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(B) For the light beam to pass through the material without undergoing total internal reflection at the interface,the condition for the critical angle

Vedclass · Accepted Answer

$53.1$

Vedclass · Answer

(B) For the light beam to pass through the material without undergoing total internal reflection at the interface,the condition for the critical angle must be satisfied at the boundary between the material and the glass.From Snell's Law at the interface,$n_1 \sin i = n_2 \sin r$. For the limiting case of total internal reflection,the angle of refraction $r = 90^{\circ}$.Here,the refractive index of glass is $n_1 = 1.5$ and the minimum refractive index of the material is $n_2 = 1.2$.Thus,$\sin i_{\max} = \frac{n_2}{n_1} = \frac{1.2}{1.5} = 0.8$.Therefore,$i_{\max} = \sin^{-1}(0.8) = 53.1^{\circ}$.

$A$ material is embedded between two glass plates. The refractive index $n$ of the material varies with thickness as shown in the figure. The maximum incident angle (in degrees) on the material for which the beam will pass through the material is

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