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WAEC Physics 2020 Paper

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Question : 16

Total: 51

A ray of light traveling from glass into ethyl alcohol is incident at the boundary at an angle of incidence

30^{\circ}

. Calculate the angle of refraction.

[Refractive index of glass

=1.5

; refractive index of ethyl alcohol

=1.36

$27^{\circ}$
$33.5^{\circ}$
$51.7^{\circ}$
$72.8^{\circ}$

Solution:

The angle of incidence

I=30^{\circ}

The angle of refraction,

r=

?

The refractive index of glass

n_g=1.5

The refractive index of ethyl alcohol,

n_e=1.36

from snell's law :

\;\; \frac{n_e}{n_g} = \frac{\sin i}{\sin r}

\;\; \frac{1.36}{1.5} = \frac{\sin 30^{\circ}}{\sin r}

\;1.36 \sin r = 1.5 \times \sin 30^{\circ}

\sin r = \frac{1.5 \times \sin 30^{\circ}}{1.36}

\sin r = 0.5515

r = \sin^{-1}(0.5515)

r = 33.5^{\circ}

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