(c) Find the maximum value of 0 for which the refracted ray undergoes total internal reflection at the point P located at the left vertical face of the carbon disulfide. (If an answer does not exist, enter DNE.) Omax < Explain.

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Ch. 34

(c) Find the maximum value of 0 for which the refracted ray undergoes total internal reflection at the point P located at the left vertical face of the block if the polystyrene block is immersed in
carbon disulfide. (If an answer does not exist, enter DNE.)
Omax <
Explain.
Transcribed Image Text:(c) Find the maximum value of 0 for which the refracted ray undergoes total internal reflection at the point P located at the left vertical face of the block if the polystyrene block is immersed in carbon disulfide. (If an answer does not exist, enter DNE.) Omax < Explain.
A light ray of wavelength 589 nm is incident at an angle 0 on the top surface of a block of polystyrene as shown in the figure below.
Р.
(a) Find the maximum value of 0 for which the refracted ray undergoes total internal reflection at the point P located at the left vertical face of the block. (If an answer does not exist, enter
DNE.)
Omax <
Explain.
(b) Find the maximum value of 0 for which the refracted ray undergoes total internal reflection at the point P located at the left vertical face of the block if the polystyrene block is immersed in
water. (If an answer does not exist, enter DNE.)
Omax <
Explain.
Transcribed Image Text:A light ray of wavelength 589 nm is incident at an angle 0 on the top surface of a block of polystyrene as shown in the figure below. Р. (a) Find the maximum value of 0 for which the refracted ray undergoes total internal reflection at the point P located at the left vertical face of the block. (If an answer does not exist, enter DNE.) Omax < Explain. (b) Find the maximum value of 0 for which the refracted ray undergoes total internal reflection at the point P located at the left vertical face of the block if the polystyrene block is immersed in water. (If an answer does not exist, enter DNE.) Omax < Explain.
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