1.

A double slit experiment produces interference fringes for sodium light (lambda = 5890 Å) that are 0.20^(@) apart What will be angular fringe separation if the entire arrangement is immersed in water (mu = 4/3)?

Answer»

`1.25^(@)`
`0.30^(@)`
`0.15^(@)`
`0.45^(@)`

SOLUTION :`beta = (lambda D)/(d)` (In air) and `THETA = (beta)/(D)`
`THEREFORE theta D = (lambda_(0)D))/(d)` or `theta = (lambda_(0))/(d)`
Now `MU = (lambda_(0))/(lambda)`
`thereforetheta. = (lambda)/(d) ("In water") & (theta.)/(theta) = (lambda)/(d) xx (d)/(lambda_(0)) = (lambda)/(lambda_(0))`
But `(lambda)/(lambda_(0)) = (1)/(mu)`
`therefore theta.= (theta)/(mu) = 0.15^(@)`.


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