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) $d = \frac{D\lambda}{\beta} = \frac{1 \times 5 \times 10^{- 7}}{5 \times 10^{- 3}} = 10^{- 4}m = 0.1\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.