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The Brewster angle for the glass-air interface is $54 . 74^{o} .$ If a ray of light going from air to glass strikes at an angle of incidence $45^{o}$ ,then the angle of refraction is $(\tan 54 . 74^{o} = \sqrt{2})$

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$60^{o}$

$30^{o}$

$25^{o}$

$54 . 74^{0}$

answer is B.

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Detailed Solution

$μ = \tan 54 . 74^{o} = \sqrt{2} ∴ \sin r = \frac{\sin i}{μ} = \frac{\sin 45^{o}}{\sqrt{2}} = \frac{1}{2} \Rightarrow r = 30^{o}$

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