1. Relate slit intensities to amplitudes

If the intensities at the two slits are in the ratio $I_1: I_2 = 4:1,$ then the amplitudes are in the ratio $a_1:a_2 = \sqrt{I_1}:\sqrt{I_2} = 2:1.$

Let $I_2 = I, \qquad I_1 = 4I.$

2. Resultant intensity in YDSE

For phase difference $\phi$, the intensity on the screen is $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi.$

So, $I = 4I + I + 2\sqrt{4I\cdot I}\cos\phi = 5I + 4I\cos\phi.$

Hence:

• Maximum intensity when $\cos\phi = 1$: $I_{\max} = 5I + 4I = 9I.$
• Minimum intensity when $\cos\phi = -1$: $I_{\min} = 5I - 4I = I.$

Thus the dark fringes are not completely dark.

3. Angular positions of maxima

In YDSE, path difference is $\Delta = d\sin\theta.$

Maxima occur when $d\sin\theta = n\lambda, \qquad n=0,\pm 1,\pm 2,\dots$

Since $|\sin\theta|\le 1$, allowed maxima satisfy $\left|\frac{n\lambda}{d}\right| \le 1.$

4. Check Option A: If $d=\lambda$, the screen will contain only one maximum

If $d=\lambda$, then maxima condition becomes $\sin\theta = n.$

Possible integer values of $n$ are:

• $n=0 \Rightarrow \sin\theta=0$ (central maximum)
• $n=\pm 1 \Rightarrow \sin\theta=\pm 1 \Rightarrow \theta = \pm 90^\circ$

These are at grazing directions and are not formed on a finite screen placed in front in the usual YDSE geometry. Hence on the screen, effectively only the central maximum is observed.

So A is correct.

5. Check Option B: If $\lambda < d < 2\lambda$, at least one more maximum (besides the central maximum) will be observed on the screen

Since $1<\frac{d}{\lambda}<2,$ we have $\frac{\lambda}{d}<1.$

So for $n=1$, $\sin\theta = \frac{\lambda}{d} <1,$ which gives a real maximum.

Similarly for $n=-1$, another symmetric maximum exists.

Thus besides the central maximum, at least one more maximum will definitely be seen.

So B is correct.

6. Check Option C: Reduce intensity at slit 1 so that it becomes equal to slit 2

Initially: $I_1=4I,\quad I_2=I$ So $I_{\max}=9I,\quad I_{\min}=I.$

Now reduce slit 1 intensity to make it equal to slit 2: $I_1'=I,\quad I_2'=I.$

Then $I_{\max}'=(\sqrt I + \sqrt I)^2 = (2\sqrt I)^2 = 4I,$ $I_{\min}'=(\sqrt I - \sqrt I)^2=0.$

Comparison:

• Bright fringe intensity changes from $9I$ to $4I$ → decreases
• Dark fringe intensity changes from $I$ to $0$ → decreases

So it is false that both bright and dark fringe intensities increase.

Hence C is incorrect.

7. Check Option D: Increase intensity at slit 2 so that it becomes equal to slit 1

Now make $I_2'=4I,\quad I_1=4I.$

Then $I_{\max}'=(2\sqrt I+2\sqrt I)^2=(4\sqrt I)^2=16I,$ $I_{\min}'=(2\sqrt I-2\sqrt I)^2=0.$

Comparison with original:

• Bright fringe intensity: $9I \to 16I$ → increases
• Dark fringe intensity: $I \to 0$ → decreases

So both do not increase.

Hence D is incorrect.

8. Final conclusion

Correct options are: $\boxed{A,\ B}$

These match the stored correct answer.