1.

The distance between the slit and biprism and screen and biprism are 50 cm each.The obtuse angle of biprism is 179^(@) and its refractive index is 1.5. If the distance between successive fringes is 0.135 mm, the wavelength of light used is :

Answer»

`5893 Å`
`11786 Å`
`2946 Å`
`6574 Å`

SOLUTION :`alpha + 179 + alpha = 180^(@)`
`alpha = (0.5 pi)/(180^(@))rad`
`d = 2a(MU - 1) alpha`

SOLVING , `d = (0.25 pi)/(180) m`
`Rightarrowbeta = (D.lambda)/(d) = 5834 Å`
Then `lambda = 11786 Å`


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