For an isosceles prism of angle A and refractive index \mu , it is found that the angle of minimum deviation \delta _m = A. Which of the following options is/are correct?

For the angle of incidence i_1 = A, the ray inside the prism is parallel to the base of the prism, At minimum deviation, the incident angle i_1 and the refracting angle r_1 at the first refracting surface are related by r_1 = ( i_1/2 ), For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is i_1 = \sin ^ - 1[ \sin A\sqrt 4\cos ^2A \over 2 - 1 - \cos A ]

For the angle of incidence i_1 = A, the ray inside the prism is parallel to the base of the prism

For this prism, the refractive index \mu and the angle of prism A are related as A = 1 \over 2\cos ^ - 1( \mu \over 2 )

At minimum deviation, the incident angle i_1 and the refracting angle r_1 at the first refracting surface are related by r_1 = ( i_1/2 )

For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is i_1 = \sin ^ - 1[ \sin A\sqrt 4\cos ^2A \over 2 - 1 - \cos A ]

Geometrical Optics PYQ — JEE Advanced 2017 · Shift 1 · Q43 with Solution

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Correct answer: C, A, D

Use the condition for minimum deviation

For a prism at minimum deviation,

\qquad r_1=r_2=r= rac{A}{2}$$ and $$\delta_m=2i-A.$$ Given $$\delta_m=A,$$ so $$2i-A=A \implies 2i=2A \implies i=A.$$ Thus at minimum deviation, $$i_1=A,\qquad r_1=\frac{A}{2}.$$ Using Snell's law at the first face, $$\sin i=\mu \sin r$$ so $$\sin A=\mu \sin \frac{A}{2}.$$ Now, $$\sin A=2\sin \frac{A}{2}\cos \frac{A}{2}$$ therefore $$2\sin \frac{A}{2}\cos \frac{A}{2}=\mu \sin \frac{A}{2}.$$ Assuming $A\neq 0$, cancel $\sin\frac A2$: $$\mu=2\cos \frac{A}{2}.$$ Hence, $$\cos \frac{A}{2}=\frac{\mu}{2} \quad\Rightarrow\quad \frac{A}{2}=\cos^{-1}\left(\frac{\mu}{2}\right).$$ So, $$A=2\cos^{-1}\left(\frac{\mu}{2}\right).$$ --- 2. **Check option B** Option B states $$A=\frac12\cos^{-1}\left(\frac{\mu}{2}\right).$$ But we derived $$A=2\cos^{-1}\left(\frac{\mu}{2}\right).$$ So **B is incorrect**. --- 3. **Check option C** At minimum deviation, we already found $$i_1=A,\qquad r_1=\frac{A}{2}.$$ Thus, $$r_1=\frac{i_1}{2}.$$ So **C is correct**. --- 4. **Check option A** In an isosceles prism, the prism angle is split equally by the perpendicular to the base, so each refracting face makes angle $$\frac{A}{2}$$ with the base. If inside the prism the ray has refracting angle $$r_1=\frac{A}{2},$$ then the ray makes angle $\frac{A}{2}$ with the normal to the first face. Therefore it makes angle $$90^\circ-\frac{A}{2}$$ with the face itself. Since the face is inclined at $\frac{A}{2}$ to the base, the ray becomes parallel to the base. Equivalently, at minimum deviation for an isosceles prism, the internal ray is symmetric and hence parallel to the base. Since from the given condition $\delta_m=A$, the corresponding minimum-deviation incidence is $i_1=A$, **A is correct**. --- 5. **Check option D** We need the incidence at first surface such that the emergent ray at second surface is tangential to the surface. That means at the second surface, the internal incidence equals the critical angle $c$: $$\sin c=\frac{1}{\mu}.$$ From above, $$\mu=2\cos\frac{A}{2}.$$ So, $$\sin c=\frac{1}{2\cos(A/2)}.$$ Let the refracted angle at first face be $r_1$. Inside prism, $$r_1+r_2=A.$$ For tangential emergence, $$r_2=c,$$ thus $$r_1=A-c.$$ At first surface, Snell's law gives $$\sin i_1=\mu\sin r_1=2\cos\frac{A}{2}\sin(A-c).$$ Expand: $$\sin(A-c)=\sin A\cos c-\cos A\sin c.$$ So, $$\sin i_1=2\cos\frac{A}{2}(\sin A\cos c-\cos A\sin c).$$ Now, $$\sin c=\frac{1}{2\cos(A/2)}.$$ Also, $$\cos c=\sqrt{1-\sin^2 c} =\sqrt{1-\frac{1}{4\cos^2(A/2)}} =\frac{\sqrt{4\cos^2(A/2)-1}}{2\cos(A/2)}.$$ Substitute: $$\sin i_1=2\cos\frac{A}{2}\left[\sin A\cdot \frac{\sqrt{4\cos^2(A/2)-1}}{2\cos(A/2)}-\cos A\cdot \frac{1}{2\cos(A/2)}\right].$$ This simplifies to $$\sin i_1=\sin A\sqrt{4\cos^2\frac{A}{2}-1}-\cos A.$$ Hence, $$i_1=\sin^{-1}\left[\sin A\sqrt{4\cos^2\frac{A}{2}-1}-\cos A\right].$$ This matches option D. So **D is correct**. --- 6. **Final selection** Correct options are: $$\boxed{A,\ C,\ D}$$ This matches the stored correct answer.

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Correct answer: C, A, D

Use the condition for minimum deviation

For a prism at minimum deviation,

\qquad r_1=r_2=r= rac{A}{2}$$ and $$\delta_m=2i-A.$$ Given $$\delta_m=A,$$ so $$2i-A=A \implies 2i=2A \implies i=A.$$ Thus at minimum deviation, $$i_1=A,\qquad r_1=\frac{A}{2}.$$ Using Snell's law at the first face, $$\sin i=\mu \sin r$$ so $$\sin A=\mu \sin \frac{A}{2}.$$ Now, $$\sin A=2\sin \frac{A}{2}\cos \frac{A}{2}$$ therefore $$2\sin \frac{A}{2}\cos \frac{A}{2}=\mu \sin \frac{A}{2}.$$ Assuming $A\neq 0$, cancel $\sin\frac A2$: $$\mu=2\cos \frac{A}{2}.$$ Hence, $$\cos \frac{A}{2}=\frac{\mu}{2} \quad\Rightarrow\quad \frac{A}{2}=\cos^{-1}\left(\frac{\mu}{2}\right).$$ So, $$A=2\cos^{-1}\left(\frac{\mu}{2}\right).$$ --- 2. **Check option B** Option B states $$A=\frac12\cos^{-1}\left(\frac{\mu}{2}\right).$$ But we derived $$A=2\cos^{-1}\left(\frac{\mu}{2}\right).$$ So **B is incorrect**. --- 3. **Check option C** At minimum deviation, we already found $$i_1=A,\qquad r_1=\frac{A}{2}.$$ Thus, $$r_1=\frac{i_1}{2}.$$ So **C is correct**. --- 4. **Check option A** In an isosceles prism, the prism angle is split equally by the perpendicular to the base, so each refracting face makes angle $$\frac{A}{2}$$ with the base. If inside the prism the ray has refracting angle $$r_1=\frac{A}{2},$$ then the ray makes angle $\frac{A}{2}$ with the normal to the first face. Therefore it makes angle $$90^\circ-\frac{A}{2}$$ with the face itself. Since the face is inclined at $\frac{A}{2}$ to the base, the ray becomes parallel to the base. Equivalently, at minimum deviation for an isosceles prism, the internal ray is symmetric and hence parallel to the base. Since from the given condition $\delta_m=A$, the corresponding minimum-deviation incidence is $i_1=A$, **A is correct**. --- 5. **Check option D** We need the incidence at first surface such that the emergent ray at second surface is tangential to the surface. That means at the second surface, the internal incidence equals the critical angle $c$: $$\sin c=\frac{1}{\mu}.$$ From above, $$\mu=2\cos\frac{A}{2}.$$ So, $$\sin c=\frac{1}{2\cos(A/2)}.$$ Let the refracted angle at first face be $r_1$. Inside prism, $$r_1+r_2=A.$$ For tangential emergence, $$r_2=c,$$ thus $$r_1=A-c.$$ At first surface, Snell's law gives $$\sin i_1=\mu\sin r_1=2\cos\frac{A}{2}\sin(A-c).$$ Expand: $$\sin(A-c)=\sin A\cos c-\cos A\sin c.$$ So, $$\sin i_1=2\cos\frac{A}{2}(\sin A\cos c-\cos A\sin c).$$ Now, $$\sin c=\frac{1}{2\cos(A/2)}.$$ Also, $$\cos c=\sqrt{1-\sin^2 c} =\sqrt{1-\frac{1}{4\cos^2(A/2)}} =\frac{\sqrt{4\cos^2(A/2)-1}}{2\cos(A/2)}.$$ Substitute: $$\sin i_1=2\cos\frac{A}{2}\left[\sin A\cdot \frac{\sqrt{4\cos^2(A/2)-1}}{2\cos(A/2)}-\cos A\cdot \frac{1}{2\cos(A/2)}\right].$$ This simplifies to $$\sin i_1=\sin A\sqrt{4\cos^2\frac{A}{2}-1}-\cos A.$$ Hence, $$i_1=\sin^{-1}\left[\sin A\sqrt{4\cos^2\frac{A}{2}-1}-\cos A\right].$$ This matches option D. So **D is correct**. --- 6. **Final selection** Correct options are: $$\boxed{A,\ C,\ D}$$ This matches the stored correct answer.

Geometrical Optics question

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