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In a Young's double - slit experiment the slit separation is 0.5 \mathrm{~mm} and the screen is 0.5 \mathrm{~m} from the slit. For a monochromatic light of wavelength 500 \mathrm{~nm} the distance of third maxima from the second minima on the other side is:

Option: 1

2.75 mm


Option: 2

2.5 mm


Option: 3

22.5 mm


Option: 4

2.25 mm


Answers (1)

best_answer

Here, \mathrm{\mathrm{d}=0.5 \mathrm{~mm}=0.5 \times 10^{-3} \mathrm{~m}}

\mathrm{ \mathrm{D}=0.5 \mathrm{~m}, \lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m} }

The distance of third maxima from the second minima on the other side is

\mathrm{ \begin{aligned} & =\frac{9}{2} \beta \text { (where } \beta \text { is the fringe width) } \\ & =\frac{9}{2} \frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{9 \times 500 \times 10^{-9} \times 0.5}{2 \times 0.5 \times 10^{-3}}=2.25 \times 10^{-3} \mathrm{~m}=2.25 \mathrm{~mm} \end{aligned} }

Posted by

HARSH KANKARIA

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