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In Young's double slit experiment carried out with wavelength `lamda=5000`Å, the distance between the slits is 0.2 mm and the screen is 2 m away from the slits. The central maxima is at n=0. the third maxima will be at a distance x (from central maxima) is equal to

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The correct Answer is:
D
For third maxima, `x=(3Dlamda)/(d)`
`=(3xx2xx5000xx10^(-10))/(0.2xx10^(-3))`
`=(6xx5xx10^(-7))/(2xx10^(-4))`
`=15mm=1.5`cm
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