Question

A mixture of light, consisting of wavelength $ 590nm $ and an unknown wavelength, illuminates Young’s double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the $ {4^{\operatorname{th} }} $ bright fringe of the unknown light. From this data, the wavelength of the unknown light is:
(A) $ 393.4nm $
(B) $ 885.0nm $
(C) $ 442.5nm $
(D) $ 776.8nm $

Answer 1

Hint
To solve this question, we have to use the formula for the position of the nth bright fringe in the case of Young’s double slit experiment. Applying the given condition for the interference of the mixture of light will give the wavelength of the unknown source.
Formula Used: The formula used in solving this question is given by
 $\Rightarrow {y_n} = \dfrac{{n\lambda D}}{d} $, where $ {y_n} $ is the distance of the $ {n^{th}} $ bright fringe from the central maximum for the Young’s double slit experiment in which the light of wavelength $ \lambda $ is used, $ d $ is the distance between the two slits, and $ D $ is the distance of the screen from the slits.

Complete step by step answer
Let the wavelength of the unknown source be $ x $.
We know that the position of the $ {n^{th}} $ bright fringe in a Young’s double slit experiment is given by
 $\Rightarrow {y_n} = \dfrac{{n\lambda D}}{d} $
Since the central maximum of both the lights coincide, so the above expression will be valid for both the lights.
The position of the third bright fringe for the known source is obtained by putting $ n = 3 $ in the above expression. Also, the wavelength of the known source is given to be $ \lambda = 590nm $. So, the position of the third bright fringe is
 $\Rightarrow {y_3} = \dfrac{{3 \times 590D}}{d} $ (1)
Now, for the unknown source we have $ \lambda = x $. So the above formula of the $ {n^{th}} $ bright fringe for the unknown source becomes
 $\Rightarrow {y_n} = \dfrac{{nxD}}{d} $
The position of the fourth bright fringe for the unknown source is obtained by putting $ n = 4 $ in the above expression. So, we have
 $\Rightarrow {y_4} = \dfrac{{4xD}}{d} $ (2)
According to the question, we have
 $\Rightarrow {y_3} = {y_4} $
From (1) and (2)
 $\Rightarrow \dfrac{{4xD}}{d} = \dfrac{{3 \times 590D}}{d} $
Cancelling $ \dfrac{D}{d} $ from both the sides, we have
 $\Rightarrow 4x = 3 \times 590 $
 $\Rightarrow x = \dfrac{{3 \times 590}}{4} $
On solving we get
 $\Rightarrow x = 442.5nm $
Thus the wavelength of the unknown source is equal to $ 442.5nm $
Hence, the correct answer is option C.

Note
The parameters of the Young’s double slit experiment in this question are not given. This is due to the fact that when the position of the third and the fourth bright fringes are equated, they get cancelled. So we could assume them suitably.