Angular width of the central maxima in the Fraunhofer diffraction for λ = 6000 Å is θ0. When the same slit is illuminated by another monochromatic light, the angular width decreases by 30%. The wavelength of this light is,

  1. 6000 Å
  2. 420 Å
  3. 1800 Å
  4. 4200 Å

Answer (Detailed Solution Below)

Option 4 : 4200 Å
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Detailed Solution

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CONCEPT:

The angular width is written as;

\(θ = \frac{2λ}{d}\)

Here we have θ is the angular width, λ is the wavelength and d is the distance.

CALCULATION:

Given: Wavelength, λ = 6000 Å 

As we know,

\(θ = \frac{2λ}{d}\)    ----(1)

and When the same slit is illuminated by another monochromatic light, the angular width decreases.

\(θ' = \frac{2λ'}{d}\)

\(θ (1- \frac{30}{100})= \frac{2λ'}{d}\)   ----(2)

Now, on dividing the equation (1) and (2) we have;

\(\frac{θ}{θ (1- \frac{30}{100})}= \frac{\frac{2λ}{d}}{\frac{2λ'}{d}}\)

\(\frac{10}{7} = \frac{λ}{λ'}\)

⇒ \(λ' = \frac{7 \times 6000}{10}\)

⇒ λ' = 4200  Å 

Hence, option 4) is the correct answer.

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