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Physics Question on Youngs double slit experiment

Two slits are made one millimeter apart and the screen is placed one metre away. The fringe separation when blue green light of wavelength 500 nm500\, nm is used is

A

5×10−4 m5 \times 10^{-4}\,m

B

2.5×10−3 m2.5 \times 10^{-3}\,m

C

2×10−4 m2 \times 10^{-4}\,m

D

10×10−4 m10 \times 10^{-4}\,m

Answer

5×10−4 m5 \times 10^{-4}\,m

Explanation

Solution

Fringe width, β=Dλd\beta = \frac{D\lambda}{d} Here, D=1m,d=1×10−3mD = 1m, d = 1\times10^{-3} m, λ=500 nm=5×10−7m \lambda = 500 \,nm = 5\times10^{-7} m ∴β=1×5×10−71×10−3\therefore\beta = \frac{1\times5\times10^{-7}}{1\times10^{-3}} =5×10−4m= 5\times10^{-4} m