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Physics Question on Wave optics

In Young's double slit experiment, fringes of width β\beta are produced on a screen kept at a distance of 1 m from the slit. When the screen is moved away by 5×10−25 \times 10^{-2} m, fringe width changes by 3×10−53 \times10^{-5} m. The separation between the slits is 1×10−3m1 \times10^{-3} m. The wavelength of the light used is

A

500 nm

B

600 nm

C

700 nm

D

400 nm

Answer

600 nm

Explanation

Solution

In YDSE, we have Wavelength λ=ΔβdΔD=3×10−5×1×10−35×10−2\lambda =\frac{\Delta \beta d}{\Delta D}=\frac{3 \times 10^{-5} \times 1 \times 10^{-3}}{5 \times 10^{-2}} =6×10−7 m=6 \times 10^{-7} \,m or λ=600 nm\lambda=600 \,nm