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Physics Question on Youngs double slit experiment

Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5?10−7 m6.5 ? 10^{-7}\, m. The interference fringes are observed on a screen placed 1 m from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to

A

0.65 mm

B

1.63 mm

C

3.25 mm

D

4.88 mm

Answer

1.63 mm

Explanation

Solution

Here, d=1 mm=10−3 m,λ=6.5×10−5 md=1\,mm=10^{-3}\,m, \lambda=6.5\times10^{-5}\,m D=1 mD=1\,m x5=nλDd=5×6.5×10−7×110−3=32.5×10−5 mx_{5}=n\lambda \frac{D}{d}=5\times6.5\times10^{-7}\times\frac{1}{10^{-3}}=32.5\times10^{-5}\,m x3=(2n−1)λ2Dd=(2×3−1)×6.5×10−7×12×10−3x_{3}=\left(2n-1\right) \frac{\lambda}{2} \frac{D}{d}=\frac{\left(2\times3-1\right)\times6.5\times10^{-7}\times1}{2\times10^{-3}} =16.25×10−4 m=16.25\times10^{-4}\,m x5−x3=(32.5−16.25)×10−4 m=16.25×10−4m=1.63mmx_{5}-x_{3}=\left(32.5-16.25\right) \times10^{-4}\,m =16.25\times10^{-4}m=1.63mm