A thin equiconvex lens of glass of refractive index `mu=3//2` & of focal length `0.3m`in air is sealed into an opeing at one end of a tank filled with water `(mu=4//3)` .Ont he oppostie side of the lens ,a mirror is placed inside the tank on the tank wall perpendicualar tot he lens axis ,as shown in figure .the seperation bertween the lens and the mirror is `0.8m`A small object is placed outside the tank in front of the lens at a distance of `0.9`m form the lens along its axis .Find the position (relative to the lens)of the image of the object fromed byt he stsyem.

Answer

`90cm`from the lens towards right

Explanation

For radius of curvature
`(1)/(f)=(._(a)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2))) rArr (1)/(f)=(._(a)mu_(g)-1)((1)/(R )-(1)/(-R)) rArr (1)/(30)=((3)/(2)-1)xx(2)/(R )`
`R=30cm`
taking refraction from first curved surface
`(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R )rArr (3)/(2v)-(1)/(-90)=(3//2-1)/(30) rArr(3)/(2v)=(1)/(60)-(1)/(90)`
`(3)/(2v)=(3-2)/(180) rArr v= 270cm`
this image will act as object for second surface.`(4)/(3v)-(3)/(2xx270)=(4//3-3//2)/-(30)rArr v=120cm`
this image will act a object for mirror ,u for mirror=`(120-80)=40cm` this image will formed by mirror infornt ata distance of `40cm`.This image wil act as object for curved surface of lens.
`u=-(80-40)cm rArr u=-40cm rArr(3)/(2v)-(4)/(3xx(-40))=(3//2-4//3)/(30)`
`(3)/(2v)+(1)/(30)=(1)/(180)rArr (1)/(180)-(1)/(30) rArr (3)/(2v)=(1-6)/(180) rArr (3)/(2v)=(-5)/(180)`
`(3)/(2v)=(1)/(-36)rArr v=-54cm`
This image will act as object for second refraction.
`(1)/(v)-(3)/(2(-54))=(1-3//2)/(-30) rArr (1)/(v)+(1)/(36)=(1)/(60)rArr (1)/(v)=(1)/(60)-(1)/(36)`
`(1)/(v)=(3-5)/(180)rArr v=(180)/(-2)`
`v=-90cm`
So image wil formed at `90cm`towards right of lens.