1.

Consider a two slit interferennce arrangements such that the distance of the screen from the slits is half the distance between the slits.That the first minima on the screen falls at a distance D from the centre O, then D is

Answer»

`(lambda)/(3)`
`(lambda)/(2)`
`(lambda)/(2 sqrt 5)`
`(lambda)/(sqrt3)`

Solution :`T_(2)P = D+ x, T_(1) P = D -x`
`S_(1)P = sqrt((s_(1)T_(1))^(2) + (PT_(1))^(2))`
=`[D^(2) + (D - x)^(2)]1/2`
`S_(2)P = [D^(2) + (D + x)^(2)]1/2`
MINIMA will occur when
`[D^(2) + (D + x)^(2)]1/2 - [D^(2) + (D - x)^(2)]1/2 = (lambda)/(2)`
If x = D
`[D^(2) + 4D^(2)]1/2 = (lambda)/(2)`
`(5D^(2))1/2 = (lambda)/(2), therefore D = (lambda)/(2 sqrt 5)`


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