In a Young's double - slit experience, the slits are 2mm apart and ar

Question

In a Young's double - slit experience, the slits are 2mm apart and are illuminated with a mixture of two wavelengths

λ_{0} = 750 n m

and

λ = 900 n m

. The minimum distance from the common central bright fringe on a screen 2 m away from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other, is

Options:

A . 1.5mm

B . 3 mm

C . 4.5 mm

D . 6 mm

Answer: Option C
:
C

y_{0} = n [\frac{D λ}{d}]

and

y_{n}^{'} = n^{'} (\frac{D λ^{'}}{d})

Equating

y_{n}

and

y_{n}^{'}

,we get

\frac{n}{n^{'}} = \frac{λ^{'}}{λ} = \frac{900}{750} = \frac{6}{5}

Hence, the first position at which overlapping occurs is

y_{6} = y_{6} = \frac{6 (2) (750 \times 10^{- 9})}{2 \times 10^{- 3}} = 4.5 m m

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