Physics - Geometrical Optics Question with Solution | TestHub

According to lens maker's formula $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ For the first lens : $n=1.5,R_1=+14\mathrm{cm},R_2=\infty$ $\therefore \frac{1}{f_1}=(1.5-1)\left(\frac{1}{14}-\frac{1}{\infty }\right)=\frac{0.5}{14}$ For the second lens : $n=1.2,R_1=\infty ,R_2=-14\mathrm{cm}$ $\therefore \frac{1}{f_2}=(1.2-1)\left(\frac{1}{\infty }-\frac{1}{-14}\right)=\frac{0.2}{14}$ The focal length of the bi-convex lens is $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{0.5}{14}+\frac{0.2}{14}=\frac{0.7}{14}=\frac{1}{20}$ According to thin lens formula; $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ $\Rightarrow \frac{1}{v}-\frac{1}{-40}=\frac{1}{20}$ or $v=40\mathrm{cm}$