A ray of light strikes a glass slab 5 cm thick, making an angle of incidence equal to $30^{\circ}$ . Construct the ray diagram showing showing the emergent ray and the refracted ray through the glass block. The refractive index of glass is 1.5. Also, measure the lateral displacement of the ray. Take $sin {19.5}^{\circ} = 1 / 3$ .

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Solution

Refractive index (n) = 1.5,
angle of incidence (i) - $30^{\circ}$
By Snell's law, $\frac{s i n i}{s i n r} = n$
$∴ s i n r = \frac{s i n i}{n} = \frac{s i n 30^{\circ}}{1.5} = \frac{1 / 2}{1.5} = \frac{1}{3}$
$\Rightarrow ∠ r = {19.5}^{\circ}$ (since sin ${19.5}^{\circ} = \frac{1}{3}$ )
By measurement, on proper scale lateral displacement = 1 cm nearly.

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Refractive index (n) = 1.5,angle of incidence (i) - 30∘By Snell's law, sinisinr=n∴sinr=sinin=sin30∘1.5=1/21.5=13⇒∠r=19.5∘ (since sin 19.5∘=13)By measurement, on proper scale lateral displacement = 1 cm nearly.