Question

In Young's double-slit experiment, the separation between the two slits is d and the wavelength of the light is $\lambda$. The intensity of light falling on slit 1 is four times the intensity of light falling on the slit 2. Choose the correct choice(s)

• A. If $d=\lambda$, the screen will contain only one maximum
• B. If $\lambda <d<2\lambda$, at least one more maximum (besides the central maximum) will be observed on the screen
• C. If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase
• D. If the intensity of light falling on slit 2, is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase

Answer 1

Answer: (A), (B)

The correct options are
A If $d=\lambda$, the screen will contain only one maximum
B If $\lambda <d<2\lambda$, at least one more maximum (besides the central maximum) will be observed on the screen

Path difference between the lights from $S_1$ and $S_2$ reaching the point P, $\Delta x=dsin\theta$
For maxima, $\Delta x=n\lambda$
So, $dsin\theta =n\lambda$
For $d=\lambda ⟹sin\theta =n(sin\theta \ne 1,-1)$
Thus $n=0$ gives $\theta =0$ means there will be only one bright fringe that is central maxima.
Similarly for $\lambda <d<2\lambda ⟹1<\frac{n}{sin\theta }<2$
Thus $sin\theta <nandsin\theta >\frac{n}{2}$
$⟹n=0,1$
Hence there will be one central maxima and first order maxima.
Resultant intensity for dark fringes,$I=(\sqrt{I_1}-\sqrt{I_2}{)}^2$
Now initially $I_2=I_o{I}_1=4I_o⟹I=I_o$
When $I_1=I_2$ gives $I=0$
Thus by making intensities of both the slits equal,then $I$ for dark fringes is always zero i-e, less than the $I$ when slits have unequal intensity.