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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Physics
Chapter
Wave Optics

JEE Advanced 2025 · Paper 1 · Physics · Wave Optics

A single slit diffraction experiment is performed to determine the...

A single slit diffraction experiment is performed to determine the slit width using the equation, bdD=mλ\frac{b d}{D}=m \lambda, where bb is the slit width, DD the shortest distance between the slit and the screen, dd the distance between the mth m^{\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 maximum absolute error (in μm\mu \mathrm{m}) in the value of bb estimated using the diffraction maximum that occurs for m=3m=3 with d=5 mmd=5 \mathrm{~mm} and D=1 mD=1 \mathrm{~m} is ______.

Correct answer

0

Step-by-step solution

Original question asked about absolute error which can be both minimum or maximum, we added word maximum to make it clear for students b=mλDd=360μ m bmax⁡=3×600×10−3×1.014×10−3μ m=454.5μ m bmin⁡=3×600×10−3×0.996×10−3μ m=297μ m\begin{aligned} & \mathrm{b}=\frac{\mathrm{m} \lambda \mathrm{D}}{\mathrm{d}}=360 \mu \mathrm{~m} \\ & \mathrm{~b}_{\max }=\frac{3 \times 600 \times 10^{-3} \times 1.01}{4 \times 10^{-3}} \mu \mathrm{~m}=454.5 \mu \mathrm{~m} \\ & \mathrm{~b}_{\min }=\frac{3 \times 600 \times 10^{-3} \times 0.99}{6 \times 10^{-3}} \mu \mathrm{~m}=297 \mu \mathrm{~m}\end{aligned} Maximum value of b gives error, Δb1=94.5μ m\Delta \mathrm{b}_1=94.5 \mu \mathrm{~m} Minimum value of bb gives error, Δb2=63μ m∴\Delta b_2=63 \mu \mathrm{~m} \therefore We always report the largest error, hence correct answer should be 94.5μ m94.5 \mu \mathrm{~m}
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