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Draw a neat labelled diagram of a compound microscope. Derive the magnifying power for it. A telescope has an objective of focal length 140cm and an eyepiece of focal length 5cm. Find the magnifying power and separation between objective and eyepiece.

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Last updated date: 13th Jun 2024
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Answer
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Hint: Magnification depends on focal length of objective and the eye lens. The least distance of distinct vision is 25 cm. The separation between the objective and eyepiece is given by the length of tube.

Complete Step by Step Solution
Real and magnified images of minuscule particles or objects can be achieved using a combination of lenses. A compound microscope is an intricate gathering of a combination of lenses that renders a highly maximized and magnified image of microscopic living entities and other complex details or tissues and cells.
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1. Foot or base: It is a U-shaped structure and supports the entire weight of the compound microscope.
2. Pillar: It is a vertical projection. This stands by resting on the base and supports the stage.
3. Arm: The entire microscope is handled by a strong and curved structure known as the arm.
4. Stage: The flat and rectangular plate that is connected to the arm’s lower end is called the stage. The specimen is placed on the stage for studying and examining the various features. The centre of the stage has a hole through which light can pass.
5. Inclination joint: It is a joint, wherein the arm is fastened to the compound microscope’s pillar. The microscope can be tilted using the inclination joint.
6. Clips: The upper part of the stage is connected to two clips. The slide can be held in its position with the help of the clips.
7. Diaphragm: The diaphragm is fastened below the stage. It controls and adjusts the intensity of light that passes into the microscope. The diaphragm can be of two types:
Disc diaphragm
Iris diaphragm
8. Nose piece: The nose piece is circular and a rotating metal part that is connected to the body tube’s lower end. The nose piece has three holes wherein the objective lenses are embedded.
9. Body tube: The upper part of the arm of the microscope comprises a hollow and tubular structure known as the body tube. The body tube can be shifted down and up using the adjustment knobs.
10. Fine adjustment knob: It is the smaller knob, which is used for sharp and fine focusing of the object. For accurate and sharp focusing, this knob can be used.
11. Coarse adjustment knob: It is a large knob that is used for moving the body tube down and up for bringing the object to be examined under exact focus.
The magnifying power of a compound microscope is the ratio of the angle subtended at the eye by the image to the angle subtended by the object at the unaided eye, when both are placed at the least distance of distinct vision It is denoted by ' $ M $ ' and is given as $ M = 1 + {D \mathord{\left/
 {\vphantom {D f}} \right.} f} $ where $ M $ is the magnification, $ D $ is the least distance of the distinct vision and is equal to image distance ' $ v $ ', $ f $ is the focal length.
Since, $ D $ is constant so magnifying power depends upon the focal length of the lens. Smaller the focal length, greater will be the magnifying power of the lens.
Let,
 $ {f_{\left( e \right)}} $ = Focal length of the eye piece
 $ {f_{\left( o \right)}} $ = Focal length of the objective = object distance $ u $ .
 $ {M_{\left( e \right)}} $ = Magnification of eyepiece
 $ {M_{\left( o \right)}} $ = Magnification of objective
 $ v = L $ = Length of the microscope tube.
Hence, $ M = {M_{\left( e \right)}} \times {M_{\left( o \right)}} $ .
Hence,
 $ M = {v \mathord{\left/
 {\vphantom {v {u\left( {1 + {D \mathord{\left/
 {\vphantom {D {{f_e}}}} \right.} {{f_e}}}} \right)}}} \right.} {u\left( {1 + {D \mathord{\left/
 {\vphantom {D {{f_e}}}} \right.} {{f_e}}}} \right)}} $
 $ M = {L \mathord{\left/
 {\vphantom {L {{F_{\left( o \right)}}}}} \right.} {{F_{\left( o \right)}}}}\left( {1 + {D \mathord{\left/
 {\vphantom {D {{f_e}}}} \right.} {{f_e}}}} \right) $
It has been given that a telescope has an objective of focal length 140cm and an eyepiece of focal length 5cm.
Thus, $ {f_o} = 5cm $ and $ {f_e} = 140cm $ .
Magnifying Power $ m = - \dfrac{{{f_o}}}{{{f_e}}} = - \dfrac{{140}}{5} = - 28cm $ when the telescope is in normal adjustment.
When the final image is formed at the least distance of distinct vision, $ m = - \dfrac{{{f_o}}}{{{f_e}}}\left( {1 + \dfrac{{{f_0}}}{D}} \right) $ .
We know that the least distance of distinct vision is 25 cm.
Given that, $ {f_o} = 5cm $ , $ {f_e} = 140cm $ and $ D = 25cm $ . Assigning the given values in the above equation, we have,
 $ = - \dfrac{{140}}{5}\left( {1 + \dfrac{5}{{25}}} \right) = - 33.6cm $
The distance between two lenses=length of tube = $ L = {f_0} + {f_e} = 140 + 5 = 145cm. $

Note
The Optical Parts of Compound Microscope include:
1. Eyepiece lens or Ocular: At the top of the body tube, a lens is planted which is known as the eyepiece. On the rim of the eyepiece, there are certain markings such as 5X, 10X, 15X, etc. Which indicates the magnification power. The object’s magnified image can be observed with the help of an eyepiece.
2. Mirror: A mirror is found attached either to the pillar or the lower end of the arm. It consists of a concave mirror on one side and a plain mirror on the other side. It can be used for reflection of light rays into the microscope.
3. Objective lenses: At the bottom of the body tube, there are two objective lenses, which are connected to the revolving nosepiece.