Question

The light of wavelength $6328A^o$ is incident on a slit of width $0.2mm$ perpendicularly, the angular fringe width will be:

• A. ${0.36}^o$
• B. ${0.18}^o$
• C. ${0.72}^o$
• D. ${0.09}^o$

Answer 1

The correct option is B ${0.18}^o$

$\sin \theta =\frac{n\lambda }{d}(forn=0,1,\mathrm{2...})$

for frige width, $n=1$

$\Rightarrow \theta =\sin \left(\frac{6328\times {10}^{-10}}{0.2\times {10}^{-3}}\right)={0.18}^o$

Hence, Angular width is ${0.18}^o$