i )An equilateral glass prism has a refractive index 1.6 in air. Calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index $\frac{4\sqrt{2}}{5}$.
ii )You are given three lenses ${\mathrm{L}}_1 {\mathrm{L}}_2$ and ${\mathrm{L}}_3$ each of focal length 15 cm. An object is kept at 20 cm in front of ${\mathrm{L}}_1$, as shown. The final real image is formed at the focus ‘I’ of ${\mathrm{L}}_3$ . Find the separations between ${\mathrm{L}}_1 {\mathrm{L}}_2$ and ${\mathrm{L}}_3$.

$$\begin{gathered} \mathrm{Given}, \mathrm{A}=60° (\mathrm{for} \mathrm{equilateral} \mathrm{prism}) \\ {\mathrm{\mu }}_1=\frac{4\sqrt{2}}{5}, {\mathrm{\mu }}_2=1.6 \\ \mathrm{The} \mathrm{refractive} \mathrm{index} \mathrm{is} \mathrm{given} \mathrm{by} \\ \frac{{\mathrm{\mu }}_2}{{\mathrm{\mu }}_1}=\frac{\sin \left({\displaystyle \frac{\mathrm{A}+\mathrm{D}}{2}}\right)}{\sin \left({\displaystyle \frac{\mathrm{A}}{2}}\right)} \\ \mathrm{where}, \mathrm{D}=\mathrm{angle} \mathrm{of} \mathrm{minimum} \mathrm{deviation}. \\ \frac{1.6\times 5}{4\sqrt{2}}=\frac{\sin \left({\displaystyle \frac{60°+\mathrm{D}}{2}}\right)}{\sin \left({\displaystyle \frac{60°}{2}}\right)} \\ \sqrt{2}\times \sin 30°=\sin \left(\frac{60°+\mathrm{D}}{2}\right) \\ \Rightarrow \frac{1}{\sqrt{2}}=\sin \left(\frac{60°+\mathrm{D}}{2}\right) \\ \Rightarrow \sin 45°=\sin \left(\frac{60°+\mathrm{D}}{2}\right) \\ \Rightarrow 45°=\frac{60°+\mathrm{D}}{2} \\ \mathrm{D}=90°-60°=30° \end{gathered}$$

ii) Let ${\text{⨍}}_1{\text{,⨍}}_2$ and ${\text{⨍}}_3$be the focal length of three lenses 

            $$\begin{gathered} \mathrm{for} \mathrm{lens} {\mathrm{L}}_1;\mathrm{u}=20\mathrm{cm} \\ \frac{1}{{\text{⨍}}_1}+\frac{1}{{\mathrm{v}}_1}-\frac{1}{{\mathrm{u}}_1}\Rightarrow \frac{1}{{\mathrm{v}}_1}=\frac{1}{15}-\frac{1}{20}\Rightarrow {\mathrm{v}}_1=60 \mathrm{cm} \\ \mathrm{Now} \mathrm{for} \mathrm{lens} {\mathrm{L}}_3, {\mathrm{v}}_3=15\mathrm{cm}, {\text{⨍}}_3{\text{=15 cm, u}}_3\text{=?} \\ \frac{1}{{\text{⨍}}_3}=\frac{1}{{\mathrm{v}}_3}+\frac{1}{{\mathrm{u}}_3}\Rightarrow \frac{1}{{\mathrm{u}}_3}=\frac{1}{15}-\frac{1}{15}=0 \therefore {\mathrm{u}}_3=\infty \end{gathered}$$

 $$\begin{gathered} \mathrm{It} \mathrm{shows} \mathrm{that} \mathrm{lens} \mathrm{infinte}. \\ \mathrm{Hnce} \mathrm{for} \mathrm{lens} {\mathrm{L}}_1, \mathrm{image} \mathrm{is} \mathrm{formed} \mathrm{at} \mathrm{a} \mathrm{distance} \mathrm{of} 15 \mathrm{cm} \mathrm{from} {\mathrm{L}}_2 \\ \therefore \mathrm{the} \mathrm{focus} \mathrm{of} {\mathrm{L}}_2 \mathrm{i}.\mathrm{e}. {\mathrm{u}}_2=15 \mathrm{cm} \\ \mathrm{Now}, \mathrm{to} \mathrm{calculate} \mathrm{the} \mathrm{distance} \mathrm{between} {\mathrm{L}}_1 \mathrm{and} {\mathrm{L}}_2, \\ {\mathrm{u}}_1+{\mathrm{H}}_2=60+15=75 \mathrm{cm} \\ \mathrm{Distance} \mathrm{between} {\mathrm{L}}_2 \mathrm{and} {\mathrm{L}}_3={\mathrm{v}}_2+{\mathrm{v}}_3=\infty \mathrm{or} \mathrm{can} \mathrm{be} \mathrm{any} \mathrm{value}. \end{gathered}$$