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Question: In Young’s double slit experiment the slits are separated by 0.28 mm and the screen is placed 1.4 m ...

In Young’s double slit experiment the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe is measured to be 1.2 cm. The wavelength of light used in the experiment is :

A

6×10−7m6 \times 10^{- 7}m

B

3×10−7m3 \times 10^{- 7}m

C

1.5×10−7m1.5 \times 10^{- 7}m

D

5×10−6m5 \times 10^{- 6}m

Answer

6×10−7m6 \times 10^{- 7}m

Explanation

Solution

: For constructive interference,

x=nλDdx = n\lambda\frac{D}{d}

Here n=4,d=0.28×10−3mn = 4,d = 0.28 \times 10^{- 3}mand D=1.4mD = 1.4m

∴λ=xdnD=1.2×10−2×0.28×10−34×1.4=6×10−7m\therefore\lambda = \frac{xd}{nD} = \frac{1.2 \times 10^{- 2} \times 0.28 \times 10^{- 3}}{4 \times 1.4} = 6 \times 10^{- 7}m