A thin equiconvex glass lens `(mu_(g)=1.5)` is beign placed on the top of a vessel of height `h=20cm` as shown figure. A luminous point source is beign placed at the bottom of vessel on the principal axis of the lens. When the air is on both the side of the lens the image of luminous source is formed at a distance of 20cm from the lens out side the vessel. When the air inside the vessel is being replaced by a liquid of refractive index `mu_(l)` the image of the same source is being formed at a distance 30cm from the lens outside the vessel. Find the `mu_(l)`.

Answer

1.11

Explanation

For the first lens,
`(1)/(v)-(1)/(u)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))`
`(1)/(20)-(1)/(-20)=(1.5-1)((1)/(R)-(1)/(-R))rArrR=10cm`
For the second case when the vessel is being fillle dwith the liquid then
`(mu_(air))/(v_(1))-(mu_(l))/(u)=(u_(g)-mu_(l))/(R)+(mu_(air)-mu_(g))/(-R)`
`(1)/(30)-(mu_(l))/(-20)=(1.5-mu_(l))/(10)+(1-1.5)/(-10)`
`(mu_(l))/(20)+(mu_(l))/(10)=(3)/(20)+(1)/(20)-(1)/(30)`
`(3mu_(l))/(20)=(1)/(6)`
`mu_(l)=(10)/(9)=1.11`