JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Wave Optics

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

]

Answer: 0.45

Solution

For first minimum:

d sin θ = λ

, so

λ = d sin θ

.

Taking logarithm and differentiating:

\frac{d λ}{λ} = \frac{Δ d}{d} + \frac{c o s θ}{s i n θ} d θ

.

With

Δ d = 0.002

mm,

d = 0.016

mm,

θ = 2°

d θ = 4 0^{'} = \frac{2}{3} \times \frac{π}{180} = \frac{π}{270}

rad:

\frac{d λ}{λ} = \frac{0.002}{0.016} + \frac{1 - s i n ^{2} ( 2° )}{s i n ( 2° )} \times \frac{π}{270} = 0.125 + \frac{1 - 0.03 5 ^{2}}{0.035} \times \frac{π}{270} \approx 0.125 + 28.55 \times 0.0116 \approx 0.125 + 0.332 = 0.457 \approx 0.45

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer

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JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Wave Optics

Solution

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