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In a Young's double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining 10th maxima of double slit within the central maximum of single slit pattern? [CBSE AIPMT 2015J

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0.2 mm

0.1 mm

0.5 mm

0.02 mm

answer is A.

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Detailed Solution

Given, d = 1 mm = I x 10^-3m, D = 1m

$λ$ = 500 nm = 5 x 10^-7 m

As, width of central maximum = width of 10th maxima

$∴ \frac{2 D λ}{a} = 10 (\frac{λ D}{d})$

$\Rightarrow a = \frac{d}{5} = \frac{10^{- 3}}{5} = 0.2 \times 10^{- 3} m$

a = 0.2 mm

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