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Q.

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
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