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Physics Question on Wave optics

Two slits in Young’s double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is 600×10−9 m600 \times 10^{-9}\, \text{m}, then the fringe separation is:

A

4×10−5 m4 \times 10^{-5}\, \text{m}

B

9×10−8 m9 \times 10^{-8}\, \text{m}

C

4×10−7 m4 \times 10^{-7}\, \text{m}

D

4×10−4 m4 \times 10^{-4}\, \text{m}

Answer

4×10−4 m4 \times 10^{-4}\, \text{m}

Explanation

Solution

Fringe separation (β) in Young's double slit experiment is given by:

β=λDd\beta = \frac{\lambda D}{d}

where:

λ is the wavelength of light.

D is the distance between the slits and the screen.

d is the distance between the slits.

Given:

λ = 600 × 10-9 m D = 1 m d = 1.5 mm = 1.5 × 10-3 m

β=(600×10−9 m)(1 m)1.5×10−3 m=4×10−4 m\beta = \frac{(600 \times 10^{-9} \text{ m})(1 \text{ m})}{1.5 \times 10^{-3} \text{ m}} = 4 \times 10^{-4} \text{ m}