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**Hint:**Use the formula of the maximum intensity and the minimum intensity to calculate it both of the light that enters the slit. Divide both to obtain the ratio of their intensities. In Young's experiment, the monochromatic light forms the different colors all over the screen.

**Useful formula:**

(1) The maximum intensity in the experiment is given by

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

Where ${I_{\max }}$ is the maximum intensity, ${I_1}$ is the intensity of the first slit and the ${I_2}$ is the intensity of the second slit.

(2) The minimum intensity is given by

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

Where ${I_{\min }}$ is the minimum intensity of the fringes.

**Complete step by step solution:**

It is given that

Reduction in the percentage of the intensity, if one slit is covered with glass, is $50\,\% $.

Let us assume that intensity of the slit one, ${I_1} = I$

So, the intensity of the slit two is ${I_2} = 0.5I$ (since the intensity reduces to $50\,\% $ )

By using the formula (1),

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

Substituting the values known,

${I_{\max }} = {\left( {\sqrt I + \sqrt {0.5I} } \right)^2}$

By simplifying,

${I_{\max }} = 2.9I$

Similarly using the formula (2),

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

${I_{\min }} = {\left( {\sqrt I - \sqrt {0.5I} } \right)^2}$

By simplification,

${I_{\min }} = 0.086I$

Ratio of the maximum and the minimum intensity is calculated as

$r = \dfrac{{{I_{\max }}}}{{{I_{\min }}}}$

$r = \dfrac{{2.9I}}{{0.086I}}$

By performing division in the above step,

$r = 33.8$

**Thus the ratio of the maximum and the minimum intensity is $33.8$.**

(b) Normally, the monochromatic light is used. But instead of this, if white light is used, the white fringe is formed at the center of the screen and the other colored fringes are positioned at various positions on the screen.

**Note:**The slits in the young’s double slit experiment are of the same distance from the center and of the same size. Hence the intensity of the light emitted by the slits $\left( I \right)$ are the same. But in this case , the second slit is covered with glass, so its intensity reduces $50\,\% $ . Hence the intensity of slit one is $I$ and slit two is $0.5I$ .

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