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Isaac Physics Optical Dipstick Question

I need help answering part b) to this Isaac Physics question on Refraction. Part a) In this diagram, the light from the lamp is returned to the photocell as long as the RI of the prism exceeds a certain value,n. I calculated n= 1.41. For part b) the RI of the prism is now n1=1.90 and immersed in liquid. The RI also varies with depth. 1.33 at its surface to 1.39 at a depth of 12.0cm. What depth will the light no longer completely reach the photocell?
https://isaacphysics.org/questions/optical_dipstick#:~:text=Use%20as%20a%20dipstick&text=The%20liquid%20is%20a%20solution,a%20depth%20of%2012.0cm.

Reply 1

Original post
by bitter-chaise
I need help answering part b) to this Isaac Physics question on Refraction. Part a) In this diagram, the light from the lamp is returned to the photocell as long as the RI of the prism exceeds a certain value,n. I calculated n= 1.41. For part b) the RI of the prism is now n1=1.90 and immersed in liquid. The RI also varies with depth. 1.33 at its surface to 1.39 at a depth of 12.0cm. What depth will the light no longer completely reach the photocell?
https://isaacphysics.org/questions/optical_dipstick#:~:text=Use%20as%20a%20dipstick&text=The%20liquid%20is%20a%20solution,a%20depth%20of%2012.0cm.
Immagine 2025-05-02 114127.png
Part A
Now, here's a key idea: how much light bounces back depends on what the prism is made of its refractive index. This whole setup looks like it's playing with something called total internal reflection. Think of it like a super-efficient mirror inside the prism.
The really interesting bit happens when the light inside the prism tries to get out into the air. That's the crucial boundary we need to focus on.
Figuring Out That Refractive Index

1.

What Makes Light Stay Inside? For that "super-efficient mirror" effect (total internal reflection) to kick in, two things need to happen. First, light has to be going from a place where it travels slower (the prism, presumably) to a place where it travels faster (air). Second, the angle at which it hits that boundary has to be pretty steep steeper than a certain angle called the critical angle.

2.

Tracking the Light's Angle: Look at the prism's shape and that 45° angle. We need to figure out the angle at which the light beam smacks into the bottom surface. When it bounces off, the angle it goes in is the same as the angle it comes out. Then, as that reflected beam heads towards the prism-air edge, its angle changes again based on the prism's geometry. We need to find the final angle of incidence at the prism-air surface.

3.

The Tipping Point: That angle we just found? It has to be at least as big as the critical angle for the light to stay trapped inside and bounce back completely.

4.

Snell's Law to the Rescue: The critical angle is a special case. It's the angle inside the prism that makes the light bend so much when it hits the air that it skims right along the surface (at 90° to the boundary). We can use Snell's Law here: nprism​×sin(θcritical​)=nair​×sin(90°)

5.

We know that nair​ is roughly 1, and sin(90°) is exactly 1. So, our equation simplifies to: nprism​×sin(θcritical​)=1

6.

Putting It All Together: Now, that θcritical in the equation is related to the angle of incidence we found in step 2. By connecting these pieces and doing a little math (you've got this!), you can solve for prism​. That value will be the minimum refractive index the prism needs for the total internal reflection. The question mentions it "exceeds a certain value," and that "certain value" is what you'll calculate. Make sure to round your final answer to three significant figures.

Take your time working through the angles and applying Snell's Law. You're on the right track to figuring this out yourself!
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Part B
Key Ideas:
Total internal reflection stops when light can't move from a denser medium (like a prism) to a less dense one (like a liquid). The critical angle can't be larger than the angle of incidence.
How to Calculate:
To find the critical angle using Snell’s Law: The critical angle, θc, is the angle at which light exits the prism at a right angle (90°) in the liquid.
Use Snell's Law:
n1 sin(θc) = n2 sin(90°)
Here, n1 = 1.90 (prism) and n2 is the refractive index of the liquid.
Index matched liquid: Refractive index is varied from n2 = 1.33 to n2 = 1.39, up on 0-12.0 cm. Construct a linear equation for n2:
n2 = md + b, m = the slope and b= the y- intercept.
Discovering the Critical Depth: When the prism's angle of incidence (kept at 45°) equals its critical angle, light stops reaching the photocell. If you had a refraction into vacuum rather than out of it that would be 0 degrees, and you would just be able to reverse the two indices of refraction in your ratio above to compute the n value. Calculate "d" which gives you the depth at which TIR (Total Internal Reflection).
By doing this, you learn about the depth at which light stops getting to the photocell, and you focus on Snell’s Law and the refractive index of the liquid as it varies with depth.
Here is my 2 cents!
(edited 10 months ago)

Reply 2

GetImage.jpeg Thank you! Here is what I did. I got the answer right!!

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