Optics Ques 126
- The angular width of the central maximum in a single slit diffraction pattern is $60^{\circ}$. The width of the slit is $1$ $ \mu m$. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance $50 $ $cm$ from the slits. If the observed fringe width is $1 $ $cm$, what is slit separation distance?
(i.e. distance between the centres of each slit.)
(2018 Main)
(a) $100 $ $\mu m$
(b) $25 $ $\mu m$
(c) $50 $ $\mu m$
(d) $75 $ $\mu m$
Show Answer
Answer:
Correct Answer: 126.(b)
Solution:
Formula:
- In single slit diffraction pattern, $\lambda=b \sin \theta$
At $\theta=30^{\circ}$,
$ \lambda=\frac{b}{2}=\frac{1 \times 10^{-6}}{2}=5 \times 10^{-7} $ $m $
In YDSE,
fringe width,$\quad \omega=\frac{\lambda D}{d} \Rightarrow d=\frac{\lambda D}{\omega}=\frac{5 \times 10^{-7} \times 0.5}{1 \times 10^{-2}}$
$ d=25 \times 10^{-6} m=25 $ $\mu m $
BETA
