How to find unknown refractive index to solve for theta

[Richie Smash]

Homework Statement

Hello, I will post a picture below.
An optical device consists of two glass blocks of different indices of refraction, arranged as shown in the picture. The refractive index from air to medium 1 is 1.2, and the refractive index from medium 1 to medium 2 is 1.25
If light is incident on the block with the smaller refractive index at 60°, determine the angles θ1,θ2,θ3,θ4 and θ5.

Homework Equations

Sin(i)/Sin(r) = ₁n₂

The Attempt at a Solution

Hello, I have figured out theta 1, by using the fact that sin60°/sinθ2= 1.2, from there I rearranged the formula to find theta 1 which is 46.2°.

From there I found theta 2 using alternate angle rule, so from there I found theta 3 using the same concept
: Sin46.2°/sinθ4 =1.25
so theta 3 = 35.3°

So from here once again i found theta 4 using the alternate angle rule, now I'm stuck trying to find theta 5.

I would have done the same thing sin35.3°/sinθ5=??

The problem is i don't know the refractive index of medium 2 to air, i only know the two refractive indices they gave me.
But would the refractive index of medium 2 to air be the same as from medium 1 to medium 2? I can figure it out immediately I'm just not sure what obvious sign I'm clearly missing.

 

[Orodruin]

How does the index of refraction say about the light speed in the different media? What does this tell you about the speed of light in medium 2 expressed in terms of the speed of light in vacuum?

Also: You can figure it out directly without ever computing the relative index of refraction between medium 2 and air and without doing a single computation.

 

[Richie Smash]

the speed of light will be slower in an optically denser medium, so it says to me that it will be slower in medium 2 than in air.

I'm trying to figure out how I can figure it out without doing a single computation however, it must be something obvious I am missing.

 

[Richie Smash]

This is just a guess but would theta 5 be the very same as the first incident angle 60°?

 

[Orodruin]

This is just a guess but would theta 5 be the very same as the first incident angle 60°?

Yes. But you will need to come up with some sort of argument for why this is so ... why did you make this guess?

 

[Richie Smash]

I guessed that it was 60 degrees because one it looked like the path of light travel returned to it's original path, like the two were lined up almost.

And after thinking about it some more, it might be to do with the fact that it's returning to air, which is the medium it started it.

 

[Orodruin]

I guessed that it was 60 degrees because one it looked like the path of light travel returned to it's original path, like the two were lined up almost.

I suggest that you do not trust illustrations to this extent. Illustrations can be misleading and you need to do the maths if you want to actually have an argument. That I draw it a way that makes it seem so does not make it so.

I also suggest that you think about it with the absolute indices of refraction rather than the relative ones. It should make things clearer.

 

[Richie Smash]

I suggest that you do not trust illustrations to this extent. Illustrations can be misleading and you need to do the maths if you want to actually have an argument. That I draw it a way that makes it seem so does not make it so.

I also suggest that you think about it with the absolute indices of refraction rather than the relative ones. It should make things clearer.

Ok, well I have looked it up a bit, and I see that the absolute refractive index is the ratio of the speed of light in a vacuum to the speed of light in a medium, which is always greater than one... so from the medium to air the light would be going back to it's original speed, yes so the angle must be increased.. to match the speed?

(sorry for my not so great guesses)

 

[Orodruin]

I suggest that you use Snell's law of refraction $n_i \sin\theta_i = n_j \sin\theta_j$ to make your argumentation.

 

[Richie Smash]

n_isinθi=n_jsinθj

θj= sin^-1(n_isinθi)/sinθj

But to be honest I'm not sure what to do at this point, may I ask for your assistance in understanding what's happening here?

 

[Orodruin]

No, do not solve for the $\theta_i$. Compare what you have in each transition. The main point is that $n_i \sin(\theta_i)$ is going to be the same number wherever you are. Thus, if you start in air and end up in the same medium with refractive index $n$ ...

 

[Richie Smash]

Oh thank you very much Oruduin, I now understand, the
n _i sinθi will be the same in air regardless of what medium the light is going in or exiting from