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Width of the principal maximum on a screen at a distance of 50 cm from the slit having width 0.02 cm is 312.5 × 10 - 3 c m . If waves were incident normally on the slit, then wavelength of the light from the source will be

Options

  1. A6000 Å
  2. B6250 Å
  3. C6400 Å
  4. D6525 Å

Correct answer

B. 6250 Å

Step-by-step solution

Width of central max. = 2 λ D a λ = 0.2 × 10 - 2 × 312.5 × 10 - 3 × 10 - 2 2 × 1 2 m = 6250 × 10 - 10 m = 6250 Å

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