1.

A Fresnel's biprism is usedto form the interference fringes.The distance between the source and the biprism is 20 cms and that between the biprism and the screen is 80 cm. Iflambda = 6563 Å and the separation between the virtual sources is 3.6 mm, then the fringe width is :

Answer»

`1.82 CM`
`0.182 cm`
`0.0182 cm`
`0.00182 cm`

SOLUTION :d = 3.6 MM = 0.36 cm
`LAMBDA = 6563 Å`
`BETA = (D)/(d)lambda = 0.0182 cm`.


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