A ray of light enters a spherical drop of a liquid of refractive index μ=72. This ray after getting reflected from back surface comes out as shown. Given sin41°=37.

1 sinϕ=72sinr1 and δ=2(ϕ−r1)+π−2r1δ is min. when dδdϕ=0⇒dr1dϕ=12Since rays enters from air to glass ⇒ r2≤θcThus no TIR from back surface

A ray of light enters a spherical drop of a liquid of refractive index μ=72. This ray after getting reflected from back surface comes out as shown. Given sin41°=37.

1 sinϕ=72sinr1 and δ=2(ϕ−r1)+π−2r1δ is min. when dδdϕ=0⇒dr1dϕ=12Since rays enters from air to glass ⇒ r2≤θcThus no TIR from back surface

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A ray of light enters a spherical drop of a liquid of refractive index $μ = \frac{\sqrt{7}}{2} .$ This ray after getting reflected from back surface comes out as shown. Given $\sin 41 ° = \sqrt{\frac{3}{7}} .$

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Total deviation suffered by ray (refraction + reflection + refraction) is minimum when $ϕ = 60 ° .$

Total internal reflection at back surface is not possible for any value of $ϕ .$

Ray suffers total internal reflection at back surface.

When angle of incidence is chosen such that total deviation is minimum, then value of total deviation is $136 ° .$

answer is B, C, D.

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Detailed Solution

$1 \sin ϕ = \frac{\sqrt{7}}{2} \sin r_{1} a n d δ = 2 (ϕ - r_{1}) + π - 2 r_{1}$

$δ$ is min. when $\frac{d δ}{d ϕ} = 0 \Rightarrow \frac{d r_{1}}{d ϕ} = \frac{1}{2}$
Since rays enters from air to glass $\Rightarrow r_{2} \leq θ_{c}$
Thus no TIR from back surface

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A ray of light enters a spherical drop of a liquid of refractive index μ=72. This ray after getting reflected from back surface comes out as shown. Given sin41°=37.

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A ray of light enters a spherical drop of a liquid of refractive index $μ = \frac{\sqrt{7}}{2} .$ This ray after getting reflected from back surface comes out as shown. Given $\sin 41 ° = \sqrt{\frac{3}{7}} .$