Solution

Using $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ for each layer and adding: $f_{net}=-100\ cm$.

$\frac{1}{f_1}=(1.6-1)\left(\frac{1}{\infty}-\frac{1}{20}\right)=-\frac{3}{100}$ $\frac{1}{f_2}=(1.5-1)\left(\frac{1}{20}-\frac{1}{-20}\right)=\frac{1}{20}$ $\frac{1}{f_3}=\frac{1}{f_1}=-\frac{3}{100}$ $\frac{1}{f_{net}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=-\frac{3}{50}+\frac{1}{20}$ $\Rightarrow f_{net}=-100\ cm$

Concept equivalent focal length of a lens combination

Ray Optics and Optical Instruments in past papers

89 questions from this chapter have appeared across 25 exam years.