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# Help me solve this A linear aperture whose width is 0.02 cm is placed immediately in front of alens of focal length 60 cm. The aperture is illuminated normally by aparallel beam of wavelength 5 x 10-5 cm. The distance of the first darkband of the dif

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
parallel beam of wavelength 5 x 10-5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is

• Option 1)

0.10 cm

• Option 2)

0.25 cm

• Option 3)

0.20 cm

• Option 4)

0.15 cm

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The screen must be placed at a distance of 60cm from lens.

Hence $O=60cm$

$b=0.02cm$

$\lambda = 5\times 10^{-5e}cm$
For first d ark b and b$\sin \theta =\frac{\lambda }{b} = \tan \theta - (1)$

$\tan \theta = \frac{y}{D}=\frac{\lambda }{b}$

$y-\frac{\lambda D}{b} = \frac{5\times 10^{-5}cm\times 0.6m}{0.02cm}$

= 1.5 mm or 0.15cm

Option 1)

0.10 cm

This solution is incorrect.

Option 2)

0.25 cm

This solution is incorrect.

Option 3)

0.20 cm

This solution is incorrect.

Option 4)

0.15 cm

This solution is correct.

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