1.

What is the relation between the refractive indices` mu,mu_(1) ` and `mu_(2)` if the behaviour of light rays is shown in Figure.

Answer» In Fig., the focal length `f` of the lens is given by
`(1)/(f) = ((mu)/(mu_(1) - 1))((1)/(R_(1)) - (1)/(R_(2)))`
As the rays pass through the lens without any deviation, therefore, `f = oo`
`:. ((mu)/(mu_(1) -1))(1)/(R_(1) - (1)/(R_(2))) = (1)/(oo) = 0`
For a double convex lens, `((1)/(R_(1)) - (1)/(R_(2))) is +`
`:. (mu)/(mu_(1)) = 1 = 0 or (mu)/(mu_(1)) = 1 or mu = mu_(1)`
In Fig.(b),
`(1)/(f) = ((mu)/(mu_(2)) - 1)((1)/(R_(1)) - (1)/(R_(2)))`
As the lens behave as a diverging lens , therefore `f` is negative
As the factor `((1)/(R_(1)) - (1)/(R_(2)))` is positive,
`:. ((mu)/(mu_(1)) - 1)` is negative
i.e., `((mu)/(mu_(2)) - 1) lt 0 or (mu)/(mu_(2)) lt 1 or mu lt mu_(2)`


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