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JAMB Physics 2023 Paper

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Question : 38

Total: 65

A beam of light travelling in water is incident on a glass which is immersed in the water. The incident beam makes an angle of

40^{\circ}

with the normal. Calculate the angle of refraction in the glass.

[Refractive index of water = 1.33, Refractive index of glass = 1.5]

$29.36^{\circ}$
$25.37^{\circ}$
$37.21^{\circ}$
$34.75^{\circ}$

Solution:

Concept:

Use Snell's law to relate the angles and refractive indices as light passes from water into glass.

Explanation:

Given:

n_{\text{water}} = 1.33

,

n_{\text{glass}} = 1.5

, angle of incidence

i = 40^{\circ}

.

Snell's law is

n_1 \sin i = n_2 \sin r

.

Substitute the values:

1.33 \times \sin 40^{\circ} = 1.5 \times \sin r

.

So,

\sin r = \dfrac{1.33 \times 0.6428}{1.5}

.

This gives

\sin r = 0.5700

.

Therefore,

r = \sin^{-1}(0.5700) = 34.75^{\circ}

.

Answer:

The angle of refraction in the glass is

34.75^{\circ}

, which is option D.

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