What is a compound microscope ? With the help of a labelled ray diagram, show the image formation by a compound microscope. Derive an expression for its magnifying power.
Compound Microscope :
A compound microscope consists of two suitable lenses to give large magnification by compounding the magnification given by the two lenses.
Construction : compound microscope consists of two convex lenses called objective and the eye piece. An objective lens is of small aperture and small focal length and faces the object to be seen. It forms a real, inverted and magnified image of an object. An eyepiece is a convex lens of large aperture and large focal length as compared to objective. It gives enlarged and virtual image by compounding the effect of the objective.
Principle : When an object is placed in front of a convex lens O at a distance between $F_o$ and $2F_o$, the real, inverted and magnified image is formed on the other side of this lens. If this image lies within the focal length of another convex lens E of large aperture, then the image acts as an object for this lens. The final image produced by this lens is virtual, inverted and highly magnified.
MAGNIFYING POWER (OR ANGULAR MAGNIFICATION) :
(a) Case 1. When final image is formed at least distance of distinct vision :
Deduce a relation for magnifying power of a compound microscope when final image is formed at least distance of distinct vision.
The magnifying power of a compound microscope is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the eye when both are at a distance of least distance of distinct vision.
STEP 1. Determination of Magnifying power:
Let $\angle PC_2A'' = \alpha = $ angle subtended by the object at the eye when placed at the distance of distinct vision.
and $\angle A''C_2B'' = \beta = $ angle subtended by the final image at the eye when placed at the distance of distinct vision.
$M.P. = \frac{\beta}{\alpha}$
Since $\alpha$ and $\beta$ are small angles, so using the relation $\theta = l/r$, we get
$\beta = \frac{A''B''}{C_2A''}$
and $\alpha = \frac{PA''}{C_2A''}$
$M.P. = \frac{A''B''/C_2A''}{PA''/C_2A''}$
$\therefore M.P. = \frac{A''B''}{PA''}$
Since $PA'' = AB$
$M.P. = \frac{A''B''}{AB}$
Multiplying and dividing Right Hand Side of eqn. (iv) by $A'B'$, we get
$M.P. = \frac{A''B''}{A'B'} \times \frac{A'B'}{AB}$
But magnification for objective lens,
$m_o = \frac{A'B'}{AB}$
and magnification for eye lens,
$m_e = \frac{A''B''}{A'B'}$
Hence eqn. (v) can be written as,
$M.P. = m_o \times m_e$
STEP 2. Applying sign conventions :
$m_o = \frac{A'B'}{AB} = -\frac{v_o}{u_o}$
where $u_o$ and $v_o$ are the distances of object and image from the objective lens.
STEP 3. Using lens formula :
For an eyepiece,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_e}$
where $v$ is distance of final image $A''B''$ from eyepiece and $u_e$ = distance of object $A'B'$ from eyepiece.
Since final image is formed at a distance of distinct vision (i.e., near point D), so $v = -D$.
Using sign convention, $u = -u_e$, $v = -D$, we get $\frac{1}{-D} - \frac{1}{-u_e} = \frac{1}{f_e}$ or $\frac{1}{u_e} = \frac{1}{f_e} + \frac{1}{D}$
Multiplying both sides by $D$, we get
$\frac{D}{u_e} = \frac{D}{f_e} + 1 \quad \text{or} \quad \frac{D}{u_e} = \frac{D}{f_e} + 1$
$\frac{D}{u_e} = 1 + \frac{D}{f_e}$
Substituting the value of eqn. (x) in eqn. (ix), we get $m_e = 1 + \frac{D}{f_e}$.
Hence eqn. (viii) becomes
$m_e = 1 + \frac{D}{f_e}$
STEP 4. Using eqns. (vii) and (xi) in eqn. (vi), we get
$M.P. = m_o \left(1 + \frac{D}{f_e}\right)$
$M.P. = -\frac{v_o}{u_o} \left(1 + \frac{D}{f_e}\right)$
Since object lies very close to $F_o$, so $u_o = f_o$. Also image $A'B'$ is formed closed to eye lens. So $v_o = L$, length of microscope's tube or distance between two lenses.
Hence,
$M.P. = -\frac{L}{f_o} \left(1 + \frac{D}{f_e}\right)$
(b) Case 2. When final image is formed at infinity
Deduce a relation for the magnifying power of a compound microscope when final image is formed as infinity.
When image A'B' formed by objective (O) is at the focus $F_e$ of the eyepiece, then final image is formed at infinity (Figure 96).
Now, from eq. (ix)
$m_e = \frac{D}{f_e}$
from eq. (vi)
$M.P. = m_o m_e$
$\therefore \quad M.P. = -\frac{v_0}{u_0} \times \frac{D}{f_e} \quad \cdots (xiv)$
Since object and image are close to lenses,
so, $u_0 = f_0$ and $v_0 = L$
$\therefore$ $M.P. = \frac{L}{f_0} \times \frac{D}{f_e}$
In this case, length of microscope,
$L = v_0 + u_e \quad \cdots (xv)$