Draw a ray diagram to show the image formation by a combination of two thin convex lenses in contact. Obtain the expression for the power of this combination in terms of the focal lengths of the lenses.

(b) A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when the angle of incidence is ${\frac{3}{4}}^{\mathrm{th}}$of the angle of prism. Calculate the speed of light in the prism.

(a) Ray diagram to show the image formation by a combination of two thin convex lenses in contact
$\frac{1}{{\mathrm{f}}_1}=\frac{1}{{\mathrm{v}}^{\mathrm{\prime }}}-\frac{1}{\mathrm{u}}.$........(1)
For second lens,
$\frac{1}{{\mathrm{f}}_2}=\frac{1}{\mathrm{v}}-\frac{1}{{\mathrm{v}}^{\mathrm{\prime }}}\dots \cdots$......(ii)
Adding (i) and (ii),
$\frac{1}{{\mathrm{f}}_1}+\frac{1}{{\mathrm{f}}_2}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\dots$...(iii)
If f is the combined focal length then
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$
From equation (iii),
$\begin{array}{l}\frac{1}{\mathrm{f}}=\frac{1}{{\mathrm{f}}_1}+\frac{1}{{\mathrm{f}}_2}\\ \mathrm{P}=\frac{1}{{\mathrm{f}}_1}+\frac{1}{{\mathrm{f}}_2}\left[\because \frac{1}{\mathrm{f}}=\mathrm{P}\right]\\ \mathrm{P}=\frac{{\mathrm{f}}_1+{\mathrm{f}}_2}{{\mathrm{f}}_1{\mathrm{f}}_2}\end{array}$

(b) Here, Angle of prism, $\mathrm{A}={60}^{\circ }$
Angle of incidence, $\mathrm{i}=\frac{3}{4}\times 60$
$\mathrm{i}+\mathrm{e}=\mathrm{A}+\mathrm{\delta }$
e is emergence angle
where δ is deviation angle
At minimum deviation δ=δmin

i = e
2i = A + δmin                          
2 × 45 = 60 + δmin                                         
δmin = 30
By prism law,
$\begin{array}{l}\mathrm{\mu }=\frac{\sin \frac{\mathrm{A}+{\mathrm{\delta }}_{min}}{2}}{\sin \frac{\mathrm{A}}{2}}\\ \mathrm{\mu }=\frac{\sin \frac{60+30}{2}}{\sin \frac{60}{2}}=\frac{2}{\sqrt{2}}=1.41\end{array}$
$\mathrm{v}=\frac{\mathrm{c}}{\mathrm{\mu }}=\frac{3\times {10}^8}{1.41}=2\times {10}^8\mathrm{m}/\mathrm{s}$
Let v be velocity of light in prism