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Q.

Light of wavelengths λ1 and λ2 is made incident successively on the surface of metal Light of wavelength λ1 produces photo-electric effect whereas light of wavelength λ2does not. Then interference is produced using these light sources, the fringe widths0 obtained are β1 and β2 respectively, then

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a

β1>β2

b

β1<β2

c

β1=β2

d

none of these

answer is B.

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Detailed Solution

According to the condition of question, λ1<λ2 we know that β=λD/(2d) where 2 d is the distance between two coherent sources and D is the distance of screen from two coherent sources. Henceβ∝λ β1<β2 ∵λ1<λ2

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