Question

The two slits are acting as coherent sources of equal amplitude $n$ and of wavelength $\lambda$, in Young's double-slit experiment. In another experiment with the identical setup, the two slits are a source of equal amplitude $n$ and wavelength $\lambda$, but they are incoherent. Find out the ratio of intensities of light at the midpoint of the screen in the first case to that in the second case?
$\begin{aligned} & A.2:1 \\\ & B.1:2 \\\ & C.3:4 \\\ & D.4:3 \\\ \end{aligned}$

Answer 1

For the coherent sources, the parameters like frequency and amplitude will be the same. But in the case of incoherent sources, there will be a phase shift of $90{}^\circ$happening. Find out the resultant amplitudes in both the cases and find the intensities. Take their ratio which will be the answer to the question.

Complete step-by-step solution
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First of all let us consider the first case. In this case the slits are acting as coherent sources. So the waves from this source can be written as,
The wavelength of the waves are mentioned as $\lambda$ and the amplitude of the waves are given as $n$. Therefore we can write that,
$\begin{aligned} & {{x}_{1}}=n\sin \left( \omega t \right) \\\ & {{x}_{2}}=n\sin \left( \omega t \right) \\\ \end{aligned}$
$\omega$ be the angular frequency of the wave.
Taking the resultant of the both of these waves will be written as,
${{x}_{r1}}={{x}_{1}}+{{x}_{2}}=2n\sin \left( \omega t \right)$
The intensity is found to be directly proportional to the square of the amplitude of the wave. Therefore we can write that,
${{I}_{r1}}={{\left( 2n \right)}^{2}}=4{{n}^{2}}$
Now let us discuss the second case. In this case, the waves are having equal parameters but the sources are incoherent. Therefore, there will be a phase shift of $90{}^\circ$occurring. This can be written as,
$\begin{aligned} & {{x}_{1}}=n\sin \left( \omega t \right) \\\ & {{x}_{2}}=n\cos \left( \omega t \right) \\\ \end{aligned}$
It has been mentioned in the question that the wavelength of the wave will be $\lambda$ and the amplitude be $n$.
Therefore the resultant of the wave will be,
${{x}_{r2}}={{x}_{1}}+{{x}_{2}}=\sqrt{2}n\sin \left( \omega t+\dfrac{\pi }{4} \right)$
The intensity of the wave will be the square of the amplitude of the resulting wave. That is,
${{I}_{r2}}={{\left( \sqrt{2}n \right)}^{2}}=2{{n}^{2}}$
Taking the ratio of the intensities of both the waves can be written as,
$\dfrac{{{I}_{r1}}}{{{I}_{r2}}}=\dfrac{2}{1}$
Therefore the correct answer is calculated as option A.

Note: Young's double-slit experiment is performed by passing monochromatic light through two narrow slits. This will create illumination on a distant screen, a characteristic pattern of bright and dark fringes is visible. This interference pattern is because of the superposition of the waves coming from the two slits.