JEE Advanced 2026 Paper 2, Physics Section 3 Q2: Diffraction

JEE Advanced2026Paper 2Physics Section 3

In a single slit diffraction experiment, a slit of width $(0.016 \pm 0.002)$ mm is used to measure the wavelength of a monochromatic light source. In the diffraction pattern, the angular distance between the central maximum and first minimum is measured to be $(2^{\circ} \pm 4 0^{'})$ . The value of the fractional error in the measurement of wavelength is: [Given: $sin (2^{\circ}) = 0.035$ ]

Your answer

Solution

Governing relation. For the first minimum, $d sin θ = λ$ .

Error propagation. Taking logarithm and differentiating,

$\frac{d λ}{λ} = \frac{Δ d}{d} + cot θ Δ θ$ .

Convert $Δ θ$ . $4 0^{'} = \frac{40}{60} \cdot \frac{π}{180} = \frac{π}{270}$ rad $\approx 0.01164$ .

Compute $cot θ$ . $sin 2^{\circ} = 0.035 \Rightarrow cos 2^{\circ} \approx 1 - 0.03 5^{2} \approx 0.9994$ , so $cot 2^{\circ} \approx \frac{0.9994}{0.035} \approx 28.55$ .

Substitute.

$\frac{Δ λ}{λ} = \frac{0.002}{0.016} + 28.55 \times \frac{π}{270} = 0.125 + 0.332 \approx 0.457$ .

Rounded to two decimal places, $\frac{Δ λ}{λ} \approx 0.46$ .

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JEE Advanced 2026 Paper 2, Physics Section 3 Q2: Diffraction

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