Question

In a Young's double slit experiment, the slit separation is $ 1\,mm $ and the screen is $ 1\,m $ from the slit. For a monochromatic light of wavelength $ 500 \,nm $ , the distance of $ 3^{\text{rd}} $ minima from the central maxima is

• A. $ 0.50 \,mm $
• B. $ 1.25 \,mm $
• C. $ 1.50 \,mm $
• D. $ 1.75 \,mm $

Answer 1

Answer: (B)

Given, $d=1 \,mm =1 \times 10^{-3} \,m , D=1 \,m$
$\lambda=500 \,nm =500 \times 10^{-9} \,m$
Distance of nth minima from central maxima
maximax $x_{n}=\frac{(2 n-1) \lambda}{2} \frac{D}{d}$
$=\frac{(2 \times 3-1) \times 500 \times 10^{-9}}{2} \times \frac{1}{1 \times 10^{-3}}$
$=2.5 \times 500 \times 10^{-6}$
$=12.5 \times 10^{-4} \,m$
$=1.25\, nm$