A glass block having index of refraction ng = 1.57 has laser light shining into

Question

A glass block having index of refraction ng = 1.57 has laser light shining into it at point P, which is 4.00 cm down from the top of the block. The transmitted light inside the glass at point P travels up to point Q, which is 6.00 cm from the left side of the block, where it is completely reflected down to point R.

(a) What is the value of the angle ? shown in the figure at point P?

(b) What is the angle of incidence of the laser light at point Q, thetaiQ? Draw in and label this angle in the figure in your solutions.

(c) What is the value of the critical angle for total internal reflection for this glass block?

(d) Explain why the light must be completely reflected at point Q.

(e) What is the value of the angle of transmission of the light into the air at the point R, thetatR?

THE SKITCH IN THE LINK BELOW

http://capone.mtsu.edu/phys2120/HW_Problems/CH24-I-HWP.pdf

Explanation / Answer

Part A)

tan(angle) = 6/4

angle = 56.3o

Part B)

Since alternate interior anges are congruent, the angle of incidence is also 56.3o

Part C)

sin(ic) = n2/n1 = 1/1.57

ic = 39.6o

Part D)

Because the angle of incidence is larger than the critical angle. Any angle at or larger than that value will create total internal reflection.

Part E)

The angle of incidence at R = 90 - 56.3 = 33.7o

Then, by Snell's Law n1sin(i) = n2sin(r)

1.57(sin 33.7) = 1(sin r)

r = 60.6o (If your answer key want the angle the other way, subtract from 90 and get 29.4o)

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