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QJEEM22Phcbbd: JEE Main 2022 Physics - Class 12 JEEM22Phcbbd Solution: The X-Y plane be taken as the boundary between two transparent…

The X-Y plane be taken as the boundary between two transparent media M1\mathrm{M}_{1} and M2\mathrm{M}_{2} . M1\mathrm{M}_{1} in Z⩾0Z \geqslant 0 has a refractive index of 2\sqrt{2} and M2M_{2} with Z<0 has a refractive index of 3\sqrt{3} . A ray of light travelling in M1\mathrm{M}_{1} along the direction given by the vector P→=43i^−33j^−5k^\overrightarrow{\mathrm{P}}=4 \sqrt{3} \hat{i}-3 \sqrt{3} \hat{j}-5 \hat{k} , is incident on the plane of separation. The value of difference between the angle of incident in M1\mathrm{M}_{1} and the angle of refraction in M2\mathrm{M}_{2} will be __________ degree.

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Normal will be −k^- \widehat k so

JEE Main 2022 (Online) 29th July Morning Shift Physics - Geometrical Optics Question 78 English Explanation

cos⁡i=P← . n^∣P←∣ . ∣n^∣\cos i = {{\overleftarrow P \,.\,\widehat n} \over {\left| {\overleftarrow P } \right|\,.\,\left| {\widehat n} \right|}}

510=12{5 \over {10}} = {1 \over 2}

⇒i=60∘\Rightarrow i = 60^\circ

and using Snells law

2sin⁡60∘=3sin⁡r\sqrt 2 \sin 60^\circ = \sqrt 3 \sin r

32=3sin⁡r{{\sqrt 3 } \over {\sqrt 2 }} = \sqrt 3 \sin r

⇒r=45∘\Rightarrow r = 45^\circ

So, i−r=15∘i - r = 15^\circ

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