Animated Solution for Physics - Optics: In a 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).

This problem is a beautiful exploration of both the spatial distribution and the intensity profile of interference fringes in Young's Double Slit Experiment (YDSE). Let's break it down step by step to understand exactly what happens on the screen.

To determine how many maxima are formed on the screen, we must look at the condition for constructive interference. The path difference $\Delta x$ between the waves from the two slits must be an integer multiple of the wavelength $\lambda$:

From this, we can express the sine of the angle $\theta$ as:

For a fringe to be physically observable on a flat screen placed at a finite distance, the angle $\theta$ must be strictly less than $90^{\circ }$. This means that the magnitude of $\sin \theta$ must be strictly less than $1$ (i.e., $-1<\sin \theta <1$). If $\sin \theta =\pm 1$, the diffracted rays travel parallel to the screen and never intersect it.

Let's evaluate Option (a). If the slit separation $d$ is exactly equal to the wavelength $\lambda$, our equation simplifies to:

The only integer values $n$ can take are $-1,0,$ and $1$. However, for $n=\pm 1$, we get $\sin \theta =\pm 1$, which corresponds to $\theta =\pm 90^{\circ }$. As discussed, these rays never reach the screen. Therefore, only the central maximum ($n=0$) is formed on the screen. Option (a) is absolutely correct.

Now, let's evaluate Option (b). If the slit separation $d$ is slightly larger, specifically $\lambda <d<2\lambda$, the ratio $\frac{\lambda }{d}$ falls between $0.5$ and $1$.

For $n=\pm 1$, the value of $\sin \theta$ will be strictly between $0.5$ and $1$, meaning $\theta$ is a valid angle less than $90^{\circ }$. For $n=\pm 2$, $\sin \theta$ would exceed $1$, which is impossible. Thus, the valid orders are $n=-1,0,$ and $1$. This gives us a total of three maxima on the screen. Since there are two additional maxima besides the central one, the statement "at least one more maximum will be observed" is true. Option (b) is correct.

Now, let's shift our focus to the brightness of these fringes. The general formula for the resultant intensity $I$ of two interfering waves is:

The maximum and minimum intensities occur when $\cos \varphi =1$ and $\cos \varphi =-1$, respectively:

Initially, we are given that $I_1=4I_0$ and $I_2=I_0$. Let's calculate the initial extreme intensities:

Notice that the minimum intensity is not zero! The dark fringes are not completely dark because the interfering waves have unequal amplitudes.

Let's test Option (c). If we reduce the intensity of slit 1 so that $I_1=I_0$ and $I_2=I_0$, the new intensities become:

The maximum intensity decreased from $9I_0$ to $4I_0$, and the minimum intensity decreased from $I_0$ to $0$. Since both decreased, Option (c) is incorrect.

Finally, let's test Option (d). If we increase the intensity of slit 2 so that $I_1=4I_0$ and $I_2=4I_0$, the new intensities become:

Here, the maximum intensity increased (from $9I_0$ to $16I_0$), but the minimum intensity decreased (from $I_0$ to $0$). Since they did not both increase, Option (d) is also incorrect.

Final Conclusion: The correct choices are (a) and (b).