Rankers Physics

Optical Instruments: Practice Problem & Solution

In compound microscope the magnification is 95, and the distance of object from objective lens 1/3.8 cm and focal length of objective is 1/4 cm. What is the magnification of eye pieces when final image is formed at least distance of distinct vision: (1999)
$5$
$10$
$100$
None

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Optical Instruments. Understanding the underlying formula is key to arriving at the correct answer below:

Objective magnification is $m_o = \frac{f_o}{f_o + u_o}$. Substituting $f_o = 0.25$ and $u_o = -1/3.8$, $m_o = \frac{1/4}{1/4 - 1/3.8} = \frac{1/4}{(3.8 - 4)/15.2} = -19$. Total magnification is $M = m_o \times m_e$, so $95 = 19 \times m_e$ (using magnitudes). This gives the magnification of the eyepiece as $m_e = 5$.

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