In Young's experiment the distance of the screen from the two slits is 4 m . When light of wavelength λ is allowed to fall on the slits, the width of the fringes obtained on the screen is 4 mm . The distance between the two slits is 0.06 cm . The wavelength λ is ....... (nm)
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a
600 nm
b
200 nm
c
100 nm
d
800 nm
answer is A.
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Detailed Solution
Distance of the screen D=4 m Fringe width β=4 mm=4×10−3 m Distance between the slits d=0.06 cm=0.06×10−2 m Wavelength λ=? Fringe width β=λDd Wavelength λ=βdD =4×10−3×6×10−44 =6×10−7 ∴ λ=600 nm