JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 15)
A ray of light entering from air into a denser medium of refractive index $${4 \over 3}$$, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle $$\theta$$ should be equal to :
_25th_July_Evening_Shift_en_15_1.png)
$${\sin ^{ - 1}}{{\sqrt 7 } \over 3}$$
$${\sin ^{ - 1}}{{\sqrt 5 } \over 4}$$
$${\sin ^{ - 1}}{{\sqrt 7 } \over 4}$$
$${\sin ^{ - 1}}{{\sqrt 5 } \over 3}$$
Explanation
_25th_July_Evening_Shift_en_15_2.png)
At maximum angle $$\theta$$ ray at point B goes in gazing emergence, at all less values of $$\theta$$, TIR occurs.
At point B
$${4 \over 3} \times \sin \theta '' = 1 \times \sin 90^\circ $$
$$\theta '' = {\sin ^{ - 1}}\left( {{3 \over 4}} \right)$$
$$\theta ' = \left( {{\pi \over 2} - \theta ''} \right)$$
At point A
$$1 \times \sin \theta = {4 \over 3} \times \sin \theta '$$
$$\sin \theta = {4 \over 3} \times \sin \left( {{\pi \over 2} - \theta ''} \right)$$
$$\sin \theta = {4 \over 3}\cos \left[ {{{\cos }^{ - 1}}{{\sqrt 7 } \over 4}} \right]$$
_25th_July_Evening_Shift_en_15_3.png)
$$\sin \theta = {4 \over 3} \times {{\sqrt 7 } \over 4}$$
$$\theta = {\sin ^{ - 1}}\left( {{{\sqrt 7 } \over 3}} \right)$$
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