Q. In a biprism experiment, by using light of wavelength$5000\ \mathring{\mathrm{A}}$, 5mm wide fringes are obtained on a screen 1.0 m away from the coherent sources. The separation between the two coherent sources is
A) 1.0 mm
B) 0.1 mm
C) 0.05 mm
D) 0.01 mm
Explanation
(b) From $\beta = \frac{\lambda D}{d}$
$5 \times 10^{- 3} = \frac{\left( 5000 \times 10^{- 10} \right) \times 1.0}{d}$
$d = \frac{5 \times 10^{- 7}}{5 \times 10^{- 3}} = 10^{- 4}m = 0.1\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.