A ray of light enters a spherical drop of water of refractive index - as shown in the figure- The angle - for which minimum deviation is produced will be given by -

The correct option is B cos^2ϕ = μ^2 13 d δd ϕ=4dαdϕ+2 d δd ϕ=0 , ⇒ d αd ϕ=12 α=sin ^ 1(1μsinϕ) cos^2ϕ=μ^2 13 #160; ...

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Standard XII

Physics

TIR in Prism

A ray of ligh...

Question

A ray of light enters a spherical drop of water of refractive index $μ$ as shown in the figure. The angle $ϕ$ for which minimum deviation is produced will be given by :

{cos}^{2} ϕ = \frac{μ^{2} + 1}{3}

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{cos}^{2} ϕ = \frac{μ^{2} - 1}{3}

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{sin}^{2} ϕ = \frac{μ^{2} + 1}{3}

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{sin}^{2} ϕ = \frac{μ^{2} - 1}{3}

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Solution

The correct option is B ${cos}^{2} ϕ = \frac{μ^{2} - 1}{3}$
$\frac{d δ}{d ϕ} = \frac{4 d α}{d ϕ} + 2$
$\frac{d δ}{d ϕ} = 0$ ,
$\Rightarrow \frac{d α}{d ϕ} = \frac{1}{2}$
$α = {sin}^{- 1} (\frac{1}{μ} sin ϕ)$
${cos}^{2} ϕ = \frac{μ^{2} - 1}{3}$

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A ray of light enters a spherical drop of water of refractive index μ as shown in the figure. The angle ϕ for which minimum deviation is produced will be given by :

The correct option is B cos2ϕ=μ2−13dδdϕ=4dαdϕ+2dδdϕ=0,⇒dαdϕ=12α=sin−1(1μsinϕ)cos2ϕ=μ2−13

A ray of light enters a spherical drop of water of refractive index $μ$ as shown in the figure. The angle $ϕ$ for which minimum deviation is produced will be given by :

The correct option is B ${cos}^{2} ϕ = \frac{μ^{2} - 1}{3}$
$\frac{d δ}{d ϕ} = \frac{4 d α}{d ϕ} + 2$
$\frac{d δ}{d ϕ} = 0$ ,
$\Rightarrow \frac{d α}{d ϕ} = \frac{1}{2}$
$α = {sin}^{- 1} (\frac{1}{μ} sin ϕ)$
${cos}^{2} ϕ = \frac{μ^{2} - 1}{3}$