Question

Find the angular spread between central maximum and first order maximum of the diffraction pattern due to single slit of width $0.20$mm,when light of wavelength $5460\mathring{A}$ is incident on it normally.Also calculate distance between first two dark bands on each side of central bright band in the diffraction pattern observed on a screen place $1.4m$ from the slit:

• A. $8.19\times {10}^{-3}rad,7.6mm$
• B. $9.19\times {10}^{-4}rad,9mm$
• C. $10\times {10}^{-3}rad,4mm$
• D. $6\times 19\times {10}^{-3}rad,8.6mm$

Answer 1

The correct option is A $8.19\times {10}^{-3}rad,7.6mm$

Angular position of first maxima $(n=1)$
$\theta =(n+\frac{1}{2})\frac{\lambda }{a}=(1+\frac{1}{2})\frac{5460\times {10}^{-10}}{0.2\times {10}^{-3}}$
$=4.095\times {10}^{-3}$ radians
$\therefore$ Angular spread $=2\times 4.095\times {10}^{-3}\times radians$
$8.19\times {10}^{-3}$ radians
Position of first darle band
$\frac{D\lambda }{a}=\frac{1.4\times 5460\times {10}^{-10}}{0.2\times {10}^{-3}}$
$3.822\times {10}^{-3}$
$\therefore$between two dark band
$=2\times 3.822\times {10}^{-3}m$
$=7.644mm$