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MCQEasyJEE 2024Diffraction & Polarisation

JEE Physics 2024 Question with Solution

The diffraction pattern of light of wavelength 400 nm400 \, \text{nm} diffracting from a slit of width 0.2 mm0.2 \, \text{mm} is focused on the focal plane of a convex lens of focal length 100 cm100 \, \text{cm}. The width of the 1st secondary maxima will be:

  • A

    2 mm2 \, \text{mm}

  • B

    2 cm2 \, \text{cm}

  • C

    0.02 mm0.02 \, \text{mm}

  • D

    0.2 mm0.2 \, \text{mm}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: wavelength of light λ=400×10−9 m\lambda = 400 \times 10^{-9} \, \text{m}, slit width a=0.2×10−3 ma = 0.2 \times 10^{-3} \, \text{m}, and focal length of lens D=f=100 cm=1 mD = f = 100 \, \text{cm} = 1 \, \text{m}.

Find: the width of the 1st secondary maxima in the single-slit diffraction pattern.

In a single-slit diffraction pattern, minima occur at

asin⁡θ=mλa \sin \theta = m\lambda

For small angles,

sin⁡θ≈tan⁡θ≈θ\sin \theta \approx \tan \theta \approx \theta

so the position of the mthm^{\text{th}} minimum on the focal plane is

ym=mλDay_m = \frac{m\lambda D}{a}

The first secondary maxima lies between the first and second minima, so its width is

W=y2−y1W = y_2 - y_1

Now,

y1=λDay_1 = \frac{\lambda D}{a} y1=(400×10−9 m)(1 m)0.2×10−3 m=2×10−3 m=2 mmy_1 = \frac{(400 \times 10^{-9} \, \text{m})(1 \, \text{m})}{0.2 \times 10^{-3} \, \text{m}} = 2 \times 10^{-3} \, \text{m} = 2 \, \text{mm}

Also,

y2=2λDa=4 mmy_2 = \frac{2\lambda D}{a} = 4 \, \text{mm}

Therefore,

W=4 mm−2 mm=2 mmW = 4 \, \text{mm} - 2 \, \text{mm} = 2 \, \text{mm}

So, the width of the 1st secondary maxima is 2 mm2 \, \text{mm}. The correct option is A.

Direct Formula

Given: λ=400×10−9 m\lambda = 400 \times 10^{-9} \, \text{m}, a=0.2×10−3 ma = 0.2 \times 10^{-3} \, \text{m}, D=1 mD = 1 \, \text{m}.

Find: width of the 1st secondary maxima.

The width of any secondary maxima in single-slit diffraction is directly

W=DλaW = \frac{D\lambda}{a}

Substituting the given values,

W=(1 m)(400×10−9 m)0.2×10−3 mW = \frac{(1 \, \text{m})(400 \times 10^{-9} \, \text{m})}{0.2 \times 10^{-3} \, \text{m}} W=2×10−3 m=2 mmW = 2 \times 10^{-3} \, \text{m} = 2 \, \text{mm}

Therefore, the width of the 1st secondary maxima is 2 mm2 \, \text{mm}, so the correct option is A.

Common mistakes

  • Using the width of the central maximum instead of the width of the first secondary maxima. The central maximum has width 2Dλa\frac{2D\lambda}{a}, whereas the first secondary maxima has width Dλa\frac{D\lambda}{a}. Identify which bright region the question asks for before applying the formula.

  • Failing to convert units into SI consistently. Here 100 cm=1 m100 \, \text{cm} = 1 \, \text{m}, 0.2 mm=0.2×10−3 m0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m}, and 400 nm=400×10−9 m400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}. Convert all quantities first to avoid order-of-magnitude errors.

  • Confusing the position of a minimum with the width of a maxima. The first minimum is at y1=Dλay_1 = \frac{D\lambda}{a}, but the first secondary maxima width is the distance between the first and second minima, that is y2−y1y_2 - y_1. Do not stop after calculating only y1y_1 unless you recognize it also equals the required width here.

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