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NVAEasyJEE 2024Diffraction & Polarisation

JEE Physics 2024 Question with Solution

In a single slit diffraction pattern, a light of wavelength 6000 A˚6000 \, \text{\AA} is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm3 \, \text{mm} when the screen is placed 50 cm50 \, \text{cm} away from the slits. The width of the slit is ×10−4 m\times 10^{-4} \, \text{m}:

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given: wavelength λ=6000 A˚=6000×10−10 m\lambda = 6000 \, \text{\AA} = 6000 \times 10^{-10} \, \text{m}, screen distance D=50 cm=0.5 mD = 50 \, \text{cm} = 0.5 \, \text{m}, and distance between the first and third minima y3−y1=3 mm=0.003 my_3 - y_1 = 3 \, \text{mm} = 0.003 \, \text{m}.

Find: width of the slit aa.

For single slit diffraction, the minima are given by

asin⁡θ=nλa \sin \theta = n\lambda

For small angles,

sin⁡θ≈tan⁡θ=yD\sin \theta \approx \tan \theta = \frac{y}{D}

So the position of the nnth minima is

yn=nλDay_n = \frac{n\lambda D}{a}

Step-by-step Calculation

For the first minima,

y1=λDay_1 = \frac{\lambda D}{a}

For the third minima,

y3=3λDay_3 = \frac{3\lambda D}{a}

Hence, the distance between them is

y3−y1=3λDa−λDa=2λDay_3 - y_1 = \frac{3\lambda D}{a} - \frac{\lambda D}{a} = \frac{2\lambda D}{a}

Direct Substitution

Equating the given separation,

2λDa=0.003\frac{2\lambda D}{a} = 0.003

Therefore,

a=2⋅6000×10−10⋅0.50.003a = \frac{2 \cdot 6000 \times 10^{-10} \cdot 0.5}{0.003} a=6000×10−100.003=2×10−4 ma = \frac{6000 \times 10^{-10}}{0.003} = 2 \times 10^{-4} \, \text{m}

Therefore, the width of the slit is 2×10−4 m2 \times 10^{-4} \, \text{m}, so the required numerical value is 2.

Common mistakes

  • Using the separation as y3y_3 instead of y3−y1y_3 - y_1 is incorrect because the question gives the distance between two minima, not the position from the central maximum. First write both positions separately, then subtract.

  • Not converting 6000 A˚6000 \, \text{\AA} into metres leads to a wrong order of magnitude. Use 1 A˚=10−10 m1 \, \text{\AA} = 10^{-10} \, \text{m} before substitution.

  • Ignoring the small-angle relation sin⁡θ≈tan⁡θ=yD\sin \theta \approx \tan \theta = \frac{y}{D} without justification can cause formula misuse. For screen diffraction at small angles, convert angular condition into linear position using this approximation.

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