A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm . The aperture is illuminated normally by a pa - Physics MCQ Physics | ExamTest.live

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A linear aperture whose width is is placed immediately in front of a lens of focal length . The aperture is illuminated normally by a parallel beam of wavelength . The distance of the first dark band of the diffraction pattern from the centre of the screen is:

A $0.15 cm$
B $0.10 cm$
C $0.25 cm$
D $0.20 cm$

Solution & Step-by-step Explanation

Shutterstock Explore For the first dark band (minimum), a sin θ = nλ = 1 λ .For small angles, sin θ \approx \frac{y}{D} . Here D = f = 60 cm . a \cdot \frac{y}{D} = λ y = \frac{λ D}{a} = \frac{( 5 \times 1 0 ^{- 5} cm ) ( 60 cm )}{0.02 cm} y = \frac{300 \times 1 0 ^{- 5}}{2 \times 1 0 ^{- 2}} = 150 \times 1 0^{- 3} = 0.15 cm .

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A linear aperture whose width is is placed immediately in front of a lens of focal length . The aperture is illuminated normally by a parallel beam of wavelength . The distance of the first dark band of the diffraction pattern from the centre of the screen is:

0.15 cm

0.10 cm

0.25 cm

0.20 cm

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A linear aperture whose width is 0.02cm is placed immediately in front of a lens of focal length 60cm. The aperture is illuminated normally by a parallel beam of wavelength 5×10−5cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is:

Solution & Step-by-step Explanation

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