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A simple microscope is an optical instrument used to view near objects. It is simply a magnifying glass or converging lens of short focal length. The lens is held slightly nearer the object than its focal length and the eye is positional close to the other side of the lens. A virtual, erect, magnified image of the object is observed as shown in the figure.
The lens is so adjusted that the image of the object OA lies at the near point, and the image distance is the least distance of distinct vision. Let α be the angle subtended at the unaided eye by the object at the near point and β be the angle subtended at the lens by the image at the same distance. The angular magnification of the simple microscope is defined as the ratio of β to α,
\begin{align*} \text {Angular magnification, M} &= \frac {\beta }{\alpha }\end{align*} From the figure, we have$$ \alpha = \frac{h}{D}$$ where h is the height object. from next figure$$\beta = h’/D \text {where} h’ = \text {the height of the image. So,} $$\begin{align*} M &= \frac {\beta }{\alpha } = \frac {h’}{D}/\frac {h}{D} = \frac {h’}{h} \\ \text {But,} \: h’/h = v/u \end{align*}is the linear magnification produced by the lens. Using lens formula \begin{align*}\frac 1f = \frac 1u + \frac 1v, \text {we get} \\ \frac uv &= \frac vf – 1 \\ \text {Since} \: v = -D, \text {we have} \\ M &= \frac vf – 1 = - \frac Df – 1 \\ \text {So, the angular magnification, M} \: &= - \left ( \frac Df + 1\right ) . \end{align*}text {So the magnitude of angular magnification is\begin{align*} M = 1 + \frac df \\ \end{align*}
When the image is formed at infinity, the object must be placed at the focus. The magnifying power of the microscope in the case is given by
\begin{align*} M &= \frac {\beta }{\alpha } \\ &= \frac {(h/f)}{(h/D)} \\ &= \frac Df \\ \end{align*}
We see that smaller the focal length, greater will be the magnifying power of the microscope.
A compound microscope magnifies more than a simple microscope. It consists of two converging lenses arranged coaxially. The lens facing to the object is called the objective and that closer to the eye is called eyepiece.
The object OA, which is to be viewed, is placed just beyond the focus, F_{o} of objective and a real, inverted, magnified image IB, is formed at the other side of the objective. This image acts as the object for eyepiece which is adjusted that AI lies within focus F_{e} of the eyepiece. The final image I’N seen by the eye is virtual, erect and very much enlarged. The position of I’ N may be anywhere between the near point and far point of the eye, but the magnification is higher when the final image lies at least distance of distinct vision. The ray diagram of a compound microscope is shown in the figure.
The compound microscope produces total magnification in two steps: the objective produces the lateral magnification is the product of two factors, m×M_{e}.
When an object OA is seen by unaided eye, the visual angle subtended by it at the eye is
\begin{align*} \alpha &= \frac hD \\ \text {where } \: h = OA, \end{align*}height of object. When the compound microscope is used to view the object, the visual angle subtended at the eye is \begin{align*}\beta &= \frac {h_2}{D} \\ \end{align*}
where h_{2} = I’N, height of the final image. This angle can be written as β = h_{1}/ u_{e} , where h_{1} is the height of image IB and u_{e} is the object distance for eyepiece. The overall magnifying power of the instrument is
\begin{align*} M &= \frac {\beta }{\alpha } &= \frac {(h_1/u_e )}{(h/D)} &= \frac {h_1}{h} \times \frac {D}{u_e} \\ \text {From figure, we have} \\ \frac {h_1}{h} &= \frac {v_o}{u_o} \\ \end{align*}
where u_{o} = object distance and v_{o} = image distance for the objective. The ratio D/u_{o} is the magnifying power of the eyepiece treated as a simple microscope, and this is equal to –D/f_{e} in the normal adjustment and –[(D/F_{e}) + 1] when the image is obtained at least distance of distinct vision. Thus, the magnifying power of the compound microscope in normal adjustment is
\begin{align*} M &= \frac {v_o}{u_o} \frac {D}{f_e} \\ \end{align*}and for the adjustment with final image at near point,\begin{align*}M &= - \frac {v_o}{u_o} \left (\frac {D}{f_e} + 1 \right ) \\ \text {from the lens formula,} \: 1/f = 1/u_o + 1/ v_o , \text {we have} \\ \frac {v_o}{u_o} &= \frac {v_o}{f_o} – 1 \\ \text {and the magnifying power} \\ M &= -\left (\frac {V_o}{f_o} + 1 \right ) \left (\frac {D}{f_e} + 1 \right ) \\ \end{align*}Hence, final the magnifying image I’N is seen \begin{align*}F_o \text {and} F_e \: \text {are small} \\ \end{align*}
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