Question

In Young's double slit experiment, how many maxima can be obtained on a screen (including central maxima), if $\text{d}=\frac{5\lambda }{2}$ (where $\lambda$ is the wavelength of light)?

• A. $\text{5}$
• B. $\text{4}$
• C. $\text{7}$
• D. $\text{1}$

Answer 1

Answer: (A)

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 $\text{d}=\frac{5\lambda }{2}$ )

$\Rightarrow sin\theta =\frac{2n}{5}.........(i)$
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.