A concave mirror has the form of a hemisphere with a radius of `R=60cm` A thin layer of an unknown transparent liquid is poured into the mirror ,The mirror-liquids system forms one real Image and another real image is formed by mirror alone ,with the sources in a certain position .One of them coincides with the source and the other is at distance of `l=30 cm` from source.find the possible value(s) refractive index `mu`of the liquid.

Answer

1.5 or `sqrt(5) - 1`

Explanation

`(1)/(f') = (1)/(u) + (1)/(v) = (1)/(-60) - (1)/(30)`
`implies f' = -(30 xx 60)/(90) implies f' = -20 cm`
`(1)/(f') = (1)/(f_(m)) - (2)/(f_(l)), (1)/(f_(l)) = (mu-1) ((1)/(60) - (1)/(oo))`
`(1)/(-20) = (1)/(-30) - (2)/(60) (mu-1) implies (2(mu-1))/(60) = (1)/(20) - (1)/(30)`
`implies mu-1 = (3)/(2) - 1 implies mu = 1.5`
Case- II

When there is liquid in the mirror u = 2f' & u =f'
when there is no liquid in the mirror
`(1)/(-u) - (1)/(u-30) = -(1)/(30) y implies (u(u-30))/(2u-30) = 30`
`implies u^(2) - 10 x + (30)^(2) = 0`
`u = (90pmsqrt((10)^(2) - 4(30)^(2)))/(2) , u = 45 pm 15sqrt(5)`
But there u is +ve so that
`implies u = 45 + 15 sqrt(5)` & ` u =2f' implies f' = (45)/(2) + (15sqrt(5))/(2)`
`(1)/f') = (1)/(f_(M)) - (2)/(f_(t)) - (1)/((15)/(2) (3 + sqrt(5))) = (1)/(-30) + (2(mu-1))/(60)`
`implies (mu-1) = 1 - (4)/(3+sqrt(5)) xx (3-sqrt(5))/(3-sqrt(5))`
`implies mu-1 = sqrt(5) -2 implies mu-1 = sqrt(5) -1`