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In the Young's double slit experiment, the intensities at two points $P_1$ and $P_2$ on the screen are respectively $I_1$ and $I_2$. If $P_1$ is located at the centre of a bright fringe and $P_2$ is located at a distance equal to a quarter of fringe width from $P_1$, then $\frac{I_1}{I_2}$ is
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16
Fringe width
$\beta=\frac{\lambda D}{d}$
Let the amplitude of that place where constructive inference takes place is $a$. The position of fringe at $p_2$ is
$x=\frac{n \lambda D}{d}$
$\begin{array}{rlrl}\text { Given, } & \beta^{\prime} & =\left(\frac{\beta}{4}\right) \\ & \therefore & \frac{\lambda D}{4 d} & =\frac{n \lambda D}{d}\end{array}$
$\begin{aligned} & \text { or } \quad n=\frac{1}{4} \\ & \therefore \quad \frac{I_1}{I_2}=\frac{a^2}{\left(\frac{a}{4}\right)^2} \\ & \text { or } \quad I_1: I_2=16: 1 \\ & \end{aligned}$
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