A ray of light from a denser medium strikes a rarer medium at an angle of incidence i. If the angle of reflection is r and the angle of refraction is r1 and the reflected and refracted rays make an angle of 900 with each other, then the critical angle will be :

1μ=sinisinr′But r′=90∘−r and i=r1μ=sin⁡isin⁡90∘−r=sin⁡icos⁡r=sin⁡rcos⁡r=tanrAgain if C= Critical angle, then C=sin−1⁡1μ=sin−1tanr

A ray of light from a denser medium strikes a rarer medium at an angle of incidence i. If the angle of reflection is r and the angle of refraction is r1 and the reflected and refracted rays make an angle of 900 with each other, then the critical angle will be :

1μ=sinisinr′But r′=90∘−r and i=r1μ=sin⁡isin⁡90∘−r=sin⁡icos⁡r=sin⁡rcos⁡r=tanrAgain if C= Critical angle, then C=sin−1⁡1μ=sin−1tanr

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A ray of light from a denser medium strikes a rarer medium at an angle of incidence i. If the angle of reflection is r and the angle of refraction is r¹ and the reflected and refracted rays make an angle of 90⁰ with each other, then the critical angle will be :

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$\sin^{- 1} (\tan r)$

$\sin^{- 1} (\tan r^{'})$

$\tan^{- 1} (\sin r)$

$\tan^{- 1} (\sin r^{1})$

answer is A.

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Detailed Solution

$\frac{1}{μ} = \frac{sini}{\sin r^{'}}$

But r′=90^∘−r and i=r

$\frac{1}{μ} = \frac{\sin i}{\sin (90^{\circ} - r)} = \frac{\sin i}{\cos r} = \frac{\sin r}{\cos r} = \tan r$

Again if C= Critical angle, then $C = \sin^{- 1} (\frac{1}{μ}) = \sin^{- 1} (\tan r)$

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A ray of light from a denser medium strikes a rarer medium at an angle of incidence i. If the angle of reflection is r and the angle of refraction is r1 and the reflected and refracted rays make an angle of 900 with each other, then the critical angle will be :

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A ray of light from a denser medium strikes a rarer medium at an angle of incidence i. If the angle of reflection is r and the angle of refraction is r¹ and the reflected and refracted rays make an angle of 90⁰ with each other, then the critical angle will be :