Target Exam

CUET

Subject

Physics

Chapter

Ray Optics

Question:

A rectangular glass slab ABCD, of refractive index $n_1$, is immersed in water of refractive index $n_2 (n_1 > n_2)$. A ray of light is incident on the surface AB of the slab as shown. The maximum value of the angle of incidence $α_{max}$, such that the ray emerges only from the surface CD is given by

Options:

$\sin^{-1}[\frac{n_1}{n_2}\cos(\sin^{-1}\frac{n_2}{n_1})]$

$\sin^{-1}[n_1\cos(\sin^{-1}\frac{1}{n_2})]$

$\sin^{-1}(\frac{n_1}{n_2})$

$\sin^{-1}(\frac{n_2}{n_1})$

Correct Answer:

$\sin^{-1}[\frac{n_1}{n_2}\cos(\sin^{-1}\frac{n_2}{n_1})]$

Explanation:

Here, total internal reflection must take place at AD Applying Snell’s law at P we obtain,

$\frac{\sin α_m}{\sin r}=\frac{n_1}{n_2}$

Geometrically $r + θ_c = π/2$

where critical angle $θ_c =\sin^{−1} (n_2/n_1)$

$⇒ α_m = \sin^{−1} [ (\sin r) (n_1/n_2)]$

Putting $\sin r = \sin (π/2 - θ_c) = \cos θ_c = \cos [\sin^{−1}(n_2/n_1) ]$ in the above equation we obtain

$α_m = \sin^{−1}[\frac{n_1}{n_2}\cos(\sin^{-1}\frac{n_2}{n_1})]$