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JEE Advanced 2026 · Paper 2, Physics question 11

In a single slit diffraction experiment, a slit of width (0.016±0.002)(0.016\pm0.002) 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∘±40′)(2^\circ\pm40'). The value of the fractional error in the measurement of wavelength is:
[Given: sin⁡(2∘)=0.035\sin(2^\circ)=0.035]

Answer and worked solution

Official answer: 0.43

First minimum: asin⁡θ=λa\sin\theta=\lambda, so
Δλλ=Δaa+cot⁡θ Δθ.\frac{\Delta\lambda}{\lambda}=\frac{\Delta a}{a}+\cot\theta\,\Delta\theta.

  • Δaa=0.0020.016=0.125\dfrac{\Delta a}a=\dfrac{0.002}{0.016}=0.125.
  • Δθ=40′=23∘=0.01164\Delta\theta=40'=\frac23^\circ=0.01164 rad, and cot⁡2∘≈0.99940.035=28.6\cot2^\circ\approx\dfrac{0.9994}{0.035}=28.6. So cot⁡θ Δθ≈0.333\cot\theta\,\Delta\theta\approx0.333 (for small angles this is just Δθθ=40′120′=13\frac{\Delta\theta}\theta=\frac{40'}{120'}=\frac13).

Δλλ≈0.125+0.333≈0.46.\frac{\Delta\lambda}\lambda\approx0.125+0.333\approx0.46.

Topics:Errors in measurementDiffraction
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