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 3rd 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)

1.25 mm

Answer 2

Distance of n^th minima from central maxima is given as $x = \frac{(2n - 1)\lambda D}{2d}$

So here $x = \frac{(2 \times 3 - 1) \times 500 \times 10^{- 9} \times 1}{2 \times 10^{- 3}} = 1.25mm$