Optics Ques 1
1 In a Young’s double slit experiment, the slits are placed 0.320 $\mathrm{mm}$ apart. Light of wavelength $\lambda=500 \mathrm{n}-\mathrm{m}$ is incident on the slits. The total number of bright fringes that are observed in the angular range $-30^{\circ} \leq \theta \leq 30^{\circ}$ is
(2019 Main, 9 Jan II)
(a) 320
(b) 321
(c) 640
(d) 641
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Answer:
Correct Answer: 1.( d )
Solution:
Formula:
- In Young’s double slit experiment, the condition of bright fringe and dark fringe are, for bright fringes (maxima), path difference $=n \lambda$
$ d \sin \theta=n \lambda $
for dark fringes (minima), path-difference $=(2 n-1) \frac{\lambda}{2}$
$ d \sin \theta=(2 n-1) \frac{\lambda}{2} $
Wavelength of light used $(\lambda)=500 \mathrm{n}-\mathrm{m}$
Angular range for bright fringe $(\theta)=-30^{\circ}$ to $30^{\circ}$ Hence, for bright-fringe,
$ \begin{aligned} n \lambda & =d \sin \theta \\ n & =\frac{d \sin \theta}{\lambda}=\frac{0.320 \times 10^{-3} \times \sin 30^{\circ}}{500 \times 10^{-9}} \\ n_{\max } & =320 \end{aligned} $
$\therefore$ Total number of maxima between the two lines are
$ \begin{aligned} & n=\left(n_{\max } \times 2\right)+1 \\ & n=(320 \times 2)+1 \\ & n=641 \end{aligned} $
BETA
