A convergent beam is incident on two slabs placed in contact as shownin figure. Where will the rays finally converge?

Explanation

A ray of light from the object undergoes refraction at three interfaces: (1) Air-medium A, (2) Medium A-mudium B, (3) Medium B-air. The coordinate system for each of the interface is shown in the following figure.
Air-medium A interface:
`d_(1)=+14cm, mu_(1)=1, mu_(2)=1.5`
As `(mu_(1))/(d_(1))=(mu_(2))/(d_(2))` or `d_(2)=(mu_(2))/(mu_(1))d_(1)` , we get
`d_(2)=+21 cm`
Medium A-medium B interface:
`d_(1)=(21-6)=15 cm, mu_(1)=1.5, mu_(2)=2`
As `(mu_(1))/(d_(1))=(mu_(2))/(d_(2))` or ` d_(2)=(mu_(2))/(mu_(1))d_(1)` , we get
`d_(2)=+20cm`
Medium B-air interface:
`d_(1)=+(20-4)=+16cm, mu_(1)=2, mu_(2)=1`
As `(mu_(1))/(d_(1))=(mu_(2))/(d_(2))` or `d_(2)=(mu_(2))/(mu_(1))d_(1)` , we get
`d_(2)=+8cm`
Thus, the final image is 8cm in front of the medium B-air interface.
Method 2: Method of shifting
Net shifting `S=6(1-(1)/(3//2))+4(1-(1)/(2))`
`=4cm`
The shifting will be in the direction of ray travelling, i.e., towards right.
Hence, rays will coverge finally `=14+4=18 cm` from left surface or 8cm from right surface.