Practicing Success
The figure shows a silvered lens. $μ_A = 1.6$ and $μ_B = 1.2$, $R_1$ = 80 cm, $R_2$ = 40 cm and $R_3$ = 20 cm. An object is placed at a distance of 12 cm from this lens. Find the image position. |
15 cm 18 cm 20 cm 24 cm |
24 cm |
Equivalent focal length of the two lenses $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$ $=(1.6-1))(\frac{1}{-80}-\frac{1}{-40})+(1.2-1)(\frac{1}{-40}-\frac{1}{-20})=\frac{1}{80}$ ∴ For this lens combination u = -12 cm, f = +80 cm $\frac{1}{v_1}-\frac{1}{-12}=\frac{1}{80}$ $⇒\frac{1}{v_1}=\frac{1}{80}-\frac{1}{12}=\frac{-68}{80×12}$ Now this image acts as an object for the mirror of $f=\frac{R_3}{2}=10cm$ $∴\frac{1}{v}+\frac{-68}{80×12}=\frac{1}{-10}$ $⇒\frac{1}{v_2}=\frac{68}{80×12}-\frac{1}{10}=\frac{68-96}{80×12}=\frac{-28}{80×12}$ This image formed by mirror again acts as an object for the lens system $∴\frac{1}{v}-\frac{-28}{80×12}=\frac{-1}{80}$ (focal length is taken negative because rays are now coming from right and the principal focus of the lens system will be on left side) $∴\frac{1}{v}=-\frac{1}{80}-\frac{28}{80×12}⇒v=-24cm$ ∴ The final image is formed at a distance of 24 cm to the left of the silvered lens. Alternate Method : $\frac{1}{F_{eq}}=\frac{2}{f_1}+\frac{2}{f_2}+\frac{1}{f_m}=\frac{1}{8}$ ∴ or the equivalent mirror $\frac{1}{v}+\frac{1}{-12}=-\frac{1}{8}⇒v = -24 cm$ |