In the Young's double slit experiment, the intensities at two points $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ on the screen are respectively $\mathrm{I}_{1}$ and $\mathrm{I}_{2} .$ If $\mathrm{P}_{1}$ is located at the centre of a bright fringe and $\mathrm{P}_{2}$ is located at a distance equal to a quarter of fringe width from $\mathrm{P}_{1}$, then $\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}$ is

• A

  2
• B

  $\frac{1}{2}$
• C

  4
• Correct answer
  D

  16

Solution

Fringe width $\beta=\frac{\lambda D}{d}$ Let be the amplitude of the place where constructive inference takes place. The position of fringe at $p_{2}$. $$ \Rightarrow x=\frac{n \lambda D}{d} $$ Given, $\beta^{\prime}=\left(\frac{\beta}{4}\right)$
$\therefore \frac{\lambda D}{4 d}=\frac{n \lambda D}{d}$ or $n=\frac{1}{4}$ $\therefore \frac{I_{1}}{I_{2}}=\frac{a^{2}}{\left(\frac{a}{4}\right)^{2}}=16: 1$

Hence, the correct answer is option: 'D'.