In a Young's double slit experiment, the ratio of amplitude of light coming from slits is $$2 : 1$$. The ratio of the maximum to minimum intensity in the interference pattern is

In Young's double slit experiment, the amplitude ratio of light from the two slits is 2:1. We need to find the ratio of maximum to minimum intensity.

To begin,

The intensity of a wave is proportional to the square of its amplitude:

$$ I \propto a^2 $$

If the amplitudes from the two slits are $$a_1$$ and $$a_2$$, then the individual intensities are $$I_1 \propto a_1^2$$ and $$I_2 \propto a_2^2$$.

Next,

In an interference pattern, the resultant amplitude depends on the phase difference $$\phi$$ between the two waves:

$$ a_{\text{resultant}} = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos\phi} $$

Maximum intensity occurs when $$\cos\phi = 1$$ (constructive interference):

$$ I_{max} \propto (a_1 + a_2)^2 $$

Minimum intensity occurs when $$\cos\phi = -1$$ (destructive interference):

$$ I_{min} \propto (a_1 - a_2)^2 $$

From this,

$$ \frac{I_{max}}{I_{min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} $$

Given $$a_1 : a_2 = 2 : 1$$, let $$a_1 = 2$$ and $$a_2 = 1$$:

$$ \frac{I_{max}}{I_{min}} = \frac{(2 + 1)^2}{(2 - 1)^2} = \frac{9}{1} = 9 $$

The ratio of maximum to minimum intensity is 9 : 1.

The correct answer is Option 4: 9 : 1.