A ray of light travels from ethanol into air. If the angle of incidence of the ray at the boundary is $30^{0}$ and the refractive index of ethanol is 1.36, what is the angle of refraction of the ray as it emerges out of ethanol?

{42.84}^{0}

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{21.5}^{0}

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{46.8}^{0}

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None of the above

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Solution

The correct option is A ${42.84}^{0}$
$∠ i = 30^{0},^{a i r} μ_{e t h} = 1.36$

$^{e t h} μ_{a i r} = \frac{s i n i}{s i n r}$

$⟹ s i n r = s i n 30^{0} \times 1.36 = 0.68$

$∴ r = s i n^{- 1} (0.68) = {42.84}^{0}$

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A ray of light travels from ethanol into air. If the angle of incidence of the ray at the boundary is 300 and the refractive index of ethanol is 1.36, what is the angle of refraction of the ray as it emerges out of ethanol?

The correct option is A 42.840∠i=300,airμeth=1.36ethμair=sinisinr⟹sinr=sin300×1.36=0.68∴r=sin−1(0.68)=42.840

A ray of light travels from ethanol into air. If the angle of incidence of the ray at the boundary is $30^{0}$ and the refractive index of ethanol is 1.36, what is the angle of refraction of the ray as it emerges out of ethanol?

The correct option is A ${42.84}^{0}$
$∠ i = 30^{0},^{a i r} μ_{e t h} = 1.36$

$^{e t h} μ_{a i r} = \frac{s i n i}{s i n r}$

$⟹ s i n r = s i n 30^{0} \times 1.36 = 0.68$

$∴ r = s i n^{- 1} (0.68) = {42.84}^{0}$