First slide

The optical instruments

Question

If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5mm, the focal length of the eye-piece, should be close to :

Moderate

A

12 mm
B

33 mm
C

22 mm
D

2 mm

Solution

Case-I
If final image is at least distance of clear vision
$\begin{array}{l} M . P . = \frac{L}{f_{0}} (1 + \frac{D}{f_{e}}) \\ \Rightarrow 375 = \frac{150}{5} [1 + \frac{25}{f_{e}}] \\ \Rightarrow \frac{375}{30} = 1 + \frac{25}{f_{e}} \\ \Rightarrow \frac{345}{30} = \frac{25}{f_{e}} \end{array}$
$\begin{array}{l} \Rightarrow f_{e} = \frac{750}{345} = 2.17 c m \\ \Rightarrow f_{e} \approx 22 m m \end{array}$
Case-II
If final image is at infinity
$\begin{array}{l} M . P = \frac{L}{f_{0}} (\frac{D}{f_{e}}) = 375 \\ \Rightarrow \frac{375}{30} = \frac{25}{f_{e}} \\ \Rightarrow f_{e} = \frac{750}{375} \\ \Rightarrow f_{e} = 2 c m \\ \Rightarrow f_{e} \approx 22 m m \end{array}$

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