Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of a slit of width 12×10−3 cm when the slit is illuminated by monochromatic light of wavelength 6000 A0.

Half angular width of the central maximum (θ), ⇒sinθ= λa Put a=12×10−3cm, λ=6000A0=6×10−5cm ∴sinθ=λa=6×10−512×10−5=12 ⇒θ=300

Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of a slit of width 12×10−3 cm when the slit is illuminated by monochromatic light of wavelength 6000 A0.

Half angular width of the central maximum (θ), ⇒sinθ= λa Put a=12×10−3cm, λ=6000A0=6×10−5cm ∴sinθ=λa=6×10−512×10−5=12 ⇒θ=300

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Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of a slit of width $12 \times 10^{- 3}$ cm when the slit is illuminated by monochromatic light of wavelength 6000 $\overset{0}{A}$ .

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$45^{0}$

$90^{0}$

$30^{0}$

$180^{0}$

answer is C.

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Detailed Solution

$H a l f a n g u l a r w i d t h o f t h e c e n t r a l m a x i m u m (θ), \Rightarrow \sin θ = \frac{λ}{a} P u t a = 12 \times 10^{- 3} c m, λ = 6000 \overset{0}{A} = 6 \times 10^{- 5} c m ∴ \sin θ = \frac{λ}{a} = \frac{6 \times 10^{- 5}}{12 \times 10^{- 5}} = \frac{1}{2} \Rightarrow θ = 30^{0}$