Q. 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)?
NTA AbhyasNTA Abhyas 2020Wave Optics
Solution:
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 $\text{d} = \frac{5 \lambda }{2}$ )
$\text{}$
$\Rightarrow sin\theta =\frac{2 n}{5} \, \, .........\left(\right.i\left.\right)$
Since, $- 1 \leq sin \theta \leq 1$ Form (i) $\Rightarrow - 1 \leq 2 n / 5 \leq 1$
So, $- \frac{5}{2} \leq \text{n} \leq \frac{5}{2}$
So, possible values of n are $\text{-2, -1, 0, 1, 2}$ . Thus a total of $\text{5}$ maxima will be obtained.
So, $- \frac{5}{2} \leq \text{n} \leq \frac{5}{2}$
So, possible values of n are $\text{-2, -1, 0, 1, 2}$ . Thus a total of $\text{5}$ maxima will be obtained.