SATHEE: Optics 4 Question 22

Optics 4 Question 22

23. A right angled prism $(45^{\circ} - 90^{\circ} - 45^{\circ})$ of refractive index $n$ has a plane of refractive index $n_{1} (n_{1} < n)$ cemented to its diagonal face. The assembly is in air. The ray is incident on $A B$ .

(1996, 3M)

(a) Calculate the angle of incidence at $A B$ for which the ray strikes the diagonal face at the critical angle.

(b) Assuming $n = 1.352$ , calculate the angle of incidence at $A B$ for which the refracted ray passes through the diagonal face undeviated.

Show Answer

Answer:

Correct Answer: 23. (a) $i_{1} = \sin^{- 1} \frac{1}{\sqrt{2}} (\sqrt{n^{2} - n_{1}^{2}} - n_{1})$

$\begin{array}{ll} (b) 73^{\circ} & 24 \cdot μ = \sqrt{3} \end{array}$

Solution:

(a) Critical angle $θ_{c}$ at face $A C$ will be given by

$\begin{aligned} θ_{c} & = \sin^{- 1} \frac{n_{1}}{n} \\ or \sin θ_{c} & = \frac{n_{1}}{n} \end{aligned}$

Now, it is given that $r_{2} = θ_{c}$

$∴ r_{1} = A - r_{2} = (45^{\circ} - θ_{c})$

Applying Snell’s law at face $A B$ , we have

$\begin{aligned} n = \frac{\sin i_{1}}{\sin r_{1}} or \sin i_{1} = n \sin r_{1} \\ ∴ i_{1} & = \sin^{- 1} (n \sin r_{1}) \end{aligned}$

Substituting value of $r_{1}$ , we get

$\begin{aligned} i_{1} & = \sin^{- 1} n \sin (45^{\circ} - θ_{c}) \\ = \sin^{- 1} n (\sin 45^{\circ} \cos θ_{c} - \cos 45^{\circ} \sin θ_{c}) \\ = \sin^{- 1} \frac{n}{\sqrt{2}} (\sqrt{1 - \sin^{2} θ_{c}} - \sin θ_{c}) \\ = \sin^{- 1} \frac{n}{\sqrt{2}} \sqrt{1 - \frac{n_{1}^{2}}{n^{2}}} - \frac{n_{1}}{n} \\ i_{1} & = \sin^{- 1} \frac{1}{\sqrt{2}} (\sqrt{n^{2} - n_{1}^{2}} - n_{1}) \end{aligned}$

Therefore, required angle of incidence ( $i_{1})$ at face $A B$ for which the ray strikes at $A C$ at critical angle is

$i_{1} = \sin^{- 1} \frac{1}{\sqrt{2}} (\sqrt{n^{2} - n_{1}^{2}} - n_{1})$

(b) The ray will pass undeviated through face $A C$ when either $n_{1} = n$ or $r_{2} = 0^{\circ}$ i.e. ray falls normally on face $A C$ .

Here $n_{1} \neq n$ (because $n_{1} < n$ is given )

$∴ r_{2} = 0^{\circ}$

or $r_{1} = A - r_{2} = 45^{\circ} - 0^{\circ} = 45^{\circ}$

Now applying Snell’s law at face $A B$ , we have $n = \frac{\sin i_{1}}{\sin r_{1}}$

$\begin{aligned} or & 1.352 & = \frac{\sin i_{1}}{\sin 45^{\circ}} \\ ∴ \sin i_{1} & = (1.352) \frac{1}{\sqrt{2}}, \\ \sin i_{1} & = 0.956 \\ ∴ & i_{1} & = \sin^{- 1} (0.956) \approx 73^{\circ} \end{aligned}$