Physics - Geometrical Optics Question with Solution | TestHub

The given combination of lensesHere, $f_1=$ focal length of equiconvex lenses of glass. $f_2=$ focal length of lens formed by water (concave). The focal length of the combination. $$\therefore \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=\frac{1}{f_1}-\frac{1}{f_2}+\frac{1}{f_1}=\frac{2}{f_1}-\frac{1}{f_2}$$ $$\begin{array}{rl}\frac{1}{F}=\frac{2f_2-f_1}{f_1{f}_2}\Rightarrow F& =\frac{f_1{f}_2}{2f_2-f_1}\\ F& =\frac{f_1}{2-\frac{f_1}{f_2}}\end{array}$$ $\left[\because f_3=f_1\right]$ Here, $\frac{1}{f_1}=\left({\mu }_g-1\right)\frac{2}{R},$ for $L_1$ and $L_3$ and $\frac{1}{f_2}=\left({\mu }_w-1\right)\left(-\frac{2}{R}\right),$ for $L_2$ $$\text{ Given, }{\mu }_w<{\mu }_g$$ Thus, $\frac{f_1}{f_2}<1$ So. $$F>\frac{f_1}{2}$$ From Eqs. (i) and (ii), we get $$\frac{f_1}{2}<F<f_1\text{ or }\frac{f}{2}<F<f$$