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QJEEM15Phe5b2: JEE Main 2015 Physics - Class 12 JEEM15Phe5b2 Solution: Monochromatic light is incident on a glass prism of angle . If…

Monochromatic light is incident on a glass prism of angle AA . If the refractive index of the material of the prism is μ\mu , a ray, incident at an angle θ\theta . on the face ABAB would get transmitted through the face ACAC of the prism provided :

JEE Main 2015 (Offline) Physics - Geometrical Optics Question 192 English

Multiple Choice Options

  • A θ>cos⁡−1[μ sin⁡(A+sin⁡−1(1μ)]\theta > {\cos ^{ - 1}}\left[ {\mu \,\sin \left( {A + {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]
  • B θ<cos⁡−1[μ sin⁡(A+sin⁡−1(1μ)]\theta < {\cos ^{ - 1}}\left[ {\mu \,\sin \left( {A + {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]
  • C θ>sin−1[μ sin⁡(A−sin⁡−1(1μ)]\theta > si{n^{ - 1}}\left[ {\mu \,\sin \left( {A - {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]
  • D θ<sin−1[μ sin⁡(A−sin⁡−1(1μ)]\theta < si{n^{ - 1}}\left[ {\mu \,\sin \left( {A - {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]

Step-by-Step Verified Solution

When r2=C, ∠N2Rc=90∘{r_2} = C,\,\angle {N_2}Rc = {90^ \circ }

Where C=C = critical angle

As sin⁡C=1v=sin⁡r2\sin C = {1 \over v} = \sin {r_2}

JEE Main 2015 (Offline) Physics - Geometrical Optics Question 192 English Explanation
Applying snell's law at ′R′'R'

μsin⁡r2=1sin⁡90∘                 ...(i)\mu \sin {r_2} = 1\sin {90^ \circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

Applying snell's law at ′Q′'Q'

1×sin⁡θ=μsin⁡r1                 ...(ii)1 \times \sin \theta = \mu \sin {r_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

But r1=A−r2{r_1} = A - {r_2}

So, sin⁡θ=μsin⁡(A−r2)\sin \theta = \mu \sin \left( {A - {r_2}} \right)

θ=μsin⁡Acos⁡r2−cos⁡ A     ...(iii)\theta = \mu \sin A\cos {r_2} - \cos \,A\,\,\,\,\,...\left( {iii} \right) [using (i)(i) ]

From (1)(1)

cos⁡ r2=1−sin⁡2r2=1−1μ2       ...(iv)\cos \,{r_2} = \sqrt {1 - {{\sin }^2}{r_2}} = \sqrt {1 - {1 \over {{\mu ^2}}}} \,\,\,\,\,\,\,...\left( {iv} \right)

By eq. (iii)(iii) and (iv)(iv)

sin⁡θ=μsin⁡A1−1μ2−cos⁡A\sin \theta = \mu \sin A\sqrt {1 - {1 \over {{\mu ^2}}}} - \cos A

on further solving we can show for ray not to transmitted through face ACAC

θ=sin⁡−1[usin⁡(A−sin⁡−1(1μ)]\theta = {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right)} \right.} \right]

So, for transmission through face ACAC

\theta &gt; {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right.} \right)} \right]

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