The width of one of the two slits in Young's double slit experiment is d while that of the other slit is $$x$$ d. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9:4 then what is the value of $$x$$ ? (Assume that the field strength varies according to the slit width.)

Let the electric field (amplitude) produced by a slit be directly proportional to its width.

Width of slit 1 = $$d$$     → amplitude $$E_1 \propto d$$

Width of slit 2 = $$x\,d$$  → amplitude $$E_2 \propto x\,d$$

Choose the proportionality constant equal for both slits and write

$$E_1 = E_0 d, \qquad E_2 = E_0 x d$$

Hence the ratio of the two amplitudes is

$$\frac{E_2}{E_1}=x.$$

For any point on the screen the resultant intensity is

$$I = \bigl(E_1 + E_2\bigr)^2 = E_1^2 + E_2^2 + 2E_1E_2\cos\phi,$$

where $$\phi$$ is the phase difference between the two waves at that point.

At an interference maximum, $$\cos\phi = +1$$, so

$$I_{\max} = (E_1 + E_2)^2.$$

At an interference minimum, $$\cos\phi = -1$$, so

$$I_{\min} = (E_1 - E_2)^2.$$

Given that

$$\frac{I_{\max}}{I_{\min}} = \frac{9}{4},$$

write this ratio in terms of the amplitudes:

$$\frac{(E_1 + E_2)^2}{(E_1 - E_2)^2} = \frac{9}{4}.$$

Substitute $$E_2 = xE_1$$:

$$\frac{(E_1 + xE_1)^2}{(E_1 - xE_1)^2} = \frac{(1 + x)^2}{(1 - x)^2} = \frac{9}{4}.$$

Take square roots on both sides (keeping the positive value for the ratio of two positive quantities):

$$\frac{1 + x}{|\,1 - x\,|} = \frac{3}{2}.$$

Now two cases arise.

Case 1: $$x \lt 1$$  ⇒ $$|1 - x| = 1 - x$$

$$\frac{1 + x}{1 - x} = \frac{3}{2} \;\; \Longrightarrow \;\; 2 + 2x = 3 - 3x \;\; \Longrightarrow \;\; 5x = 1 \;\; \Longrightarrow \;\; x = 0.2.$$

This contradicts the statement that the second slit is wider than the first (because $$x d$$ would then be narrower). Therefore discard this case.

Case 2: $$x \gt 1$$  ⇒ $$|1 - x| = x - 1$$

$$\frac{1 + x}{x - 1} = \frac{3}{2} \;\; \Longrightarrow \;\; 2 + 2x = 3x - 3.$$

Rearrange: $$2 + 2x - 3x = -3 \;\; \Longrightarrow \;\; 2 - x = -3 \;\; \Longrightarrow \;\; x = 5.$$

The physically acceptable value is therefore

$$x = 5.$$

Hence the width of the second slit must be five times the width of the first slit.

Option B is correct.