Question

The first minimum of the interference pattern of a monochromatic light of wavelength $\lambda $ for a single slit of width $a$, occurs at an angle of $\dfrac{\lambda }{a}$. At the same angle of $\dfrac{\lambda }{a}$ we get a maximum for two narrow slits separated by the distance $a$. Explain.

Answer 1

Hint:-Find the path difference between two secondary wavelets. When the $\theta $ is very small, the diffraction pattern becomes minimum. Then, we have to find the resultant intensity.

Complete step by step answer:
According to the question, it is given that the width of the slit is $a$.
Now, the path difference between two secondary wavelets can be given by –
$n\lambda = a\sin \theta \cdots \left( 1 \right)$
Since, the $\theta $ is very small –
$\therefore \sin \theta \approx \theta $
For the first order diffraction, $n = 1$
So, putting this value of $n$ and $\sin \theta \approx \theta $ in equation $\left( 1 \right)$, we get –
$\lambda = a\theta $
So, the above equation can be written in terms of angle as –
$\theta = \dfrac{\lambda }{a}$
We know that, $\theta $ is very small so, -
$\theta = 0$
Therefore, the diffraction pattern becomes minimum.
In the case of interference, for two interfering waves of intensity ${I_1}$ and ${I_2}$. We should have two slits separated by the distance.
Therefore, we have to calculate the resultant intensity. So, this can be done as –
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \theta } \cdots \left( 2 \right)$
We know that, $\theta = 0\left( {nearly} \right)$ which corresponds to an angle $\dfrac{\lambda }{a}$
$\therefore \cos \theta = 1$
Putting this value in equation $\left( 2 \right)$, we get –
$
  I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos 0} \\
  I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \\
 $
From the above equation, we can conclude that the resultant intensity is the sum of two intensities, there is a maxima which corresponds to the angle $\dfrac{\lambda }{a}$.
This is the reason why at the same angle $\dfrac{\lambda }{a}$ we get the maximum for two narrow slits which are separated by distance $a$.

Note: When the coherent sources of waves have the same wavelength, frequency and same phase difference. Then, the resultant intensity can be expressed as –
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \theta } $
When the waves are superimposed and interference is constructive then, resultant intensity becomes maximum and can be expressed as –
$I = \sqrt {{{\left( {{I_1} + {I_2}} \right)}^2}} $
For destructive interference, the resultant intensity can be expressed as –
$I = \sqrt {{{\left( {{I_1} - {I_2}} \right)}^2}} $