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Ray Optics and Optical Instruments — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsRay Optics and Optical Instruments5 Marks
Question
A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an objective placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
✓

Answer

Given that:
D = 25cm
Focal length of objective lens $=f_o=8.0 mm$
∴ $f_o=0.8 cm$
Focal length of eyepiece $=f_e=2.5 cm$
$u_o=-9.0 mm=-0.9 cm$
$\begin{array}{r}\text { Distance between both lenses }= L =? \\ m=?\end{array}$
Lens formula for eyepiece :
$\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}$
or $-\frac{1}{u_e}=\frac{1}{f_e}-\frac{1}{v_e} \because v_e=- D =-25$
$=\frac{1}{2.5}-\left(-\frac{1}{25}\right)=\frac{10}{25}+\frac{1}{25}$
$=\frac{2}{5}+\frac{1}{25}=\frac{10+1}{25}$
$u_e=-\frac{25}{11} cm$
$\therefore \quad\left|u_e\right|=\left|-\frac{25}{11}\right|=\frac{25}{11} cm=2.27 cm$
For objective lens :
$\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$
or $\frac{1}{v_o}=\frac{1}{f_o}+\frac{1}{u_o} \Rightarrow-\frac{1}{0.9}+\frac{1}{0.8}=\frac{-10}{9}+\frac{10}{8}$
$=\frac{-0.8+0.9}{072}=\frac{0.1}{072}$
or $\quad v_0=\frac{0.72}{0.1}=7.2 cm$
$\therefore \quad L =v_o+\left|u_e\right|$
$=7.2+2.27=9.47 cm$
Now magnification power of compound microscope :
$M =\frac{v_o}{\left|u_o\right|}\left(1+\frac{ D }{f_e}\right)=\frac{7.2}{0.9}\left(1+\frac{25}{2.5}\right)$
$=8 \times(1+10)=8 \times 11=88$

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