Question

In a Young's double slit experiment, the slit separation is 1 mm and the screen is 1 m from the the slit. For a monochromatic light of wavelength $5000\mathring{A}$, the distance of $3^{rd}$ minima from the central maxima,on the screen, is

• A. 0.50 mm
• B. 1.25 mm
• C. 1.50 mm
• D. 1.75 mm

Answer 1

The correct option is B 1.25 mm

Distance of $n$ minima from central bridge fringe
$x_n=\frac{(2n-1)\lambda D}{2d}$
For $n=3$ i.e., $3$ minima
$x_3=\frac{(2\times 3-1)\times 500\times {10}^{-9}}{2\times 1\times {10}^{-3}}$
$=\frac{5\times 500\times {10}^{-6}}{2}=1.25\times {10}^{-3}m=1.25mm$