Question

In the YDSE, the point source is placed slightly off the central axis, as shown in the figure. Find the nature & order of the interference at point $\text{P}$ ($\lambda =500\text{nm}$).

• A. Maxima, $70$
• B. Minima, $120$
• C. Maxima, $120$
• D. Minima, $70$

Answer 1

The correct option is C Maxima, $120$



When source is off-centered from the central line, then path difference is given by,

$\Delta x=(S{S}_2+S_{2}P)-(S{S}_1+S_{1}P)$

$=(S{S}_2-S{S}_1)+(S_{2}P-S_{1}P)$

$=d\sin {\theta }_1+d\sin {\theta }_2$

$\therefore \Delta x=\frac{y_{1}d}{D_1}+\frac{y_{2}d}{D_2}$

$\Delta x=\frac{{10}^{-3}\times 10\times {10}^{-3}}{1}+\frac{5\times {10}^{-3}\times 10\times {10}^{-3}}{1}$

$=60\times {10}^{-6}\text{m}$

As we know, for maxima $\Delta x=n\lambda$

For minima $\Delta x=(2n-1)\frac{\lambda }{2}$

If the point $\text{P}$ to be maxima, then $n=$integer,

$\Rightarrow \Delta x=n\lambda$

$\Rightarrow n=\frac{\Delta x}{\lambda }=\frac{60\times {10}^{-6}}{500\times {10}^{-9}}=\frac{60}{5}\times 10=120$

$\therefore$ A maxima of order $120$ occurs at $\text{P}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{C}\right)$ is the correct answer.