Question

A ray of light from a denser medium strikes a rarer medium at an angle of incidence i. The reflected and refracted rays make an angle of 90o with each other. The angles of reflection and refraction are r and r', respectively. The critical angle is

Answer 1

μ = sin i/ sin R

∠r + ∠R + 90° = 180° ( angle on a straight line)

so, ∠r + ∠R = 90°

∴ ∠R = 90° - ∠r           - (1)

Also,   ∠i =  ∠r ( by laws of reflection)

∴ μ = sin r/ sin R

putting (1) in the above equation we get,

μ = sin r/ sin (90-r)

⇒ μ = sin r/ cos r = tan r

now, sin ( critical angle) = μ

and μ = tan r

combining these two,

tan r = sin (critical angle)

∴ critical angle = sin ∧ -1 (tan r)

Answer 2

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$\large\bf\underline{Given:-}$

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• A ray of light from a denser medium strikes a rarer medium at an angle of incidence i

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• The reflected and refracted rays make an angle of 90o with each other

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• The angles of reflection and refraction are r and r', respectively

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$\large\bf\underline {To \: find:-}$

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• The critical angle

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$\large\bf\underline{Formula \: Used:-}$

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• $\huge \underline{\boxed{ \blue{ \bf \: Snell \: ' s \: law \: }}}$

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$\huge\bf\underline{Solution:-}$

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$\sf \longrightarrow u = \frac{ \sin \: r }{ \sin \: i}$

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$\sf \longrightarrow \frac{ \sin {(90}^{o} - r) }{ \sin \: r } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (i = r)$

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$\sf \longrightarrow \frac{ \cos \: r}{ \sin \: r}$

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$\sf \longrightarrow \frac{1}{ \tan \: r }$

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$\sf \longrightarrow u = \frac{1}{\sin \: i_c }$

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$\sf \longrightarrow \sin \: i_c = \tan \: r$

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$\sf \longrightarrow i_c = { \sin}^{ - 1} ( \tan \: r)$

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