JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 28)
Cross-section view of a prism is the equilateral triangle ABC in the figure. The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle. The time taken by light to travel from P (midpoint of BC) to A is ______________ $$\times$$ 10$$-$$10 s. (Given, speed of light in vacuum = 3 $$\times$$ 108 m/s and cos30$$^\circ$$ = $${{\sqrt 3 } \over 2}$$)
_31st_August_Evening_Shift_en_28_1.png)
Answer
5
Explanation
i = A = 60$$^\circ$$
$${{ \delta } _{\min }}$$ = 2i $$-$$ A
= 2 $$\times$$ 60$$^\circ$$ $$-$$ 60$$^\circ$$ = 60$$^\circ$$
$$\mu = {{{{\sin }^{ - 1}}\left( {{{{\delta _{\min }} + A} \over 2}} \right)} \over {{{\sin }^{ - 1}}\left( {{A \over 2}} \right)}}$$
$$ = \sqrt 3 $$
Vprism $$ = {{3 \times {{10}^8}} \over {\sqrt 3 }}$$
AP = 10 $$\times$$ 10$$-$$2 $$\times$$ $${{\sqrt 3 } \over 2}$$
time $$ = {{5 \times {{10}^{ - 2}}} \over {3 \times {{10}^8}}} \times \sqrt 3 \times \sqrt 3 $$
= 5 $$\times$$ 10$$-$$10 sec
Ans. = 5
$${{ \delta } _{\min }}$$ = 2i $$-$$ A
= 2 $$\times$$ 60$$^\circ$$ $$-$$ 60$$^\circ$$ = 60$$^\circ$$
$$\mu = {{{{\sin }^{ - 1}}\left( {{{{\delta _{\min }} + A} \over 2}} \right)} \over {{{\sin }^{ - 1}}\left( {{A \over 2}} \right)}}$$
$$ = \sqrt 3 $$
Vprism $$ = {{3 \times {{10}^8}} \over {\sqrt 3 }}$$
AP = 10 $$\times$$ 10$$-$$2 $$\times$$ $${{\sqrt 3 } \over 2}$$
time $$ = {{5 \times {{10}^{ - 2}}} \over {3 \times {{10}^8}}} \times \sqrt 3 \times \sqrt 3 $$
= 5 $$\times$$ 10$$-$$10 sec
Ans. = 5
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