A convex lens is held `45 cm` above the bottom of an empty tank. The image of a point at the bottom of a tank is formed `36 cm` above the lens. Now, a liquid is poured into the tank to a depth of `40 cm.` It is found that the distance of the image of the same point on the bottom of the tank is `48 cm` above the lens. Find the refractive index of the liquid.

Explanation

`(1)/(f)=(1)/(v)-(1)/(u)=(1)/(36)-(1)/(-45)=(3+4)/(180)`
`f=20cm`
Shift of object due to liquid `=40(1-(1)/(mu))=40-(40)/(mu)`
`u=-[45-(40-(40)/(mu))]=-5-(40)/(mu)`
`v=48cm,f=20cm`
`(1)/(v)-(1)/(u)=(1)/(f)`
`(1)/(u)=(1)/(v)-(1)/(f)=(1)/(48)-(1)/(20)=(5-12)/(240)`
`u=-(240)/(7)cm`
`u=-5-(40)/(mu)`
`(-240)/(7)=-5-(40)/(mu)`
`(40)/(mu)=-5+(240)/(7)=(205)/(7)`
`mu=(280)/(205)=(56)/(41)`