Question

In the young's double slit experiments, 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

• A. 2
• B. 3
• C. 4
• D. none of these

Answer 1

The correct option is A 2

$\begin{array}{c}\text{Given,}\\ P_1\text{and}{P}_2\text{are two points on screen and}\\ {\text{I}}_{1}and{I}_2\text{are their intensities respectively.}\\ P_2\text{is located at quarter of fringe width}\end{array}$

$Y_{p_2}=\frac{\beta }{4},\beta =\text{fringe width}$

$y_{p_2}=\frac{\lambda D}{4d}$

$\begin{array}{c}\text{and}{p}_1\text{is located at centre of bright fringe}\\ y_{p_1}=0\end{array}$

$\begin{array}{c}\text{now,}\\ \begin{array}{cc}\frac{Y_{2}d}{D}& =\frac{\lambda }{4}\\ \Delta x_{p_2}& =\frac{\lambda }{4}\end{array}\end{array}$

$\begin{array}{cc}\text{Now,}& \\ \text{phase difference}\left(\varphi \right)& =\frac{2\pi }{\lambda }\times \text{Path difterence}\\ & =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}\\ {\varphi }_{P_2}& =\frac{\pi }{2}\end{array}$

$\text{So,}\begin{array}{cc}{I}_2& =4I_0{\cos }^2\left(\frac{\varphi }{2}\right)\\ & =4I_0{\cos }^2\left(45\right)\\ I_2& =\frac{4I_0}{2}\end{array}$

$\text{Now for}{P}_1\text{as it lies at centre of fringe}$

$\begin{array}{cc}& {\varphi }_{p_1}=0\\ I_1& =4I_0{\cos }^2\left(0\right)\\ I_1& =4I_0\\ \text{Now,}& \frac{I_1}{I_2}=\frac{4I_0}{2I_0}=2\end{array}$