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Q.A parallel beam of light of wavelength 500nm (Nanometre) falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 metre away. It is observed that the first minimum is at a distance of 2.5 mm (millimetre) from the centre of the screen. Find the width of the slit.

Punjab PsebPSEB Punjab Class 12 Board 2019Subjective· 2mImportance★★★★★
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Using the single-slit diffraction condition for the first minimum and the small-angle approximation, the slit width works out to 0.2 mm.

For single-slit diffraction, the position of the first minimum satisfies:

asin⁡θ=λa\sin\theta = \lambda

For small angles (screen far from slit), sin⁡θ≈tan⁡θ=xD\sin\theta \approx \tan\theta = \dfrac{x}{D}, where xx is the distance of the first minimum from the centre and DD is the slit-to-screen distance.

Given: λ=500 nm=500×10−9 m\lambda = 500\text{ nm} = 500\times10^{-9}\text{ m}, D=1 mD = 1\text{ m}, x=2.5 mm=2.5×10−3 mx = 2.5\text{ mm} = 2.5\times10^{-3}\text{ m}.

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