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  • The figure shows a silvered lens. <img alt="\mu_{\mathrm{A}}=1.6$ and $\mu_{\mathrm{B}}=1.2, \mathrm{R}_{1}=80 \mathrm{~cm}, \mathrm{R}_{2}=40 \mathrm{~cm}$ and $\mathrm{R}_{3}=20 \mathrm{~cm}" src

The figure shows a silvered lens. \mu_{\mathrm{A}}=1.6$ and $\mu_{\mathrm{B}}=1.2, \mathrm{R}_{1}=80 \mathrm{~cm}, \mathrm{R}_{2}=40 \mathrm{~cm}$ and $\mathrm{R}_{3}=20 \mathrm{~cm}. An object is placed at a distance of 12 \mathrm{~cm} from this lens. Find the image position. 

Option: 1

15 \mathrm{~cm}


Option: 2

18 \mathrm{~cm}


Option: 3

20 \mathrm{~cm}


Option: 4

24 \mathrm{~cm}


Answers (1)

best_answer

Equivalent focal length of the two lenses

\mathrm{\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}} }
\mathrm{=(1.6-1)\left(\frac{1}{-80}-\frac{1}{-40}\right)+(1.2-1)\left(\frac{1}{-40}-\frac{1}{-20}\right)=\frac{1}{80}}

\therefore \quad For this lens combination

\mathrm{u}=-12 \mathrm{~cm}, \mathrm{f}=+80 \mathrm{~cm}
\frac{1}{\mathrm{v}_{1}}-\frac{1}{-12}=\frac{1}{80}
\Rightarrow \quad \frac{1}{\mathrm{v}_{1}}=\frac{1}{80}-\frac{1}{12}=\frac{-68}{80 \times 12}

Now this image acts as an object for the mirror of \mathrm{f}=\frac{\mathrm{R}_{3}}{2}=10 \mathrm{~cm}
\therefore \quad \frac{1}{v}+\frac{-68}{80 \times 12}=\frac{1}{-10}
\Rightarrow \quad \frac{1}{v_{2}}=\frac{68}{80 \times 12}-\frac{1}{10}=\frac{68-96}{80 \times 12}=\frac{-28}{80 \times 12}

This image formed by mirror again acts as an object for the lens system
\mathrm{\therefore \quad \frac{1}{v}-\frac{-28}{80 \times 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)
\mathrm{\therefore \quad \frac{1}{v}=-\frac{1}{80}-\frac{28}{80 \times 12} \quad \Rightarrow \quad v=-24 \mathrm{~cm}}
\mathrm{\therefore \quad}  The final image is formed at a distance of \mathrm{24 \mathrm{~cm}} to the left of the silvered lens.

Alternate Method :

\mathrm{\frac{1}{F_{\text {eq }}}=\frac{2}{f_{1}}+\frac{2}{f_{2}}+\frac{1}{f_{m}}=\frac{1}{8}}

\therefore \quad  or the equivalent mirror
\frac{1}{\mathrm{v}}+\frac{1}{-12}=-\frac{1}{8} \quad \Rightarrow \quad \mathrm{v}=-24 \mathrm{~cm}

Posted by

Suraj Bhandari

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