Question

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

Answer 1

Width of the slit is $a$

The path difference between two secondary wavelets is given by,

$N\lambda =asin\theta$

Since, $\theta$ is very small, $sin\theta =\theta$

So, for the first order diffraction $n=1$, the angle is $\frac{\lambda }{a}$

Now, we know that $\theta$ must be very small $\theta =0$ (nearly) because of which the diffraction pattern is minimum.

Now for interference case, for two interfering waves of intensity $l_1$ and $l_2$ we must have two slits separated by a distance.

We have the resultant intensity, $l=l_1+l-2+2\sqrt{l_1{l}_{2}cos\theta }$

Since, $\theta =0$ (nearly) corresponding to angle $\frac{\lambda }{a}$, so $cos\theta =1$ (nearly)

So,

$l=l_1+l_2+2\sqrt{l_1{l}_{2}cos\theta }$

$l=l_1+l_2+2\sqrt{l_1{l}_{2}cos0}$

$l=l_1+l_2+2\sqrt{l_1{l}_2}$

We see the resultant intensity is sum of the two intensities, so there is a maxima corresponding to the angle $\frac{\lambda }{a}$.

This is why at the same angle $\frac{\lambda }{a}$ we get a maximum for two narrow slits separated by a distance $a$.