Question

An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind 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 on film
1.7.2 cm
2.2.4cm
3.3.2 cm
4.5.6 cm

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{shift}\mathrm{in}\mathrm{image}\mathrm{position}\mathrm{due}\mathrm{to}\mathrm{glass}\mathrm{plate}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \mathrm{S}=(1-\frac{1}{\mathrm{\mu }})\mathrm{t}=(1-\frac{1}{1.5})\times 1=\frac{1}{3}\mathrm{cm} \\ \mathrm{For}\mathrm{focal}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{lens}, \\ \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{12}-\frac{1}{(-240)} \\ \mathrm{or}\mathrm{f}=\frac{240}{21}\mathrm{cm} \\ \mathrm{Now},\mathrm{to}\mathrm{get}\mathrm{back}\mathrm{image}\mathrm{on}\mathrm{the}\mathrm{film},\mathrm{lens}\mathrm{has}\mathrm{to}\mathrm{form}\mathrm{image}\mathrm{at}(12-\frac{1}{3})\mathrm{cm}=\frac{35}{3}\mathrm{cm}\mathrm{such}\mathrm{that}\mathrm{the}\mathrm{glass}\mathrm{plate}\mathrm{will}\mathrm{shift}\mathrm{the}\mathrm{image}\mathrm{on}\mathrm{the}\mathrm{film}. \\ \mathrm{Using}\mathrm{relation}, \\ \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\mathrm{or}\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{f}} \\ \frac{1}{\mathrm{u}}=\frac{3}{35}-\frac{21}{240}=-\frac{1}{560} \\ \mathrm{or}\mathrm{u}=-560\mathrm{cm}\mathrm{or}-5.6\mathrm{m} \end{gathered}$$