Question

In young's double slit experiment , the interference pattern is found to have an intensity ratio between the bright and dark fringes as $9$. This implies that

• A. The intensities at the screen due to the two slits are $5$ units and $4$ units respectively.
• B. The intensities at the screen due to the slits are $4$ units and $1$ unit respectively.
• C. The amplitude ratio is $3$.
• D. The amplitude ratio is $2$.

Answer 1

Answer: (B), (C)

The amplitude ratio is $3$.

We know that, $\text{Intensity}\propto {\text{amplitude}}^2$

Also, $\frac{I_{max}}{I_{min}}=\frac{(a+b{)}^2}{(a-b{)}^2}$

Where, $I_1\propto a^2$ and $I_2\propto b^2$

Given data, $\frac{I_{max}}{I_{min}}=\frac{9}{1}$

$\Rightarrow \frac{(a+b)}{(a-b)}=\frac{3}{1}$

$\Rightarrow \frac{a}{b}=\frac{1}{2}$

Ratio, $\frac{I_1}{I_2}=\frac{1}{4}$