In Young's double slit experiment the slits are 0.5 mm apart and the interference is observed on a screen at a distance of 100 cm from the slit. It is found that the 9th bright fringe is at a distance of 7.5 mm from the second dark fringe from the centre of the fringe pattern. The wavelength of the light used is

\frac{2500}{7} \circ A

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2500 \circ A

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5000 \circ A

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\frac{5000}{7} \circ A

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Solution

The correct option is C $5000 \circ A$
$9 β - 1.5 β = 7.5 m m 7.5 β = 7.5 m m \Rightarrow β = 1 m m \frac{λ D}{d} = 10^{- 3} \Rightarrow= \frac{0.5 \times 10^{- 3} \times 10^{3}}{1} = 5000 \circ A (c)$

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The correct option is C 5000∘A9β−1.5β=7.5mm7.5β=7.5mm⇒β=1mmλDd=10−3⇒=0.5×10−3×1031=5000∘A(c)

The correct option is C $5000 \circ A$
$9 β - 1.5 β = 7.5 m m 7.5 β = 7.5 m m \Rightarrow β = 1 m m \frac{λ D}{d} = 10^{- 3} \Rightarrow= \frac{0.5 \times 10^{- 3} \times 10^{3}}{1} = 5000 \circ A (c)$