1.

In a Young' sdouble slit experiment the slit separation is 0.5 m and distance of screen is 5m. For a monochromatic light of wavelength 500nm, the distance of 3^(rd) maxima from 2^(nd) minima on the other side is :

Answer»

`2.5 mm`
`2.25 mm`
`2.75 mm`
`22.5 mm`

Solution :`y _(n1)(max) = (n_(1)lambda D)/(d) , n_(1) = 3`
`y_(N2)(min) = (n_(2) - 1)/(2)(lambda D)/(d), n_(2) = 2`
`therefore y_(n1) + y_(n2) = (n_(1) + n_(2) - 1/2)(lambdaD)/(d)`
` = ( 3 + 2 - 1/2)(lambda D)/(d)`
` = (9)/(2) xx (lambdaD)/(d)`
`(9)/(2) xx (500 xx 10^(-6) xx 5)/(0.5)`
` = 22.5 xx 10^(-3)m`.


Discussion

No Comment Found

Related InterviewSolutions