A light ray is incident to diamond at 0^∘ to the normal from surrounding water. All angles are p

Step 1: The light ray is incident to diamond at o to the normal from surrounding water.

A light ray is incident to diamond at 0^∘ to the normal from...

A light ray is incident to diamond at $0^\circ$ to the normal from surrounding water. All angles are provided in degrees. 1. Draw a schematic representing the light propagation in these media and label each relevant component accordingly and calculate the reflection and refraction angles of that ray. 2. How many surfaces (A, B, C, D, and/or E) of the diamond does the light reflect off of? 3. What is the first surface (A, B, C, D, or E) the light passes through after entering the diamond? 4. If the surrounding media was changed from water to air, would the answers to parts 2 and 3 change? Explain with equations.

Transcript

00:02 For part a of our question, it says to find the critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and the outside air.

00:13 Okay, so i have the incident or the index of refraction for air in sub a as 1 .0.

00:19 The index of refraction for diamond, n sub d, as 2 .419, and the index of refraction for water, which will use later as 1 .333.

00:28 So the critical angle is when the index of refraction is at least equal to 90 degrees.

00:37 So using snell's law, we find that the critical angle then is defined as theta c equal to the inverse sign of the medium that it's traveling into n sub a divided by n sub d.

00:59 This is inverse sign here, so sine of the minus one.

01:02 We find then that the angle is equal to 24 .4 degrees, which can be boxed in as our solution for part a.

01:18 Part b says, consider the light ray incident normally on the top surface of the diamond as shown in figure p 22 .42, and show that the light traveling between point p and the diamond is totally reflected.

01:32 Okay? so the angle of incident in the image is 35 degrees, and the critical angle, theta c, we just found, is 24 .4 degrees.

01:44 Here, the critical angle is less than the angle of incident.

01:47 And when the angle of incident is greater than the critical angle, the light undergoes total internal reflection.

01:52 Therefore, the light undergoes total internal reflection.

01:56 So for part b, we can say that 35 degrees is greater than the critical angle, theta c.

02:13 And then we'll type this out.

02:19 Thus total internal reflection.

02:30 That's our solution for part b.

02:39 Part c says if the diamond is immersed in water, find the critical angle of the diamond water interface.

02:45 So again, we're going to use the exact same formula that we use for part c, except now our index of refractions are going to be diamond in water.

02:57 So this is inverse sign of the ratio of the second medium that it travels into.

03:04 Which is the water divided by the index of refraction of the medium it's leaving, which is the diamond.

03:13 In sub -d.

03:18 Let's see here.

03:19 Okay.

03:21 So calculating this, we find that this is equal to 33 .4 degrees.

03:31 And that can be boxed in as a solution for part c.

03:37 Part d asks us to do something similar that we did to part b, in which it wants us to figure out, or it wants us to figure out if the ray is totally internally reflected...

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