Physics - Wave Optics Question with Solution | TestHub

In YDSE, path difference,$\Delta \text{x}=\text{d sin}\theta$

For maxima,$\Delta \text{x}=\text{n}\lambda$, where$\text{n}=0\text{, }\pm 1\text{, }\pm 2\text{....}$

$\frac{5\lambda }{2}\text{sin}\theta =\text{n}\lambda$(given that$\text{d}=\frac{5\lambda }{2}$)

$$

$\begin{array}{l}\Rightarrow \sin \theta =\frac{2n}{5}\mathrm{.........}(i)\end{array}$

Since,$-1\le \sin \theta \le 1$ Form (i)$\Rightarrow -1\le 2n/5\le 1$

So, $-\frac{5}{2}\le \text{n}\le \frac{5}{2}$

So, possible values of n are $\text{-2, -1, 0, 1, 2}$. Thus a total of $\text{5}$ maxima will be obtained.