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JEE Advanced · Physics · 25. Wave Optics

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

  1. A 94.5
  2. B 95.6
  3. C 94.2
  4. D 97.6
Verified Solution

Answer & Solution

Correct Answer

(A) 94.5

Step-by-step Solution

Detailed explanation

Original question asked about absolute error which can be both minimum or maximum, we added word maximum to make it clear for students
\(\mathrm{b}=\frac{\mathrm{m} \lambda \mathrm{D}}{\mathrm{d}}=360 \mu \mathrm{~m} \)
\( \mathrm{~b}_{\max }=\frac{3 \times 600 \times 10^{-3} \times 1.01}{4 \times 10^{-3}} \mu \mathrm{~m}\) \(=454.5 \mu \mathrm{~m} \)
\( \mathrm{~b}_{\min }=\frac{3 \times 600 \times 10^{-3} \times 0.99}{6 \times 10^{-3}} \mu \mathrm{~m}=297 \mu \mathrm{~m}\)
Maximum value of b gives error, \(\Delta \mathrm{b}_1=94.5 \mu \mathrm{~m}\)
Minimum value of \(b\) gives error, \(\Delta b_2=63 \mu \mathrm{~m}\)
\(\therefore\) We always report the largest error, hence correct answer should be \(94.5 \mu \mathrm{~m}\)
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