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QJEEM21Phd20b: JEE Main 2021 Physics - Class 12 JEEM21Phd20b Solution: Cross-section view of a prism is the equilateral triangle ABC…

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 = 32{{\sqrt 3 } \over 2} )

JEE Main 2021 (Online) 31st August Evening Shift Physics - Geometrical Optics Question 107 English

Step-by-Step Verified Solution

i = A = 60 ∘^\circ

δmin⁡{{ \delta } _{\min }} = 2i −- A

= 2 ×\times 60 ∘^\circ −- 60 ∘^\circ = 60 ∘^\circ

μ=sin⁡−1(δmin⁡+A2)sin⁡−1(A2)\mu = {{{{\sin }^{ - 1}}\left( {{{{\delta _{\min }} + A} \over 2}} \right)} \over {{{\sin }^{ - 1}}\left( {{A \over 2}} \right)}}

=3= \sqrt 3

Vprism =3×1083= {{3 \times {{10}^8}} \over {\sqrt 3 }}

AP = 10 ×\times 10 −- 2 ×\times 32{{\sqrt 3 } \over 2}

time =5×10−23×108×3×3= {{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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