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A source of light of wavelength 5000 Å is placed at one end of a mirror of 200 cm long and 5 mm above its reflecting surface. Find the fringe width of the interference pattern formed on the screen.

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0.04 cm

0.01 cm

0.02 cm

0.03 cm

answer is A.

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Detailed Solution

$\begin{array}{l} B = \frac{λD}{d^{1}} = \frac{5 \times 10^{- 7} \times 2}{10 \times 10^{- 3}} \\ here d^{1} = 2 d \\ = 10^{- 4} m \\ = 0.01 cm \end{array}$

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