Question

A single slit of width $0.1mm$ is illuminated by a parallel beam of light of wavelength $6000\mathring{A}$ and diffraction bands are observed on a screen $0.5m$ from the slit. The distance of the third dark band from the central bright band is :

• A. $3mm$
• B. $4.5mm$
• C. $1.5mm$
• D. $9mm$

Answer 1

The correct option is D $9mm$

Minima for a single slit diffraction is given by:

$\Rightarrow sin\theta =\frac{m\lambda }{a}$

$\Rightarrow \theta ={\sin }^{-1}\left(\frac{3\times 6\times {10}^{-7}}{0.1\times {10}^{-3}}\right)$

$\sin \theta$ is very small

$\Rightarrow \sin \theta \simeq \tan \theta =\frac{y}{0.5}$

$\Rightarrow y=\frac{0.5\times 3\times 6\times {10}^{-7}}{0.1\times {10}^{-3}}$$y=9mm$

$\Rightarrow y=9mm$