Q. A single slit is located effectively at infinity in front of a lens of focal length 1 m and it is illuminated normally with light of wavelength 600 nm. The first minima on either side of central maximum are separated by 4 mm. Width of the slit is ……
A) 0.1 mm
B) 0.2 mm
C) 0.3 mm
D) 0.4 mm
Explanation
(c) $e\sin{\theta = \lambda\ \Rightarrow}\sin{\theta \approx \theta = \frac{\lambda}{e}}$
$x = f,\ \theta = \frac{f\lambda}{e}$
$\therefore 2x =$ separation between minima on either side of central maximum’
$= \frac{2f\lambda}{e}$
Hence, $e = \frac{2f\lambda}{2x}$
$= \frac{{2 \times 1 \times 600 \times 10}^{- 9}}{{4 \times 10}^{- 3}} = 0.3\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.