In the diffraction pattern, the first maximum is at 3 0 ∘ , when a monochromatic light of wavelength λ is incident on a slit of width a . For the same wavelength, if the slit width is changed, so that

Question

In the diffraction pattern, the first maximum is at 3 0 ∘ , when a monochromatic light of wavelength λ is incident on a slit of width a . For the same wavelength, if the slit width is changed, so that first maximum is at 4 5 ∘ . The slit width is changed by ( sin 3 0 ∘ = 2 1 ​ , sin 4 5 ∘ = 2 ​ 1 ​ )

TestHub · Accepted Answer

The condition for the first secondary maximum in a single slit diffraction pattern is given by:

$a sin θ = \frac{3 λ}{2}$

For the initial case, $θ_{1} = 3 0^{\circ}$ :

$a_{1} sin 3 0^{\circ} = \frac{3 λ}{2}$

$a_{1} (\frac{1}{2}) = \frac{3 λ}{2} \Rightarrow a_{1} = 3 λ$

For the final case, $θ_{2} = 4 5^{\circ}$ :

$a_{2} sin 4 5^{\circ} = \frac{3 λ}{2}$

$a_{2} (\frac{1}{2}) = \frac{3 λ}{2} \Rightarrow a_{2} = \frac{3 2 λ}{2} = \frac{3 λ}{2}$

The change in slit width is:

$Δ a = a_{1} - a_{2} = 3 λ - \frac{3 λ}{2} = 3 (\frac{2 - 1}{2}) λ$

Answer: $3 (\frac{2 - 1}{2}) λ$

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