There is a spherical glass shell of refractive index 1.5, inner radius 10 cm and outer radius 20 cm. Inside th espherical cavity, there is air. A point object is placed at a point O at a distance of 30cm from the outer spherical surface. Find the final position of th eimage as seen eye.

Answer

`18.75` cm from centre.

Explanation

`(1.5)/(v)-(1)/(-30)=(1.5-1)/(20)rArrv=-180cm`
For second surface, `u_(1)=-190cm`
`rArr (1)/(v)-(1.5)/(-190)=(1-1.5)/(10)rArr v=(190)/(11)`
For third surface, `u=-((190)/(11)+20)=(-410)/(11)`
`rArr(1.5)/(v)-(11)/(410)=(1.5-1)/(-10)=-(1)/(20)`
`rArr v=-(410)/(21)`
For fourth surface , `u=-((410)/(21)+10)=-(620)/(21)cm`
`mu_(1)=1.5, mu_(2)=1,R=-20cm`
`rArr (1)/(v)+(1.5xx21)/(620)=(1-1.5)/(-20)`
`rArr v=-38.75 cm`
The final image is at a distance of `18.75cm` from the center.