Optics Ques 7
- In a double-slit experiment, green light ( $5303 \AA$ ) falls on a double slit having a separation of $19.44 \mu-\mathrm{m}$ and a width of $4.05 \mu-\mathrm{m}$. The number of bright fringes between the first and the second diffraction minima is
(2019 Main, 11 Jan II)
(a) 5
(b) 10
(c) 9
(d) 4
Show Answer
Answer:
Correct Answer: 7.( a )
Solution:
Formula:
- Here, wavelength of light used $(\lambda)=5303 \AA$.
Distance between two slit $(d)=19.44 \mu$-m
Width of single slit (a) $=4.05 \mu \mathrm{m}$
Here, angular width between first and second diffraction minima
$ \theta \simeq \frac{\lambda}{a} $
and angular width of a fringe due to double slit is
$ \theta^{\prime}=\frac{\lambda}{d} $
$\therefore$ Number of fringes between first and second diffraction
$ \text { minima, } n=\frac{\theta}{\theta^{\prime}}=\frac{\frac{\lambda}{a}}{\frac{\lambda}{d}}=\frac{d}{a}=\frac{19.44}{4.05}=4.81 \text { or } n \simeq 5 $
$\therefore 5$ interfering bright fringes lie between first and second diffraction minima.
BETA
