SATHEE: Optics 4 Question 21

Optics 4 Question 21

22. A prism of refractive index $n_{1}$ and another prism of refractive index $n_{2}$ are stuck together with a gap as shown in the figure. The angles of the prism are as shown. $n_{1}$ and $n_{2}$ depend on $λ$ , the wavelength of

light according to :

$\begin{aligned} n_{1} = 1.20 + \frac{10.8 \times 10^{4}}{λ^{2}} and \\ n_{2} = 1.45 + \frac{1.80 \times 10^{4}}{λ^{2}} \end{aligned}$

where $λ$ is in $n m$ .

(1998, 8M)

(a) Calculate the wavelength $λ_{0}$ for which rays incident at any angle on the interface $B C$ pass through without bending at that interface.

(b) For light of wavelength $λ_{0}$ , find the angle of incidence $i$ on the face $A C$ such that the deviation produced by the combination of prisms is minimum.

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Answer:

Correct Answer: 22. (a) $600 n m$

(b) $\sin^{- 1} \frac{3}{4}$

Solution:

$n_{1} = 1.20 + \frac{10.8 \times 10^{4}}{λ^{2}}$ and $n_{2} = 1.45 + \frac{1.80 \times 10^{4}}{λ^{2}}$

Here, $λ$ is in $n m$ .

(a) The incident ray will not deviate at $B C$ if $n_{1} = n_{2}$

$\begin{aligned} \Rightarrow 1.20 + \frac{10.8 \times 10^{4}}{λ_{0}^{2}} = 1.45 + \frac{1.80 \times 10^{4}}{λ_{0}^{2}} \\ \Rightarrow \frac{9 \times 10^{4}}{λ_{0}^{2}} = 0.25 or λ_{0} = \frac{3 \times 10^{2}}{0.5} \end{aligned}$

$λ_{0} = 600 n m$

(b) The given system is a part of an equilateral prism of prism angle $60^{\circ}$ as shown in figure.

At minimum deviation,

$\begin{array}{ll} r_{1} = r_{2} = \frac{60^{\circ}}{2} = 30^{\circ} = r (say) \\ ∴ & n_{1} = \frac{\sin i}{\sin r} \\ ∴ \sin i = n_{1} \cdot \sin 30^{\circ} \\ \sin i = 1.20 + \frac{10.8 \times 10^{4}}{(600)^{2}} \frac{1}{2} = \frac{1.5}{2} = \frac{3}{4} \\ or i = \sin^{- 1} (3 / 4) \end{array}$