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JEE Advanced Physics Wave Optics 2025 JEE Advanced 2025 (Paper 1)

JEE Advanced Physics Question (2025) — Solution

Question

A single slit diffraction experiment is performed to determine the slit width using the equation, b d D =m , where b is the slit width, D the shortest distance between the slit and the screen, d the distance between the m^ th diffraction maximum and the central maximum, and is the wavelength. D and d are measured with scales of least count of 1 cm and 1 mm, respectively. The values of and m are known precisely to be 600 nm and 3, respectively. The maximum absolute error (in m ) in the value of b estimated using the diffraction maximum that occurs for m=3 with d=5 ~mm and D=1 ~m is ______.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

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 aligned & b = m D d =360 ~m \\ & ~b _ = 3 600 10^ -3 1.01 4 10^ -3 ~m =454.5 ~m \\ & ~b _ = 3 600 10^ -3 0.99 6 10^ -3 ~m =297 ~m aligned Maximum value of b gives error, b _1=94.5 ~m Minimum value of b gives error, b_2=63 ~m We always report the largest error, hence correct answer should be 94.5 ~m

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