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 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 lights 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)

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 $P$,

$\Delta x=d\sin \theta$

For maxima, $\Delta x=n\lambda$

So, $d\sin \theta =n\lambda$

For $d=\lambda \Rightarrow \sin \theta =n(\sin \theta \ne 1,-1)$

Thus, $n=0$ gives $\theta =0$ means there will be only one bright fringe that is central maxima.

Similarly, for $\lambda <d<2\lambda \Rightarrow 1<\frac{n}{\sin \theta }<2$

Thus, $\sin \theta <n$ and $\sin \theta >\frac{n}{2}$

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 $I_2=I_0\text{and}{I}_1=4I_0$

$\Rightarrow I=I_0$

When $I_1=I_2$ then $I=0$.

Thus, by making intensities of both the slits equal, intensity of dark fringes is always less than the intensity when slits have unequal intensity.

Hence, options $\left(\text{A}\right)$ and $\left(\text{B}\right)$ are correct.