Light ray passing thru plastic block

[herbally]

Homework Statement

A light ray passes downward into a block of transparent plastic with an angle of incidence of θ₁ = 77.7°.

If total internal reflection is to occur when the light strikes the left edge of the block at interface B, what is the index of refraction of the plastic?

Homework Equations

Snell's Law - n₁sin(θ₁)=n₂sin(θ₂)

sin(θ_crit)=n₂/n₁

The Attempt at a Solution

I thought that I could use the law of refraction at interface A and the equation for the critical value at interface B (since total internal refraction occurs there), and use those two equations to solve for n₂. I've done all manners of substitutions, but nothing that yields a reasonable answer.

I would really appreciate a nudge in the right direction.

Thanks in advance!

 

[TSny]

Hello.
Your outline of how to approach the problem sounds good. Can you show your work so we can see what you've done?

 

[herbally]

I solved Snell's law for sin(θ₂) which yielded n₁sin(θ₁)/n₂.

I set that equal to n₂/n₁. Then I solved for n₂ which gave me n₂=+-n₁√sin(θ₁).

When I solve I get some value like .98755 which is less than one making it obviously incorrect.

 

[TSny]

I think you're using the same symbol θ₂ for two different angles.

 

[Nickie]

I would first confirm the given information and make sure it is clear and accurate. Is the angle of incidence given in degrees or radians? Is the block of plastic truly transparent, or is it partially opaque? These details can affect the calculations and results.

Assuming that the angle of incidence is given in degrees and the block of plastic is fully transparent, I would proceed with solving for the index of refraction using Snell's Law and the equation for the critical angle. It is important to note that the angle of incidence at interface A is equal to the angle of refraction at interface B, since the light ray is passing through the block without changing direction. This means that θ1 = θ2.

Using Snell's Law, we can rewrite it as n1sin(θ1) = n2sin(θ1) and solve for n2. This yields n2 = n1/sin(θ1). Since we want to find the index of refraction of the plastic, we can substitute the given angle of incidence (θ1 = 77.7°) and the known index of refraction of air (n1 = 1) to get n2 = 1/sin(77.7°). This gives us an index of refraction of approximately 1.22 for the plastic block.

To verify this result, we can also use the equation for the critical angle, sin(θcrit) = n2/n1, and solve for n2. Plugging in the known values, we get n2 = sin(θcrit) = sin(77.7°) = 0.975. This is close to our previous result, confirming that the index of refraction of the plastic is indeed around 1.22.

In summary, as a scientist, I would first confirm the given information and then use the appropriate equations to solve for the index of refraction of the plastic block. It is important to double check the calculations and make sure they are reasonable and consistent with the given information.