Q. In a Young’s double slit experiment, the separation between the two slits is 0.9 mm and the fringes are observed 1 m away. If it produces the second dark fringes at a distance of 1 mm from the central fringe, the wavelength of the monochromatic source of light used is
A) 450 nm
B) 400 nm
C) 5002 nm
D) 600 nm
Explanation
(d) Given, spacing between second dark fringe and central fringe
$= \beta + \frac{\beta}{2}$
Or $\frac{3\beta}{2} = 1\ mm$
or $\beta = \frac{2}{3}\ \times 1\ mm$
$\frac{\lambda D}{d} = \frac{2}{3}mm$
$\therefore\ \lambda = \frac{2}{3} \times 10^{- 3} \times \frac{0.9 \times 10^{- 3}}{\mathbf{1}}$
$\therefore\lambda = 0.6 \times 10^{- 6}\ m\ $
$\therefore\ \lambda = 600 \times 10^{- 9}\ m$
$= 600\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.