One end of a cylindrical rod is grounded to a hemispherical surface of radius`R=20mm`.It is immersed in water`(mu=4//3)`.If the refractive index of the rod is`1.5`and an object is placed in water on the axis at a distance of`10cm`form the pole ,then the position of the image is ................(determine)

Explanation

Using sign convention,
`u=-10cm, R=+0.2cm`
`mu_(1)=1.33,mu_2 =1.50`
Therfore,using
`-(mu_(1))/(u)+(mu_(2))/(v)=(mu_(2)-mu_(1))/(R )`
`(-1.33)/(-10)+(1.5)/(v)=(1.5-1.33)/(0.2)`
`(1.5)/(v)=-(1.33)/(10)+(0.17)/(0.2)`
`=-0.133+0.085`
`=-0.048`
`v=-(1.5)/(0.048)=-31.25cm`
Thus the image is formed at a distance of`31.25cm` from the pole on the axis,inside the water.