Physics - Geometrical Optics Question with Solution | TestHub

The focal length $\left(f_L\right)$ of the left plano-concave lens can be calculated as follows-

$\begin{array}{rcl}\frac{1}{f_L}& =& \left(n-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\ & =& \left(1.6-1\right)\left(\frac{1}{\infty }-\frac{1}{20 \mathrm{cm}}\right)\\ & =& -\frac{0.6}{20 \mathrm{cm}} ...\left(1\right)\end{array}$

The focal length $\left(f_M\right)$ of the middle bi-convex lens can be calculated as follows-

$\begin{array}{rcl}\frac{1}{f_M}& =& \left(n-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\ & =& \left(1.5-1\right)\left(\frac{1}{20 \mathrm{cm}}-\frac{1}{-20 \mathrm{cm}}\right)\\ & =& \frac{0.5}{10 \mathrm{cm}} ...\left(2\right)\end{array}$

The focal length $\left(f_R\right)$ of the right plano-concave lens can be calculated as follows-

$\begin{array}{rcl}\frac{1}{f_R}& =& \left(n-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\ & =& \left(1.6-1\right)\left(\frac{1}{-20 \mathrm{cm}}-\frac{1}{\infty }\right)\\ & =& -\frac{0.6}{20 \mathrm{cm}} ...\left(3\right)\end{array}$

Hence, the effective focal length $\left(f\right)$ of the combination of lenses can be calculated as follows-

$\begin{array}{rcl}\frac{1}{f}& =& \frac{1}{f_L}+\frac{1}{f_M}+\frac{1}{f_R}\\ & =& -\frac{0.6}{20 \mathrm{cm}}+\frac{{\displaystyle 0.5}}{{\displaystyle 10 \mathrm{cm}}}-\frac{{\displaystyle 0.6}}{{\displaystyle 20 \mathrm{cm}}}\\ & =& -\frac{0.1}{10 \mathrm{cm}}\\ & \Rightarrow & f= -100 \mathrm{cm}\end{array}$