COMEDK · Physics · 26. Wave Optics
A source of light of wavelength \(5000 Å\) is placed at one end of a table \(2 \mathrm{~m}\) long and \(5 \mathrm{~mm}\) above its flat well polished top. The fringe width of the interference bands seen on a screen located at the end of the table is
- A \(2 \times 10^{-5} \mathrm{~m}\)
- B \(2 \times 10^{-4} \mathrm{~m}\)
- C \(2 \times 10^{-3} \mathrm{~m}\)
- D \(2 \times 10^{-2} \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(2 \times 10^{-4} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Given, wavelength of light used, \[ \begin{aligned} \lambda &=5000 Å=5 \times 10^{-7} \mathrm{~m} \\ D &=2 \mathrm{~m} \\ d &=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m} \end{aligned} \] \(\therefore\) Fringe width, \(\beta=\frac{D \lambda}{d}\)…
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