A beam of light is incident on a liquid of refractive index 1.40. The reflected rays are completely polarized. The angle of refraction of the beam is ( $tan {54.5}^{\circ} = 1.4$ ):

30^{\circ}

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{35.5}^{\circ}

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45^{\circ}

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60^{\circ}

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Solution

The correct option is A ${35.5}^{\circ}$
Refractive index is $μ = 1.4$
We have $tan i = μ$ for complete polarisation.
From the given data in question, $i = {54.5}^{o}$
From snell's law,
$\frac{sin i}{sin r} = μ$
$\frac{sin i}{tan i} = s i n r$
$sin r = cos i$
$sin r = sin (90 - i)$
$r = 90 - i = {35.5}^{o}$

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A beam of light is incident on a liquid of refractive index 1.40. The reflected rays are completely polarized. The angle of refraction of the beam is (tan54.5∘=1.4):

The correct option is A 35.5∘Refractive index is μ=1.4We have tani=μ for complete polarisation.From the given data in question, i=54.5oFrom snell's law,sinisinr=μsinitani=sinrsinr=cosisinr=sin(90−i)r=90−i=35.5o

A beam of light is incident on a liquid of refractive index 1.40. The reflected rays are completely polarized. The angle of refraction of the beam is ( $tan {54.5}^{\circ} = 1.4$ ):