An object $$2.4\,m$$  in front of a lens forms a sharp image on a film $$12\,cm$$  behind the lens. A glass plate $$1\,cm$$  thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film?

Solution :
The focal length of the lens
$$\eqalign{ & \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \cr & = \frac{1}{{12}} + \frac{1}{{240}} \cr & = \frac{{20 + 1}}{{240}} \cr & = \frac{{21}}{{240}} \cr & f = \frac{{240}}{{21}}cm \cr} $$
Shift $$ = t\left( {1 - \frac{1}{\mu }} \right)$$
$$\eqalign{ & \Rightarrow \,\,1\left( {1 - \frac{1}{{\frac{3}{2}}}} \right) \cr & = 1 \times \frac{1}{3} \cr} $$
Now, $$v' = 12 - \frac{1}{3}$$
$$ = \frac{{35}}{3}cm$$
Now the object distance $$u.$$
$$\eqalign{ & \frac{1}{u} = \frac{3}{{35}} - \frac{{21}}{{240}} \cr & = \frac{1}{5}\left[ {\frac{3}{7} - \frac{{21}}{{48}}} \right] \cr & \frac{1}{u} = \frac{1}{5}\left[ {\frac{{48 - 49}}{{7 \times 16}}} \right] \cr & u = - 7 \times 16 \times 5 \cr & = - 560\,cm \cr & = - 5.6\,m \cr} $$