Principle Construction, Working and Angular Magnification of Compound Microscope

Principle: The principle of the compound microscope is based on the magnification of an image by using two lenses.

Construction: A compound microscope consists of two convergent lenses (i.e. objective lens $O$ and eye-piece lens $e$) placed coaxially in a double tube system. The objective lens is an achromatic convergent lens system of short focal length and short aperture. The other eye-piece lens $e$ is also an achromatic convergent lens system of large focal length and large aperture. The observation is taken through the eye-piece lens by the observer. The eye-piece lens is fitted outer side of a movable tube and the inner side connects with a non-movable tube in which the objective lens is fitted on another side of the non-movable tube. The separation between the objective or eye-piece lens can be changed by an arrangement, this is known as rack and pinion arrangement.

Working: Suppose a small object $ab$ is placed slightly away from the first focus $f_{\circ}$ of the objective lens which forms a real, inverted, and magnified image $a_{1}b_{1}$. Now adjust the eye-piece lens by moving like this, that the image $a_{1}b_{1}$ lies in between the optical center and the second focal length $f_{e}$ of the eye-piece lens. This image $a_{1}b_{1}$ works as an object for the eye-piece lens which forms a magnified, virtual, and final image $a_{2}b_{2}$. The final image $a_{2}b_{2}$ is generally formed at the least distance $D$ of distinct vision, although it can be formed anywhere between this position and infinity.

Angular Magnification Or Magnifying Power($M$):

The angular magnification or magnifying power can be defined as the ratio of the angle subtended by the image at the eye ($\beta$) to the angle subtended by the object at the eye when placed at least distance of distinct vision ($\alpha$)

$M= \frac{Angle \: subtended \: by \: the \: image \: at \: the \: eye \: (\beta)}{Angle \: subtended \: by \: the \: object \: at \: the \: eye \: when \\ placed \: at \: least \: distance \: of \: distinct \: vision \: (\alpha)}$

$M=\frac{\beta}{\alpha} \approx \frac{tan \beta}{tan \alpha} \quad (1)$

From figure

$tan \beta = \frac{a_{2}b_{2}}{a_{2} e}$

$tan \alpha = \frac{a_{2}a_{3}}{a_{2}e}$

Now subtitute these values in equation $(1)$, then

$M=\frac{\frac{a_{2}b_{2}}{a_{2} e}}{\frac{a_{2}a_{3}}{a_{2}e}}$

$M=\frac{a_{2}b_{2}}{a_{2}a_{3}}$

Here $a_{2}a_{3} = ab$

So the above equation can be written as

$M=\frac{a_{2}b_{2}}{ab}$

$M=\frac{a_{2}b_{2}}{ab} \frac{a_{1}b_{1}}{a_{1}b_{1}}$

$M=\frac{a_{2}b_{2}}{a_{1}b_{1}} \frac{a_{1}b_{1}}{ab}$

Here $m_{e}=\frac{a_{2}b_{2}}{a_{1}b_{1}}$ and $m_{\circ}= \frac{a_{1}b_{1}}{ab}$

Now substitute the values of $m_{e}$ and $m_{\circ}$ in the above equation

$M=m_{e} \times m_{\circ} \qquad(1)$

Where
$m_{e} \rightarrow$ The linear magnification produced by eye-piece lens system
$m_{\circ} \rightarrow$ The linear magnification produced by the object lens system

So now again from the figure

The linear magnification produced by object lens system $m_{\circ} = -\frac{v_{\circ}}{u_{\circ}}$

The linear magnification produced by eye-piece lens system $m_{e} = \frac{D}{u_{e}}$

Substitute the value of $m_{\circ}$ and $m_{e}$ in equation $(1)$

$M= -\frac{v_{\circ}}{u_{\circ}} \left( \frac{D}{u_{e}} \right) \qquad(2)$

Adjustment of a Compound Microscope:

1.) Adjustment for Clear Vision: In this final image an object is formed at least a distance of distinct vision $D$. For this configuration,

On substitution $u=-u_{e}$, $v=-D$ and $f=f_{e}$ in the lens formula for eyepiece lens

$\frac{1}{-D}+\frac{1}{u_{e}}=\frac{1}{f_{e}}$

$\frac{1}{u_{e}}=\frac{1}{f_{e}} + \frac{1}{D}$

$\frac{D}{u_{e}}= \left( \frac{D}{f_{e}} + 1 \right)$

Now substitute the value of $\frac{D}{u_{e}}$ in equation $(2)$

$M= -\frac{v_{\circ}}{u_{\circ}} \left(1+ \frac{D}{f_{e}} \right)$

The length of the microscope tube in this setup

$L=$ Distance between the object and the eye-piece lenses

$L= v_{\circ} + |u_{e}|$

2.) Adjustment for Relaxed Eye: In this configuration, the final image of an object is formed in a relaxed eye position i.e. at $\infty$. In this setup, the eye-piece lens system is moved back until the image of object $ab$, formed by object lens,i.e., $a_{1}b_{1}$ fall at (coincide with)second focus $f'_{e}$ of the eye-piece lens system. Mathematically this situation comes when $u_{e} = f_{e}$

Thus, the magnifying power in this position,

$M=-\frac{v_{\circ}}{u_{\circ}} \left( \frac{D}{f_{e}} \right)$

For this set the length of the microscope tube,

$L=v_{\circ}+f_{e}$

Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater then the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less then the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of light w

Fraunhofer diffraction due to a single slit

Let $S$ be a point monochromatic source of light of wavelength $\lambda$ placed at the focus of collimating lens $L_{1}$. The light beam is incident normally from $S$ on a narrow slit $AB$ of width $e$ and is diffracted from it. The diffracted beam is focused at the screen $XY$ by another converging lens $L_{2}$. The diffraction pattern having a central bright band followed by an alternative dark and bright band of decreasing intensity on both sides is obtained. Analytical Explanation: The light from the source $S$ is incident as a plane wavefront on the slit $AB$. According to Huygens's wave theory, every point in $AB$ sends out secondary waves in all directions. The undeviated ray from $AB$ is focused at $C$ on the screen by the lens $L_{2}$ while the rays diffracted through an angle $\theta$ are focussed at point $p$ on the screen. The rays from the ends $A$ and $B$ reach $C$ in the same phase and hence the intensity is maximum. Fraunhofer diffraction due to

Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ \$\frac{d^{2} \psi(x)}{d x