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प्रश्न
Trace the path of a ray of light incident at an angle of 45° on a rectangular glass slab. Write the measure of the angle of refraction, the angle of emergence and the lateral displacement suffered by the ray as it passes through the slab.
उत्तर
In the figure, angle i is he angle of incidence, angle r is the angle of refraction and angle e is the angle of emergence.
Using snell's law, angle of refraction can be found out as follows:
\[\frac{\sin i}{\sin r} = \frac{\mu_2}{\mu_1} = \frac{1 . 5}{1}\]
\[\sin r = \sin i \times \frac{10}{15}\]
\[since, \angle i = {45}^o (\text{given})\]
\[\sin r = \sin {45}^o \times \frac{10}{15}\]
\[\sin r = \frac{1}{\sqrt{2}} \times \frac{2}{3}\]
\[r = 28 . {13}^o \]
and we know
\[\angle e = \angle i\]
\[ \Rightarrow \angle e = {45}^o\]
Note: We cannot measure lateral displacement suffered by the ray because to calculate it we should know the thickness of the glass slab as formula to calculate lateral displacement is
\[\text{Lateral displacement }= \frac{t \sin(i - r)}{\cos r}\]
where, t = thickness of the glass slab.
But thickness of the slab is not mentioned in the question.
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