A thin equi-convex glass lens (refractive index = 1.5) is being placed on the top of a vessel of height h = 20 cm as shown in the figure. A luminous point source is being placed at the bottom of the vessel on the principle axis of the lens. When air is on both sides of the lens, the image of luminous point source is found at a distance 20 cm from the lens outside the vessel. When the air inside the vessel is being replaced by a source is fomed at a distance 30 cm from the lens outside the vessel. The value of `mu` is

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`