the minimum index of refraction?

[mbnMecha]

Homework Statement

We look at the center of one face of a solid cube of glass on a line of sight making 55° with the normal to the cube face. What is the minimum refractive index of glass for which you will see through the opposite face of the cube? (Hint: see through will be possible if the ray refracted on the first face can emerge from the opposite face)

Homework Equations

Snells law: n1sin(x1) =n2sin(x2)
Where n1 and n2 are the indices of refraction and x1 if the angle of incidence while x2 is the angle of refraction.
Case of total internal reflection:
N1sin(x1) = n2sin90

The Attempt at a Solution

I tried finding the critical angle of incidence but I got no where.. been working on this for 12 hours, i feel like I am seriously missing something! Help!

 

[.Scott]

So your N1sin(x1) is the emerging ray. N1 is for air, so what is N1?
x1 is the angle that it must emerge from, so what is x1?

 

[mbnMecha]

So your N1sin(x1) is the emerging ray. N1 is for air, so what is N1?
x1 is the angle that it must emerge from, so what is x1?

N1=1 , x1 = 55°... sin55° = 0.82, no?

 

[blue_leaf77]

Are you not given a picture? If this cube of glass is immersed in air, there is no way one cannot see through the other face of the glass.

 

[mbnMecha]

Are you not given a picture? If this cube of glass is immersed in air, there is no way one cannot see through the other face of the glass.

I am not given a picture, and yes the cube is inmersed in air with n1= 1

 

[.Scott]

I' sorry. I misread the problem. The problem is with the geometry of the cube. If the cube had an index of refraction of 1, then looking at the center of the face would allow us to see through the opposite side at angles up to 26.565° (atan(0.5)) before we reached the edge of the opposite face or 35.264° (atan(sqrt(0.5))) before we reached a corner of the opposite face.

So those are the maximum internal angles (x2) you can use. Which you use is a matter of how you interpret the question. I would use 26.6 with a note that 35.3 would also work if the cube was positioned correctly.

So, yes, your N1 is 1 and your x1 is 55°.

 

[mbnMecha]

I' sorry. I misread the problem. The problem is with the geometry of the cube. If the cube had an index of refraction of 1, then looking at the center of the face would allow us to see through the opposite side at angles up to 26.565° (atan(0.5)) before we reached the edge of the opposite face or 35.264° (atan(sqrt(0.5))) before we reached a corner of the opposite face.

So those are the maximum internal angles (x2) you can use. Which you use is a matter of how you interpret the question. I would use 26.6 with a note that 35.3 would also work if the cube was positioned correctly.

So, yes, your N1 is 1 and your x1 is 55°.

Thank you so much! Therefore can I assume that for all the light to be refracted and the cube to be seethrough, the light has to hit the corner of the cube and not allow for any internal reflection of light? Therefore Id get x2 through having tan-1(d/2/d) = tan^-1(0.5)...? Or are there any laws of sine and cosine i ahould use to get x2?

 

[.Scott]

Internal reflection is OK - it is not part of the problem. The problem is with the geometry of the cube.
The part that is easy to misinterpret is "emerging from the opposite face". The problem with emerging from the opposite face is commonly internal reflection - but not with this problem. For this problem, it is the geometry of the cube. At too great an angle, all the light will reach adjoining faces before it reaches the opposite face.