1.

The new angular fringe separation when apparatus is dipped in water is

Answer»

`0.135^(@)`
`0.12^(@)`
`0.60^(@)`
`0.40^(@)`

Solution :In WATER `d = (LAMBDA)/(THETA)` in AIR
`d = (lambda)./(theta^(1))` in water
`therefore (theta^(1))/(theta)= (lambda.)/(lambda) therefore theta^(1) = (lambda.)/(lambda) theta`
But `(lambda.)/(lambda) = (C_(w))/(C_(a)) = 1/(mu_(w)) = 3/4`
then `theta. = 3/4theta = 0.135^(@)`.


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