Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is:

In Young's double slit experiment, the intensity from each slit is proportional to the slit width. If one slit has width $$w$$ and the other has width $$w/2$$, then $$I_1 = I$$ and $$I_2 = I/2$$ (say).

The maximum intensity is $$I_{max} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 = \left(\sqrt{I} + \sqrt{I/2}\right)^2 = I\left(1 + \dfrac{1}{\sqrt{2}}\right)^2$$

The minimum intensity is $$I_{min} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 = \left(\sqrt{I} - \sqrt{I/2}\right)^2 = I\left(1 - \dfrac{1}{\sqrt{2}}\right)^2$$

The ratio is:

$$\dfrac{I_{max}}{I_{min}} = \dfrac{\left(1 + \dfrac{1}{\sqrt{2}}\right)^2}{\left(1 - \dfrac{1}{\sqrt{2}}\right)^2} = \dfrac{\left(\sqrt{2} + 1\right)^2}{\left(\sqrt{2} - 1\right)^2}$$

$$= \dfrac{2 + 2\sqrt{2} + 1}{2 - 2\sqrt{2} + 1} = \dfrac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$$

Hence, the correct answer is Option B.