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In Young's double slit experiment, wavelength \mathrm{\lambda=5000\, \AA}, the distance between the slits is 0.4 \mathrm{~mm} and the screen is at 200 \mathrm{~cm} from the slits. The central maximum is at \mathrm{x}=0. The third maximum (taking the central maximum as zeroth maximum) will be at x equal to:

Option: 1

1.67 cm


Option: 2

0.75 cm


Option: 3

0.5 cm


Option: 4

5.0 cm


Answers (1)

best_answer

\mathrm{\text { Position of third maxima, } \mathrm{x}_3=\frac{3 \lambda \mathrm{D}}{\mathrm{d}}=\frac{3\left(5 \times 10^{-5}\right)(200)}{0.04} \mathrm{~cm}=0.75 \mathrm{~cm}}

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Ajit Kumar Dubey

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