Question

21. Refraction and Total Internal Reflection: A ray of light traveling in air strikes a rectangular block of silicon nitride (whose index of refraction is: n=2.0) at an angle of 60° from the normal (at point ?A”). The light enters the block and refracts towards point B, where it might refract back into the surrounding air, or it might suffer total internal reflection. a. Determine the angle of refraction at point A b. Determine the angle of incidence at point B c. Calculate the critical angle for light trying to cross the silicon nitride/air boundary d. Determine if at point B whether total internal reflection will occur or instead refraction back into the surrounding air. air B A 60° Silicon Nitride

          21. Refraction and Total Internal Reflection:  A ray of light traveling in air strikes a rectangular block of silicon nitride (whose index of refraction is: n=2.0)  at an angle of 60° from the normal (at point ?A”).  The light enters the block and refracts towards point B, where it might refract back into the surrounding air, or it might suffer total internal reflection.
a.  Determine the angle of refraction at point A
b.  Determine the angle of incidence at point B
c.  Calculate the critical angle for light trying to cross the silicon nitride/air boundary
d.  Determine if at point B whether total internal reflection will occur or instead refraction back into the surrounding air.

air
B
A
60°
Silicon Nitride

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$21. Refraction and Total Internal Reflection: A ray of light traveling in air strikes a rectangular block of silicon nitride (whose index of refraction is: n=2.0) at an angle of 60° from the normal (at point ?A”). The light enters the block and refracts towards point B, where it might refract back into the surrounding air, or it might suffer total internal reflection. a. Determine the angle of refraction at point A b. Determine the angle of incidence at point B c. Calculate the critical angle for light trying to cross the silicon nitride/air boundary d. Determine if at point B whether total internal reflection will occur or instead refraction back into the surrounding air. air B A 60° Silicon Nitride$

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University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Rajesh Kumar

11/05/2021

Step 1

To find the angle of refraction (\(\theta_2\)) when the light enters the silicon nitride block, we use Snell's Law, which is given by \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\), where \(n_1\) and \(n_2\) are the indices of refraction of the first and second Show more…

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21. Refraction and Total Internal Reflection: A ray of light traveling in air strikes a rectangular block of silicon nitride (whose index of refraction is: n=2.0) at an angle of 60° from the normal (at point ‐A”). The light enters the block and refracts towards point B, where it might refract back into the surrounding air, or it might suffer total internal reflection. a. Determine the angle of refraction at point A b. Determine the angle of incidence at point B c. Calculate the critical angle for light trying to cross the silicon nitride/air boundary d. Determine if at point B whether total internal reflection will occur or instead refraction back into the surrounding air.

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00:01 In this question we are given that there is a transparent material here which is silicon nitrite and there is a ray incident here at an angle 60 degree and this point name is a.

00:22 So it is going to hit from here to here.

00:24 This is the angle of reflection r.

00:28 This is point b.

00:31 Now what we need to do is in the part a we have to actually find the angle of reflection r.

00:39 Which we can see from the diagram.

00:41 So that is very simple.

00:42 I think everyone can do that.

00:44 It is, as we know that, there is air here, whose refractive index we know is one.

00:52 And here it is silicon nitrite whose refractive index is given to us as two.

00:57 So now we can apply snells law here, which says 1 times of sine 60 is equal to 2 times of sine r.

01:09 So which implies sine 60, we know it is root 3.

01:13 2 is equal to 2 times of sine r.

01:17 So the value of r can be calculated from here, which is sine inverse of root 3 by 4, which if you solve will come out to be approximately equal to 25 .66 degrees.

01:35 So now in the second part of the question, we need to find this angle of incidence here.

01:42 So we can see in this triangle, this is 90 degree, this is r, this is i...

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