Ray Optics and Optical Instruments Part 2

Given, focal length of first lens, $f_{1}=+10\text{ cm}$ Focal length of second lens, $f_{2}=-10\text{ cm}$ Focal length of third lens, $f_{3}=+30\text{ cm}$ Object distance from first lens, $u_{1}=-30\text{ cm}$ Using lens formula for first lens, we get $\frac{1}{f_{1}}=\frac{1}{v_{1}}-\frac{1}{u_{1}}$ $\Rightarrow \frac{1}{-10}=\frac{1}{v_{1}}-\frac{1}{-30}$ $\Rightarrow \frac{1}{v_{1}}=\frac{1}{10}-\frac{1}{30}=\frac{2}{30}$ $v_{1}=15\text{ cm}$ Thus, the image formed by first lens is 15 cm to the right of first lens. The object distance for second lens will be $u_{2}=15\text{ cm}-5\text{ cm}=10\text{ cm}$ Using lens formula for first lens, we get $\frac{1}{f_{2}}=\frac{1}{v_{2}}-\frac{1}{u_{2}}$ $\Rightarrow \frac{1}{-10}=\frac{1}{v_{2}}-\frac{1}{10}$ $\Rightarrow \frac{1}{v_{2}}=\frac{1}{-10}+\frac{1}{10}=0$ $\Rightarrow v_{2}=\infty$ Thus, the object distance for third lens, $u_{3}$ = ∞. Using lens formula for third lens, we get $\frac{1}{f_{3}}=\frac{1}{v_{3}}-\frac{1}{u_{3}}$ $\Rightarrow \frac{1}{30}=\frac{1}{v_{3}}-\frac{1}{\infty}$ $\frac{1}{v_{3}}=\frac{1}{30}$ $v_{3}=30\text{ cm}$ The final image will form 30 cm from the third lens. The distance of final image from object will be $30+10+5+30=75\text{ cm}$ Thus, the distance of final image formed from object O is 75 cm.