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Light enters a solid tube made of plastic having an index of refraction of 1.60. The light travels parallel to the upper part of the tube. You want to cut the face AB so that all the light will reflect back onto the tube after it first strikes that face.
(a) what is the largest that Theta can be if the tube is in air?
(b) if the tube is immersed in water of refractive index 1.33, what is the largest that Theta can be?

Answer :

Answer:

The answer is "[tex]\bold{51.31^{\circ} \ and\ 33.77^{\circ}}[/tex]"

Explanation:

Please find the graph in the attachment.

For point a:

Calculating the total internal reflection:

[tex]\sin(90-\theta_c) = \frac{n_1}{n_2}\\\\\cos?_c = \frac{n_1}{n_2}\\\\[/tex]

Therefore the [tex]n_1[/tex] is the refractive index in the air medium that is = 1.00

[tex]n_2[/tex] = refractive index of medium = 1.60

[tex](90 - ?_c )[/tex] =incident angle

 So,

[tex]?_c = 51.31^{\circ}[/tex]  

For point b:

When the pipe is immersed into the water then the equation is:

[tex]1.6\times \sin(90-?)=1.33\times \sin 90\\\\?_c = 33.77^{\circ}[/tex]

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