A fly `F` is sitting an a glass `S 45 cm` thick `&` of refractive index `3//2`. The slab covers the top of a container `C` containing water `(R.I.4//3)` upto a height of `20 cm`. Bottom of container is closed by a concave mirror `M` fo radius of curvature `40 cm`. Lacate the final image formed by all refractions `&` reflection assuming paraxial rays.

Answer

`(45)/(2)` cm below F

Explanation

`x=45((4)/(3)) xx (2)/(3)= 40 cm`
So total distance = 60 cm

f = -20 cm
`(1)/(v) +(1)/(u) = (1)/(f)`
`implies (1)/(v) + (1)/((-60)) = (1)/(-20)`
`implies (1)/(v) = (1)/(60) - (1)/(20)= (-40)/(60 xx 20) implies v = -30`
So image after reflection from mirror will be at d=30 cm. So 10 cm above 'A'
After refraction at A : It will be at `(10)/((4//3))(3//2) = (90)/(8) = (45)/(4)` cm
so distance from 'B' layer is `(45 - (45)/(4)) = 45 xx (3)/(4)`
So final image after refraction at 'A' will be at :
`45 xx (3)/(4) xx (2)/(3) = (45)/(2)` from 'A' surface