Question
i. Draw a ray diagram showing the image formation by a compound microscope. Hence obtain the expression for total magnification when the image is formed at least distance of distinct vision.
iii. A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.0 cm. If they are separated by a distance of 24 cm, find the total magnification when the image is formed at infinity.

Answer

Image

Magnification due to objective
$m_0=\frac{h^{\prime}}{h}=\frac{L}{f_0}\left(\because \tan \beta=\frac{h}{f_0}=\frac{h}{L}\right)$
Magnification due to eye piece when final image is formed at the near point
$m _{ e }=1+\frac{D}{f_{ e }}$
Total magnification
$\begin{aligned} m & = m _0 m_{ e } \\
m & =\frac{L}{f_0}\left(1+\frac{D}{f_{ e }}\right)\end{aligned}$
ii.
$\begin{array}{l}m=\frac{L D}{f_o f_e} \\ m=\frac{24 \times 25}{2 \times 6}=\frac{600}{12}=50\end{array}$
Hence, the total magnification when the image is formed at infinity is 50.

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