SATHEE: Optics 2 Question 12

Optics 2 Question 12

12. A rectangular glass slab $A B C D$ of refractive index $n_{1}$ is immersed in water of refractive index $n_{2} (n_{1} > n_{2})$ . A ray of light is incident at the surface $A B$ of the slab as shown. The maximum value of the angle of incidence $α_{max}$ , such that the ray comes out only from the other surface $C D$ , is given by

$(2000, 2 M)$

(a) $\sin^{- 1} [\frac{n_{1}}{n_{2}} \cos (\sin^{- 1} \frac{n_{2}}{n_{1}})]$

(b) $\sin^{- 1} [n_{1} \cos (\sin^{- 1} \frac{1}{n_{2}})]$

(d) $\sin^{- 1} [\frac{n_{2}}{n_{1}}]$

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Answer:

Correct Answer: 12. (a)

Solution:

Rays come out only from $C D$ , means rays after refraction from $A B$ get total internally reflected at $A D$ . From the figure

$\begin{matrix} r_{1} + r_{2} = 90^{\circ} \\ r_{1} = 90^{\circ} - r_{2} \\ {(r_{1})}_{max} = 90^{\circ} - {(r_{2})}_{min} and {(r_{2})}_{min} = θ_{C} \end{matrix}$

(for total internal reflection at $A D$ )

where,

$\sin θ_{C} = \frac{n_{2}}{n_{1}}$

$\begin{aligned} or & θ_{C} & = \sin^{- 1} \frac{n_{2}}{n_{1}} \\ ∴ & {(r_{1})}_{max} & = 90^{\circ} - θ_{C} \end{aligned}$

Now, applying Snell’s law at face $A B$

$\begin{aligned} \frac{n_{1}}{n_{2}} & = \frac{\sin α_{max}}{\sin {(r_{1})}_{max}} \\ = \frac{\sin α_{max}}{\sin (90^{\circ} - θ_{C})} \end{aligned}$

$\begin{aligned} = \frac{\sin α_{max}}{\cos θ_{C}} \\ or \sin α_{max} & = \frac{n_{1}}{n_{2}} \cos θ_{C} \\ ∴ α_{max} & = \sin^{- 1} [\frac{n_{1}}{n_{2}} \cos θ_{C}] \\ = \sin^{- 1} [\frac{n_{1}}{n_{2}} \cos (\sin^{- 1} \frac{n_{2}}{n_{1}})] \end{aligned}$

Optics 2 Question 12

Answer:

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Optics 2 Question 12

12. A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive index n2(n1>n2). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence αmax, such that the ray comes out only from the other surface CD, is given by

Answer:

Engineering Preparation

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Syllabus

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Sample Mock Test

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