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 souces ina certain position .One of them coincides with the source and the other is at distance of `l=30cm`from source.find the possible value(s) refractive index `mu`of the liquid.

Answer

`1.5` or`sqrt(5)-1)`

Explanation

Case I:When real Image due to mirror alone consides with object .Since both lens and mirror are conversing so Imag due to combination will shift towards mirror.
`f_(m)=(R )/(2) rArr f_(m)=-30cm`
`P_(m)=(100)/(-f_(m)) rArr P_(m)=(10)/(3)D`
For combination applying mirror formula
`(1)/(v)+(1)/(u)=(1)/(f)`
`(1)/(-30)+(1)/(-60)=(1)/(f_(eq)) rArr (1)/(f_(eq))=(-2-1)/(60)`
`P_(eq)=2P_(l)+P_(m) rArr 5=2P_(l)+(10)/(3)`
`2P_(l)=5-(10)/(3) rArr P_(l)=(5)/(6)rArr f_(l)=(100)/(P_(l)) rArr f_(l)=(100)/(5)xx6`
`f_(l)=120cm rArr (1)/(f_(l))=(mu-1)((1)/(R_(1))-(1)/(R_(2)))`
` (1)/(120)=(mu-1)((1)/(oo)-(1)/(-60)) rArr mu-1=(1)/(2)`
`mu=(3)/(2)`
Case(II) :Real image due to combination conside with object.Now image due to mirror will be away from objects.

`f_(eq)=-(x)/(2) rArr P_(eq)=100/(-f_(eq))`
`P_(eq)=(200)/(x)D....(1)`
Applying mirror formula due to mirror alone
` (1)/(v)+(1)/(u)=(1)/(f)` gives that `f_(m)=(R )/(2)=-30cm`
`(1)/(-(x+30))+(1)/(-x)=(1)/(-30) rArr (x+(x+30))/(x(x+30))=(1)/(30)`
`30(2x+30)=x^(2)+30x rArr x^(2)-30x-900=0`
`x=(30+-sqrt(900+3600))/(2) rArr x=(30+-30sqrt(5))/(2)`
`x=-15(sqrt(5)+1).....(2)`
from eq.(1)&(2) `P_(eq)=(200)/(x)D rArr P_(eq)=(200)/(15(sqrt(5)+1))`
`P_(eq)=(40)/(3(sqrt(5+1))) rArr P_(eq)=(10sqrt(5)-1)/(3)`
`P_(eq)=2P_(l)+P_(m) rArr (10)/(3)(sqrt(5)-1)=2P_(l)+(10)/(3)`
`P_(l)=(5)/(3)(sqrt(5)-2) rArr ((mu-1))/(0.6) =(5)/(3)(sqrt(5)-2)`
`(5)/(3)(mu-1)=(5)/(3)(sqrt(5)-2) rArr mu-1=sqrt(5)-2`
` mu=sqrt(5)-1`