The width of one of the two slits in Young's double slit experiment is d while that of the other slit is xd. 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.)

Step 1: Understanding the Intensity Ratio Formula

In Young’s double-slit experiment, the intensity at any point in the interference pattern is given by:

$$I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \delta$$

where:

$I_1$ and $I_2$ are the individual intensities from the two slits,

$\delta$ is the phase difference between the two waves.

The maximum intensity occurs when $\cos \delta = 1$:

$$I_{\text{max}} = ( \sqrt{I_1} + \sqrt{I_2} )^2$$

The minimum intensity occurs when $\cos \delta = -1$:

$$I_{\text{min}} = ( \sqrt{I_1} - \sqrt{I_2} )^2$$

We are given:

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

Step 2: Express Intensities in Terms of Slit Widths

Since intensity is proportional to the square of the electric field amplitude, and the field amplitude is proportional to the slit width, we can write:

$$I_1 \propto d^2, \quad I_2 \propto (xd)^2 = x^2 d^2$$

Thus,

$$I_1 = k d^2, \quad I_2 = k x^2 d^2$$

for some proportionality constant $k$.

Step 3: Use the Given Intensity Ratio

$$\frac{( \sqrt{I_1} + \sqrt{I_2} )^2}{( \sqrt{I_1} - \sqrt{I_2} )^2} = \frac{9}{4}$$

Substituting $\sqrt{I_1} = \sqrt{k} d$ and $\sqrt{I_2} = \sqrt{k} x d$,

$$\frac{( d + x d )^2}{( d - x d )^2} = \frac{9}{4}$$

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

Step 4: Solve for $x$

Taking the square root on both sides:

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

Cross multiplying:

$2(x+1)=3(x-1)$
$\Rightarrow$ 3x – 3 = 2x + 2
x = 5

Final Answer:

$x=5$