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Q.A parallel beam of light of wavelength 500 nm is incident on a narrow slit, and the resulting diffraction pattern is observed on a screen placed 1 m away. The first minimum is seen at a distance of 2.5 mm from the central region on the screen. Find the width of the slit.

Tripura TbseHigher Secondary (+2 Stage) Examination 2024Subjective· 3mImportance★★★★★
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Use the single-slit first-minimum condition asin⁡θ=λa\sin\theta = \lambda with the small-angle approximation sin⁡θ≈y/D\sin\theta \approx y/D.

For single-slit (Fraunhofer) diffraction, the first minimum occurs at an angle θ\theta satisfying:

asin⁡θ=mλ,m=1 (first minimum)a\sin\theta = m\lambda, \quad m=1\text{ (first minimum)}

asin⁡θ=λa\sin\theta = \lambda

Since the screen distance DD is much larger than the fringe position yy, we use sin⁡θ≈tan⁡θ=y/D\sin\theta \approx \tan\theta = y/D:

ayD=λ  ⟹  a=λDya\frac{y}{D} = \lambda \implies a = \frac{\lambda D}{y}

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