In Young's double slit experiment, the width of one of the two slits is four times the width of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern will be

9 : 1

The correct answer is Option (3) → 9 : 1

Let the amplitude from the smaller slit be $A$.

Since intensity $\propto$ width, the amplitude $\propto \sqrt{\text{width}}$.

So, amplitude from the larger slit = $\sqrt{4}A = 2A$.

Maximum intensity:

$I_{max} \propto (A + 2A)^2 = (3A)^2 = 9A^2$

Minimum intensity:

$I_{min} \propto (2A - A)^2 = (A)^2 = A^2$

Ratio:

$\frac{I_{max}}{I_{min}} = \frac{9A^2}{A^2} = 9$