Physics - Geometrical Optics Question with Solution | TestHub

Biconvex lens is cut perpendicularly to the principal axis, it will become a plano-convex lens. Focal length of biconvex lens$\begin{array}{c}\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\ \frac{1}{f}=(n-1)\frac{2}{R}\left(\because R_1=R,R_2=-R\right)\end{array}$ $\Rightarrow f=\frac{R}{2(n-1)}\dots (i)$ For plano-convex lens $\begin{array}{c}\frac{1}{f_1}=(n-1)\left(\frac{1}{R}-\frac{1}{\infty }\right)\\ f_1=\frac{R}{(n-1)}\dots (ii)\end{array}$ Comparing Eqs. (i) and (ii), we see that focal length becomes double. As power of lens $P\propto \frac{1}{\text{ focal length }}$ Hence, power will become half. $\text{ New power }=\frac{4}{2}=2\mathrm{D}$