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The width of a slit is 0.012 mm. Monochromatic light is incident on it. The angular position of first bright line is 5.2°. The wavelength of incident light is ______. [sin 5.2° = 0.0906]. -

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Question

The width of a slit is 0.012 mm. Monochromatic light is incident on it. The angular position of first bright line is 5.2°. The wavelength of incident light is ______. [sin 5.2° = 0.0906].

Options

  • 6040 Å

  • 4026 Å

  • 5890 Å

  • 7248 Å

MCQ
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Solution

The width of a slit is 0.012 mm. Monochromatic light is incident on it. The angular position of first bright line is 5.2°. The wavelength of incident light is 7248 Å. [sin 5.2° = 0.0906].

Explanation:

It is a unique case of a single-slit Fraunhoffer diffraction. Therefore, for bright fringe, where an is the slit's width.

a sin θ = (2n + 1)`lambda/2`

λ = `(2"a" sin theta)/(2"n"+1)`

= `(2xx1.2xx10^-5xx0.0906)/(2xx1+1)`

= 7248 Å

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