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Q.

In young’s double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890 $\overset{o}{A}$ is:

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a

${1178×10}^{- 9} m$

b

$1178 \times 10^{- 6} m$

c

$1178 \times 10^{- 12} m$

d

$5890 \times 10^{- 7} m$

answer is B.

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Detailed Solution

$d = 0 .5 mm = 0 .5 \times 10^{- 3} m and D = 0.5 m Fringe width β = \frac{λD}{d} = \frac{5890 \times 10^{- 10} \times 0 .5}{0 .5 \times 10^{- 3}} = 589 \times 10^{- 6} m$
Hence, distance between the first and third bright fringe
$\begin{array}{l} = 2 β = 2 \times 589 \times 10^{- 6} m \\ = 1178 \times 10^{- 6} m \end{array}$

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