1.

The minimum distance from centre of zero order maxima to where the intensity in half that at the centre is

Answer»

`0.15 mm`
`0.20 mm`
`0.30 mm`
`0.40 mm`

Solution :`I = a^(2) + b^(2) + 2ab COS theta , a = b`
`therefore I = 2a^(2) (1 + cos theta) = 4a^(2) cos^(2)(theta)/(2)`
`Rightarrow I = I_(0)cos^(2)(theta)/(2)`
`therefore cos^(2) (theta)/(2) = (I)/(I_(0)) = 1/2`,"`cos theta/2 = 1/sqrt2`
`therefore theta = 90^(@) = (pi)/(2)` radian.
`therefore` path difference `x = (pi)/(2) xx (lambda)/(2pi) = (lambda)/(4)`
DISTANCE of point from CENTRE = `(lambda )/(4).(D)/(2d)`
` = 1/4(Dlambda)/(2d) = 1/4 xx 6 mm = 0.15 mm`


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