Question

In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern.

• A. The intensities of both the maxima and the minima increase
• B. The intensity of the maxima increases and the minima has zero intensity
• C. The intensity of the maxima decrease and that of the minima increases
• D. The intensity of the maxima decreases and the minima has zero intensity

Answer 1

The correct option is A The intensities of both the maxima and the minima increase

The intensities due to slits are proportional to their width.
When we takes slits of equal width
$\frac{I_1}{I_2}=\frac{W_1}{W_2}=1$
$\frac{I_1}{I_2}=\frac{a^2}{b^2}=1$
or $\frac{a}{b}=1$
$\frac{I_{max}}{I_{min}}=\frac{(a+b{)}^2}{(a-b{)}^2}=\frac{{(\frac{a}{b}+1)}^2}{{(\frac{a}{b}-1)}^2}=\frac{(1+1{)}^2}{(1-1{)}^2}=\frac{2}{0}$
This implies maximum intensity $I_{max}$ is $2$ and minimum intensity $I_{min}$ is zero.
When we take the width of one slit is twice of the other
$\frac{I_1^{\prime }}{I_2^{\prime }}=\frac{W_1^{\prime }}{W_2^{\prime }}=2$
$\frac{I_1^{\prime }}{I_2^{\prime }}=\frac{a^{\prime 2}}{b^{\prime 2}}=\frac{2}{1}$
or $\frac{a^{\prime }}{b^{\prime }}=\frac{\sqrt{2}}{1}$
$\frac{I_{max}^{\prime }}{I_{min}^{\prime }}=\frac{(a^{\prime }+b^{\prime }{)}^2}{(a^{\prime }-b^{\prime }{)}^2}=\frac{{(\frac{a^{\prime }}{b^{\prime }}+1)}^2}{{(\frac{a^{\prime }}{b^{\prime }}-1)}^2}=\frac{(\sqrt{2}+1{)}^2}{(\sqrt{2}-1{)}^2}$.
This implies maximum intensity $I_{max}^{\prime }$ is $(\sqrt{2}+1{)}^2$ and minimum intensity $I_{min}$ is $(\sqrt{2}-1{)}^2$.
Therefore, the intensities of both the maxima and minima increase.