A light ray is incident on a glass slab of thickness cm and refractive index ?. The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is ____ cm.

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Solution

Q.
A light ray is incident on a glass slab of thickness $4 \sqrt{3}$ cm and refractive index $\sqrt{2}$ . The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is ____ cm.

(Given sin $15^{o}$ = 0.25) [2024]

Ans.
(2)

$i = θ_{c} \Rightarrow i = \sin^{- 1} (\frac{1}{μ})$

$\Rightarrow i = 45 °$

According to Snell’s law,

$1 \sin 45 ° = \sqrt{2} \sin r \Rightarrow \frac{1}{\sqrt{2}} = \sqrt{2} \sin r \Rightarrow \frac{1}{\sqrt{2}} = \sqrt{2} \sin r$

$\Rightarrow r = 30 °$

Now, $Δ x = t \sec 30 ° \sin 15 ° = 4 \sqrt{3} \times \frac{2}{\sqrt{3}} \times \frac{1}{4} = 2 cm$

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