Skip to content
dronacharya.ai
← JEE Advanced 2025 · Paper 1
PhysicsNumerical+4Question 28 of 48

JEE Advanced 2025 · Paper 1, Physics question 12

A single slit diffraction experiment is performed to determine the slit width using the equation, bdD=mλ\dfrac{bd}{D} = m\lambda, where bb is the slit width, DD the shortest distance between the slit and the screen, dd the distance between the mthm^{\text{th}} diffraction maximum and the central maximum, and λ\lambda is the wavelength. DD and dd are measured with scales of least count of 1 cm and 1 mm, respectively. The values of λ\lambda and mm are known precisely to be 600 nm and 3, respectively. The absolute error (in μ\mum) in the value of bb estimated using the diffraction maximum that occurs for m=3m = 3 with d=5d = 5 mm and D=1D = 1 m is _____.

Answer and worked solution

Official answer: 75

Slit width.
b=mλDd=3×600×10−9×15×10−3=3.6×10−4 m=360 μm.b=\frac{m\lambda D}{d}=\frac{3\times600\times10^{-9}\times1}{5\times10^{-3}}=3.6\times10^{-4}\ \text{m}=360\ \mu\text{m}.

Error. Taking each least count as the absolute error:
Δbb=ΔDD+Δdd=0.011+15=0.21,\frac{\Delta b}{b}=\frac{\Delta D}{D}+\frac{\Delta d}{d}=\frac{0.01}{1}+\frac{1}{5}=0.21,
so Δb=0.21×360≈75.6 μ\Delta b=0.21\times360\approx75.6\ \mum.

The key also accepts 94-95.