The width of one of the two slits in Young's double slit experiment is d while that of the other slit is $x \mathrm{~d}$. 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.)

<p>Let the amplitude from the slit of width $d$ be proportional to $E$ and from the slit of width $xd$ be proportional to $xE$ (with $x>1$, as is evident from the given options).</p> <p>For two coherent waves with amplitudes $E_1$ and $E_2$, the resultant intensity when added in phase (maximum) and out of phase (minimum) is given by</p> <p>$$ I_{\max} \propto (E_1+E_2)^2 \quad \text{and} \quad I_{\min} \propto (E_2-E_1)^2. $$</p> <p>Here, setting</p> <p><p>$E_1 = E$ (from the narrower slit of width $d$) and </p></p> <p><p>$E_2 = xE$ (from the wider slit of width $xd$),</p></p> <p>we have</p> <p>$I_{\max} \propto (E+xE)^2 = E^2 (1+x)^2,$</p> <p>and since $x > 1$ the subtraction in the destructive case gives</p> <p>$I_{\min} \propto (xE-E)^2 = E^2 (x-1)^2.$</p> <p>The ratio of maximum to minimum intensity is therefore</p> <p>$\frac{I_{\max}}{I_{\min}} = \frac{(1+x)^2}{(x-1)^2}.$</p> <p>We are given that</p> <p>$\frac{(1+x)^2}{(x-1)^2} = \frac{9}{4}.$</p> <p>Taking the square root of both sides (noting that all quantities are positive) leads to</p> <p>$\frac{1+x}{x-1} = \frac{3}{2}.$</p> <p>To solve for $x$, cross-multiply:</p> <p>$2(1+x) = 3(x-1).$</p> <p>Expanding both sides:</p> <p>$2 + 2x = 3x - 3.$</p> <p>Rearrange the equation to isolate $x$:</p> <p>$2 + 2x + 3 = 3x \quad \Longrightarrow \quad 5 + 2x = 3x,$</p> <p>which implies</p> <p>$x = 5.$</p> <p>Thus, the value of $x$ is</p> <p>$\boxed{5}.$</p>