A parallel beam of monochromatic light of wavelength $6000 ˚ A$ is incident normally on a single narrow slit of width $0.003 mm$ . The light is focused by a convex lens on a screen placed on the focal plane. The second maximum will be formed for the angle of diffraction equal to,

30^{\circ}

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15^{\circ}

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0^{\circ}

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60^{\circ}

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Solution

The correct option is A $30^{\circ}$
For the second maximum, $a sin θ = \frac{5 λ}{2}$

$\Rightarrow sin θ = \frac{5 λ}{2 a}$

$= (\frac{5 \times 6000 \times 10^{- 10}}{2 \times 0.003 \times 10^{- 3}}) = 0.5$

$∴ θ = 30^{\circ}$

Hence, $(C)$ is the correct answer.

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A parallel beam of monochromatic light of wavelength 6000 ˚A is incident normally on a single narrow slit of width 0.003 mm. The light is focused by a convex lens on a screen placed on the focal plane. The second maximum will be formed for the angle of diffraction equal to,

The correct option is A 30∘For the second maximum, asinθ=5λ2 ⇒sin θ=5λ2a =(5×6000×10−102×0.003×10−3)=0.5 ∴ θ=30∘ Hence, (C) is the correct answer.

A parallel beam of monochromatic light of wavelength $6000 ˚ A$ is incident normally on a single narrow slit of width $0.003 mm$ . The light is focused by a convex lens on a screen placed on the focal plane. The second maximum will be formed for the angle of diffraction equal to,

The correct option is A $30^{\circ}$
For the second maximum, $a sin θ = \frac{5 λ}{2}$

$\Rightarrow sin θ = \frac{5 λ}{2 a}$

$= (\frac{5 \times 6000 \times 10^{- 10}}{2 \times 0.003 \times 10^{- 3}}) = 0.5$

$∴ θ = 30^{\circ}$

Hence, $(C)$ is the correct answer.