Question
(a) When a telescope is turned upside down and looked at the objective, it appears very small, why?
(b) Why does this not happen in a microscope?
(b) Why does this not happen in a microscope?
✓
Answer
(a) In a telescope, the focal length $\left(f_o\right)$ of the objective lens is much more than the focal length $\left(f_e\right)$ of the eyepiece and its magnification power is $f_o / f_e$. On looking back, the magnification power will become $f_e / f_o$, because $f_o \ll f_e$ hence now the object will appear very small.
(b) The formula for the magnifying power of a compound microscope is $\frac{v_o}{u_o} \times \frac{ D }{f_e}$, because the value of $v _0$ is only slightly more than the focal length $f_o$ of the objective of the microscope, hence the magnification can be considered as $\frac{v_o}{f_o} \times \frac{ D }{f_e}$; because $f_o$ and $f_e$ both have low values. Therefore, even after turning the microscope, the magnification power remains almost unchanged due to there being no significant difference in the value of $v_0$.
(b) The formula for the magnifying power of a compound microscope is $\frac{v_o}{u_o} \times \frac{ D }{f_e}$, because the value of $v _0$ is only slightly more than the focal length $f_o$ of the objective of the microscope, hence the magnification can be considered as $\frac{v_o}{f_o} \times \frac{ D }{f_e}$; because $f_o$ and $f_e$ both have low values. Therefore, even after turning the microscope, the magnification power remains almost unchanged due to there being no significant difference in the value of $v_0$.
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