Question
$i$. Draw a ray diagram showing the image formation by a compound microscope. Obtain the expression for total magnification when the image is formed at infinity.
$ii$. How does the resolving power of a compound microscope get affected, when
$1$. focal length of the objective is decreased.
$2$. the wavelength of light is increased ? Give reasons to justify your answer.

Answer

$i$. The ray diagram, showing image formation by a compound microscope, is given below ; $-$
Image
$ii.$ Linear Magnification due to objective lens is given by $=\frac{\tan \beta}{\tan \alpha}$
$\tan \beta=\frac{h^{\prime}}{L}=\frac{h}{f_o}$
$\frac{h^{\prime}}{h}=\frac{L}{f_0}$
$($where the distance between the second focal point of the objective and the first focal point of the eyepiece is called the tube length of the compound microscope and is denoted by $L)$
The eyepiece will act as a simple microscope, hence we may use the formula of magnification by a simple microscope for normal adjustment.
$m_e=\frac{D}{f_e}$
Total magnification, $m = m _{ o } \times m _{ e }$
$=\frac{L}{f_o} \times \frac{D}{f_e} d_{\min }$
$a$. From the equation, it is clear that resolving power increases when the focal length of the objective is decreased.
This is because the minimum separation, $d _{\min }$ decreases when f is decreased.
$b$. Resolving power decreases when the wavelength of light is increased.
​​​​​​​This is because the minimum separation, $d _{\text {min }}$ increases when $\lambda$ is increased.

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