In a compound microscope, the magnified virtual image is formed at a distance of 25cm from the eye-piece. The focal length of its objective lens is 1cm. If the magnification is 100 and the tube length of the microscope is 20cm, then the focal length of eyepiece lens (in cm) is:________(Give answer up to 2 decimal points)
Option: 1 4.48
Option: 2 100
Option: 3 90
Option: 4 80

Answers (1)

As Lens formula is given as

\frac{1}{v}-\frac{1}{u}=\frac{1}{f}

\begin{aligned} &\text { For first lens }\\ &\Rightarrow \frac{1}{\mathrm{v}_{1}}-\frac{1}{-\mathrm{x}}=\frac{1}{1} \Rightarrow \mathrm{v}_{1}=\frac{\mathrm{x}}{\mathrm{x}-1}\\ &\text { also magnification }\left|\mathrm{m}_{1}\right|=\left|\frac{\mathrm{v}_{1}}{\mathrm{u}_{1}}\right|=\frac{1}{\mathrm{x}-1} \end{aligned}

\begin{aligned} &\text { for } 2^{\text {nd }} \text { lens this is acting as object }\\ &\text { so } \mathrm{u}_{2}=-\left(20-\mathrm{v}_{1}\right)=-\left(20-\frac{\mathrm{x}}{\mathrm{x}-1}\right)\\ &\text { and } v_{2}=-25 \mathrm{~cm}\\ &\text { angular magnification =}\left|\mathrm{m}_{\mathrm{2}}\right|=\left|\frac{\mathrm{D}}{\mathrm{u}_{2}}\right|=\frac{25}{\left|\mathrm{u}_{2}\right|} \end{aligned}

\begin{array}{l} \text { Total magnification } \mathrm{m}=\mathrm{m}_{1} \mathrm{m}_{\mathrm{2}}=100 \\ \\ \Rightarrow \left(\frac{1}{\mathrm{x}-1}\right)\left(\frac{25}{20-\frac{\mathrm{x}}{\mathrm{x}-1}}\right)=100 \\ \\ \frac{25}{20(\mathrm{x}-1)-\mathrm{x}}=100 \Rightarrow 1=80(\mathrm{x}-1)-4 \mathrm{x} \\ \\ \Rightarrow 76 \mathrm{x}=81 \Rightarrow \mathrm{x}=\frac{81}{76} \end{array}

\Rightarrow \mathrm{u}_{2}=-\left(20-\frac{81 / 76}{81 / 76-1}\right)=\frac{-19}{5}

Now apply the lense formula for the eyepiece

\frac{1}{-25}-\frac{1}{-19 / 5}=\frac{1}{\mathrm{f}_{\mathrm{e}}} \\ \\ \Rightarrow \mathrm{f}_{\mathrm{e}}=\frac{25 \times 19}{106} \approx 4.48 \mathrm{~cm}

 

 

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