Q. Light of wavelength 6000 $\mathring{\mathrm{A}}$ is incident on a single slit. The first minimum of the diffraction pattern is obtained at 4 mm from the centre. The screen is at a distance of 2 m from the slit. The slit width will be
A) 0.3 mm
B) 0.2 mm
C) 0.15 mm
D) 0.1 mm
Explanation
(a) From $a\ \sin\theta = n\lambda$
$a\frac{x}{D} = n\lambda,\ a = \frac{n\lambda D}{x} = \frac{1 \times 6000 \times 10^{- 10} \times 2}{4 \times 10^{- 3}}\ $
$= 3 \times 10^{- 4} = 0.3\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.