Compound Microscope - Practice Questions & MCQ

Compound Microscope: 

A compound microscope is of two converging lenses called objective and eye lens. It is used to view magnified images of small objects on a glass slide. It can achieve higher levels of magnification than stereo or other low power microscopes and reduce chromatic aberration.

$f_{\text {eyelens }}>f_{\text {objective }}$ and $(\text { diameter })_{\text {eyelens }}>(\text { diameter })_{\text {objective }}$
Intermediate image is real and enlarged. The final image is magnified, virtual and inverted.
Here in the diagram
$u_o=$ Distance of object from objective (0),
$v_o=$ Distance of image ( $A^{\prime} B^{\prime}$ ) formed by objective from objective,
$u_e=$ Distance of $A^{\prime} B^{\prime}$ from eye lens,
$v_e=$ Distance of final image from eye lens,
$f_{o=}$ Focal length of objective,
$f_e=$ Focal length of eye lens.

Case 1: Final image is formed at $\mathbf{D}$ : Magnification

$
m_D=\frac{v_o}{u_o}\left(1+\frac{D}{f_e}\right) \text { and length of the microscope }
$

tube (distance between two lenses) is $L_D=v_e+u_e$.
Generally, object is placed very near to the principal focus of the objective hence $u_o \equiv f_o$. The eye piece is also of small focal length and the image formed by the objective is also very near the eyepiece.

So $v_o \equiv L_D$, the length of the tube.
Hence, we can write

$
m_D=\frac{L}{f_o}\left(1+\frac{D}{f_e}\right)
$

Case 2: Final image is formed at $\infty$ : Magnification
$m_{\infty}=\frac{v_0}{u_0} \cdot \frac{D}{f_e}$ and length of tube $L_{\infty}=v_0+f_e$

$
m_{\infty}=\frac{\left(L_{\infty}-f_o-f_e\right) D}{f_o f_e}
$

In terms of length
- For large magnification of the compound microscope, both $f_o$ and $f_e$ should be small.
- If the length of the tube of microscope increases, then its magnifying power increases.
- The magnifying power of the compound microscope may be expressed as $M=m_o \times m_e$ where $m_0$ is the magnification of the objective and $m_e$ is magnifying the of eyepiece.