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Q.

In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength λ1 is found to be coincident with the third maximum at λ2 . So

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a

3λ1=0.3λ2

b

3λ1=λ2

c

λ1=3.5λ2

d

0.3λ1=3λ2

answer is C.

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Detailed Solution

Position of first minima = position of third maxima i.e., 1×λ1Dd=(2×3+1)2λ2Dd ⇒λ1=3.5λ2
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