Question

(a) Why are coherent sources necessary to produce a sustained interference pattern?
(b) In Young's double-slit experiment using monochromatic light of wavelength $\lambda $, the intensity of light at a point on the screen where the path difference is $\lambda $, is K units. Find the intensity of light at a point where path difference is $\dfrac{\lambda }{3}$

Answer 1

Hint: Here the question is asked in two parts. Young’s double-slit experiment is important to study to answer both. Coherent sources have the same wavelength and ensure the place of maxima and minima. In the second part, we first write the equation when the path difference is $\lambda $, and then $\dfrac{\lambda }{3}$, then the phase difference with it, and then intensity.

Complete step by step answer:
Step 1:
Before we start let’s see the working of Young’s double-slit experiment:
For better understanding and working of the experiment, there is a diagram below.

A labeled diagram provides the distance between the slit is ‘d’ and the distance between the screens is $D$. An electron beam or beam of light or beam of photons in bombarded on the slits and according to their nature they produce interference pattern and fringes dark and bright obtain on screen
Working:
(a) Young’s double-slit experiment uses two coherent sources of light placed at a small distance d slightly greater or equal to the wavelength of light. It helps in understanding the wave theory of light. The screen is placed at distance D.
At any point on the screen at a distance ‘y’ from the center, the waves travel distance ${l_1}$ and ${l_2}$ from holes 1 and 2. The path difference of $\Delta l$ is created by the points. The point subtends at an angle $\theta $ but D is very large so it is a very small difference of angle subtended at the source.
Coherent sources have the same wavelength. Coherent sources are necessary to ensure that the position of maxima and minima do not change with time that is the reason they produce a sustainable interference pattern.

Step 2:
Now coming to the second part of the question:
For monochromatic light the intensity is defined as I=${I_1} + {I_2}$
Or $I'$ =${I_1} + {I_2}$$2\sqrt {{I_1}{I_2}\cos \phi } $ where,$\phi $ is the phase difference…….. (1)
When the intensity I=K, then the wavelength is $\lambda $ and when intensity is K’ the $\lambda $=$\dfrac{{\lambda '}}{3}$
The wavelength of monochromatic light (path difference) =$\lambda $…….. (2)
Now phase difference is given by: $\phi $=$2\pi \times \dfrac{{path{\text{ }}difference}}{\lambda }$
Putting (2) in above phase difference this gives $\phi $=$2\pi $ …… (3)
Putting (3) in (1) then intensity I’=4${I_1}$
Now I’=K then ${I_1}$ =$\dfrac{K}{4}$ (from above), and this time path difference is $\dfrac{\lambda }{3}$
Putting in the formula of phase difference we get $\phi $=$\dfrac{{2\pi }}{3}$
We are now placing the value of phase in equation (1)
Or I =${I_1} + {I_2}$$2\sqrt {{I_1}{I_2}\cos \phi } $ $ \Rightarrow $ I=${I_1} + {I_2}$$2\sqrt {{I_1}{I_2}} \cos \dfrac{{2\pi }}{3}$
This gives I’=$\dfrac{K}{4}$ which is a new intensity of light.

The intensity of light at a point where path difference is $\dfrac{\lambda }{3}$ is $\dfrac{K}{4}$.

Note:
The whole question is purely based on the experiment of a double slit. Point to remember: when intensity was I which is equal to K the path difference is $\lambda $ and when intensity is I’ or K’ the $\lambda $=$\dfrac{{\lambda '}}{3}$. This is the point where there is more of a chance of mistake. Both will have different phase differences. We need the second one and hence new intensity will be the answer.