ray-optics-and-optical-instruments Question 47
Question: Q. 7. How is the working of a telescope different from that of a microscope? The focal lengths of the objective and eyepiece of a microscope are $1.25 \mathrm{~cm}$ and $5 \mathrm{~cm}$ respectively. Find the position of the object relative to the objective lens in order to obtain an angular magnification of 30 in normal adjustment. U] [Delhi I, II, III 2012]
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Solution:
Ans. Working differences :
(i) The objective of a telescope forms the image of a very far off object at, or within, the focus of its eyepiece. The microscope does the same for a small object kept just beyond the focus of its objective.
(ii) The final image formed by a telescope is magnified relative to its size as seen by the unaided eye, while the final image formed by a microscope is magnified relative to its absolute size.
(iii) The objective of a telescope has large focal length and large aperture, while the corresponding for a microscope have very small values.
Given : $f_{o}=1.25 \mathrm{~cm}, f_{e}=5 \mathrm{~cm}$
Angular magnification, $m=30$
Now,
$$ m=m_{e} \times m_{o} $$
In normal adjustment, the angular magnification of an eyepiece
$$ m_{e}=\frac{d}{f_{e}}=\frac{25}{5}=5 $$
$$ m_{0}=6 $$
$$ m_{0}=\frac{v_{0}}{u_{0}} \Rightarrow-6=\frac{v_{0}}{u_{0}} $$
$$ v_{o}=-6 u_{o} $$
Applying lens equation to the objective lens :
$$ \begin{align*} \frac{1}{f_{0}} & =\frac{1}{v_{0}}-\frac{1}{u_{0}} \ \frac{1}{1.25} & =\frac{1}{-6 u_{0}}-\frac{1}{u_{0}} \ \frac{1}{1.25} & =\frac{-1-6}{6 u_{0}} \ 6 u_{o} & =-1.25 \times 7 \ u_{0} & =\frac{-1.25 \times 7}{6} \mathrm{~cm} \end{align*} $$
$\left|u_{\mathrm{o}}\right|=1.46 \mathrm{~cm}$
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