A thin biconvex lens of refractive index `3//2` is placed on a horizontal plane mirror as shown in Figure . The space between the lens and the mirror is then fille with water or refractive index `4//3`. It is found that when a point object is placed 15 cm above the lens on its principal axis, the object coincides with its own image. On representing with another liquid, the object and the image again coincide at a distance 25 cm from the lens. Calculate the refractive index of the liquid.

Explanation

`RI` of lens material `=mu_(2)` , `RI` of first liquid `=mu_(3)`, `RI` of air `=mu_(1)`, `RI` of second liquid `=mu'_(3)`
In general `(mu_(3)//v)-(mu_(1)//u)={(mu_(2)-mu_(1))//R}+{(mu_(3)-mu_(2))//R}`
Here `u= -15cm`, `v=oo`
Now, `1//(+15)=[(3//2)-1}//R]-[{(4//3)-(3//2)//R]`
`rArrR=5cm`
For the second liquid, similarly,
`(1//25)=(0.5//R)-[{mu'_(3)-(3//2)}//R]`
`rArr mu'_(3)=1.8`