Optics 6 Question 11
13. 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
Solution:
- 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,$\quad \omega=\frac{\lambda D}{d} \Rightarrow d=\frac{\lambda D}{\omega}=\frac{5 \times 10^{-7} \times 0.5}{1 \times 10^{-2}}$
$$ d=25 \times 10^{-6} m=25 \mu m $$
