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Let the refractive index of a denser medium with respect to a rarer medium be n₁₂ and its critical angle be θ_C. At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90⁰. Angle A is given by :

Option 1)
$\frac{1}{\cos^{-1} \left ( \sin \Theta c \right )}$

Option 2)
$\frac{1}{\tan ^{-1}\left ( \sin \Theta c \right )}$

Option 3)
$\cos ^{-1}\left ( \sin \Theta c \right )$

Option 4)
$\tan ^{-1}\left ( \sin \Theta c \right )$

Answers (1)

best_answer

As we learnt in

Relation between angle of incidence and angle of refaction -

$\mu _{1}\sin i= \mu _{2}\sin r$

- wherein

$\mu _{1}=$ refractive index of medium of incidence.

$\mu _{2}=$ refractive index of medium where rays is refracted.

$i=$ angle of incidence.

$r=$ angle of refraction.

From Snell's law

$\frac{\mu_{R}}{\mu_{D}}=\frac{sini}{sinr}$ (i)

Where $\angle i=A$ and $\angle r=(90-A)$

We know that $sin\theta_{c}=\frac{\mu_{R}}{\mu_{D}}$

$\Rightarrow\ \; sin\theta_{c}=\frac{sinA}{sin(90-A)}=\frac{sinA}{cosA}$

$sin\theta_{c}=tanA$

$\Rightarrows\ \;tanA^{-1}(sin\theta_{c})$

Correct option is 4.

Option 1)

$\frac{1}{\cos^{-1} \left ( \sin \Theta c \right )}$

This is an incorrect option.

Option 2)

$\frac{1}{\tan ^{-1}\left ( \sin \Theta c \right )}$

This is an incorrect option.

Option 3)

$\cos ^{-1}\left ( \sin \Theta c \right )$

This is an incorrect option.

Option 4)

$\tan ^{-1}\left ( \sin \Theta c \right )$

This is the correct option.

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prateek

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