1.

The distance between the tew slits in a Young's double slit experiment is d and the distance of the screen from the plane of the slits is b, P is a point on the screen directly infront of one of the slits. The path difference between the waves arriving at P from the two slits is

Answer»

`(d^(2))/(b)`
`(d^(2))/(2b)`
`(2d^(2))/(b)`
`(d^(2))/(4B)`

SOLUTION :Distance between 2 slit(b) is dDistance from screen from plane (D) is BWHEN wave will reach at POINT P on the screen then path difference isPath difference=Dyd​Here y is slit distance from Centre Therefore y=d/2Substitute all value in above equationwe GET d^2/2b


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