Consider light travelling from a medium A to medium B separated by a plane interface. [2026]

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Q.
Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in medium A and B are $2.4 \times 10^{8} m/s$ and $2.7 \times 10^{8} m/s$ respectively, then the value of critical angle is: [2026]

1 $\tan^{- 1} (\frac{8}{\sqrt{17}})$

2 $\cot^{- 1} (\frac{3}{\sqrt{13}})$

3 $\sin^{- 1} (\frac{9}{8})$

4 $\cos^{- 1} (\frac{8}{9})$

Ans.
(1)

$μ_{A} \sin c = μ_{B} \sin 90$

$\Rightarrow \sin c = \frac{μ_{B}}{μ_{A}} = \frac{v_{A}}{v_{B}}$

$∴$ $\sin c = \frac{2.4 \times 10^{8}}{2.7 \times 10^{8}} = \frac{8}{9}$

$\Rightarrow \tan c = \frac{8}{\sqrt{81 - 64}} = \frac{8}{\sqrt{17}}$

$c = \tan^{- 1} (\frac{8}{\sqrt{17}})$

Solution

1 $\tan^{- 1} (\frac{8}{\sqrt{17}})$

2 $\cot^{- 1} (\frac{3}{\sqrt{13}})$

3 $\sin^{- 1} (\frac{9}{8})$

4 $\cos^{- 1} (\frac{8}{9})$

Ans.
(1)

$μ_{A} \sin c = μ_{B} \sin 90$

$\Rightarrow \sin c = \frac{μ_{B}}{μ_{A}} = \frac{v_{A}}{v_{B}}$

$∴$ $\sin c = \frac{2.4 \times 10^{8}}{2.7 \times 10^{8}} = \frac{8}{9}$

$\Rightarrow \tan c = \frac{8}{\sqrt{81 - 64}} = \frac{8}{\sqrt{17}}$

$c = \tan^{- 1} (\frac{8}{\sqrt{17}})$

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Solution

1 tan-1(817)

2 cot-1(313)

3 sin-1(98)

4 cos-1(89)

Ans. (1) μAsinc=μBsin90 ⇒sinc=μBμA=vAvB ∴ sinc=2.4×1082.7×108=89 ⇒tanc=881-64=817 c=tan-1(817)

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1 $\tan^{- 1} (\frac{8}{\sqrt{17}})$

2 $\cot^{- 1} (\frac{3}{\sqrt{13}})$

3 $\sin^{- 1} (\frac{9}{8})$

4 $\cos^{- 1} (\frac{8}{9})$

Ans.
(1)

$μ_{A} \sin c = μ_{B} \sin 90$

$\Rightarrow \sin c = \frac{μ_{B}}{μ_{A}} = \frac{v_{A}}{v_{B}}$

$∴$ $\sin c = \frac{2.4 \times 10^{8}}{2.7 \times 10^{8}} = \frac{8}{9}$

$\Rightarrow \tan c = \frac{8}{\sqrt{81 - 64}} = \frac{8}{\sqrt{17}}$

$c = \tan^{- 1} (\frac{8}{\sqrt{17}})$