Contents
Many modern technologies, such as computers and smartphones, are built on the principles of Physics Topics such as quantum mechanics and information theory.
What are the Factors on Which Magnification of a Compound Microscope Depends?
The instrument helps us see very small objects, which are otherwise not visible to the naked eye. It is of two types-
- Simple microscope or magnifying glass and
- Compound microscope.
Simple Microscope or Magnifying Glass
Description and working principle: A simple micro-scope or a magnifying glass is actually a convex lens of short focal length. We know that if an object is placed within the focal length of a convex lens then an erect, virtual and magnified image is formed at the same side where the object is placed. In Fig. L is a convex lens. Object PQ is placed perpendicular to the principal axis within the focal length of the lens. In this case a virtual, erect and magnified image pq is formed which will be seen by an eye placed close behind the lens. The distance of the object from the lens is so adjusted that the image is formed at the least distance of distinct vision (D = 25 cm) from the eye.
Magnification: In Fig., object PQ is placed such that its image pq is formed at the least distance of distinct vision. If the lens is thin and the eye is placed very close to the lens then the visual angle β will be given by,
β = ∠pOq = tan∠pOq = \(\frac{p q}{O q}\)
[since β is very small, tanβ = β]
Now, to observe the object distinctly without lens it must be kept at the least distance of distinct vision i.e., at P1q. In that case if the object subtends the visual angle α at the eye, then
α = ∠P1Oq = tan∠P1Oq = \(\frac{P_1 q}{O q}\)
∴ Angular magnification,
[D = least distance of distinct vision or the image distance and u = object distance]
From the equation of the lens we have,
\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{D}{-v}\) – \(\frac{D}{-u}\) = \(\frac{D}{f}\) ….. (2)
[Here D and v are negative and f is positive]
or, \(\frac{D}{u}\) = \(\frac{D}{v}\) + \(\frac{D}{f}\)
∴ m = \(\frac{D}{u}\) = \(\frac{D}{v}\) + \(\frac{D}{f}\) ….. (3)
Hence, it is observed that value of magnification is not constant. Magnification depends on the image distance.
i) If image is formed at near ppoint, in that case, v = D
∴ Magnification, m = \(\frac{D}{D}\) + \(\frac{D}{f}\) = 1 + \(\frac{D}{f}\) …. (4)
ii) If image is formed at infinity, in that case, v = ∞
∴ Magnification, m = \(\frac{D}{\infty}\) + \(\frac{D}{f}\) = \(\frac{D}{f}\) …. (5)
∴ Maximum value of m = 1 + \(\frac{D}{f}\) and minimum value of m = \(\frac{D}{f}\)
Hence, according to the position of image, the magnification lies between (1 + \(\frac{D}{f}\)) and \(\frac{D}{f}\).
For maximum magnification image forms at near point.
For normal eye D is equal to 25 cm. In that case,
m = 1 + \(\frac{25}{f}\)
But D is not equal to 25 cm for all. So, for different observers the magnifying power of the same microscope may be different.
From equation (3) it is seen that magnification may be increased by diminishing the focal length although not decreasing it much. In that case the lens will be very thick and the image will be indistinct and distorted. Hence, the magnification of a simple microscope is limited.
Uses: A magnifying glass is used by
- persons suffering from presbyopia to read small letters.
- biology students to see specimen slides,
- watch repairers to locate defects,
- detective department to match finger prints.
Numerical Examples
Example 1.
If an object is placed at a distance of 5 cm from a convex lens, a real image of the object is formed at a distance of 20 cm from the lens. If the lens is used as a magnifying glass what maximum magnification can be obtained from it? Least distance of distinct vision is 24 cm.
Solution:
Here u = -5 cm and v = 20 cm.
If f be the focal length of the lens we have,
\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1}{20}\) – \(\frac{1}{-5}\) = \(\frac{1}{f}\) or, f = 4 cm
So, focal length of the lens is 4 cm.
Maximum magnification of the magnifying glass,
∴ m = 1 + \(\frac{D}{f}\) = 1 + \(\frac{24}{4}\) = 7
Example 2.
A watch repairer kept a magnifying glass very close to his eyes and found that the magnifying power of the glass was 8. If the least distance of distinct vision of the eye is 25 cm, calculate the focal length of the lens.
Solution:
Here m = 8 and D = 25 cm
∴ m = 1 + \(\frac{D}{f}\) or, 8 = 1 + \(\frac{25}{f}\)
∴ f = \(\frac{25}{7}\) = 3.57 cm
Example 3.
A convex lens of power 10 m-1 is used as a simple magnifying glass. What are the maximum and minimum magnification of the lens?
Solution:
Power of lens,
P = \(\frac{100}{f}\) or, f = \(\frac{100}{P}\)
In this case, P = 10 m-1
∴ f = \(\frac{100}{10}\) = 10 cm
Maximum magnification = 1 + \(\frac{D}{f}\) = 1 + \(\frac{25}{10}\) = 3.5
Minimum magnification = \(\frac{D}{f}\) = \(\frac{25}{10}\) = 2.5
Compound Microscope
The magnification produced by a simple microscope is limited. If the object is very small, a simple microscope cannot sufficiently magnify it. For larger magnification compound microscope is to be used. Galelio invented this instrument in the seventeenth century.
Description: In this instrument two convex lenses of short focal lengths are placed in a tube at a certain distance apart so as to have a common axis[Fig.]. The convex lens O which faces the object to be viewed is known as the objective. The other lens E near the eye is known as the eyepiece. The objective has smaller focal length and aperture than those of the eyepiece. The tube with the lenses can be moved parallel to the axis towards the object or away from it. There is an arrangement to change the distance between the objective and the eyepiece.
Working principle: Fig. illustrates the working principle of the compound microscope. Fo and Fe are the foci of the objective and the eyepiece.
PQ is a small object. The distance of PQ from O is slightly greater than OF0. The objective forms a real, inverted and magnified image P1Q1 in front of the eyepiece. The position of the eyepiece is such that the image P1Q1 is formed within its first principal focus Fe ᐧ P1Q1 acts as an object to the eyepiece and it forms a magnified virtual image pq of P1Q1.
This pq is the final image which is virtual, inverted and magnified with respect to the object. The distance between the objective and the eyepiece is so adjusted that the final image is formed at the least distance of distinct vision from the eye. This process is called focussing of the microscope. In this case a highly magnified image is seen without any strain in the eye. The magnification becomes maximum if the final image is formed at the near point of the eye.
Magnification: In the compound microscope magnification takes place in two steps; first by the objective and then by the eyepiece. If the magnification produced by the objective be m0 and the produced by the eyepiece be me then magnification of the microscope,
m = m0 ᐧ me ….. (1)
If the distance of the object PQ from the objective be u, and the distance of the image P1Q1 formed by the objective be v, then
m0 = \(\frac{P_1 Q_1}{P Q}\) = \(\frac{v}{u}\) …… (2)
As the eyepiece acts as a magnifying glass and the final image is formed at the near point,
me = 1 + \(\frac{D}{f_e}\) ….. (3)
[D = least distance of distict vision and fe = focal length of the eyepiece]
From equations (1), (2) and (3) we have,
m = \(\)(1 + \(\frac{D}{f_e}\)) …. (4)
Now applying the general equation of lens for the objective we have,
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f_o}\) [f0 = focal length of the objective]
or, 1 + \(\frac{v}{u}\) = \(\frac{v}{f_o}\)
or, \(\frac{v}{u}\) = \(\frac{v}{f_o}\) – 1 ….. (5)
So, magnification of compound microscope,
m = (\(\frac{v}{f_0}\) – 1)(1 + \(\frac{D}{f_e}\)) …. (6)
Since the image P1Q1 of the compound microscope is formed at the end of the tube, v = length of the tube = L (approx.)
∴ m = (\(\frac{L}{f_o}\) – 1)(1 + \(\frac{D}{f_e}\)) …. (7)
Since f0 \(\ll\), fe \(\ll\) D
so from equation(7) we have,
m = \(\frac{L}{f_o} \cdot \frac{D}{f_e}\) …. (8)
Dependence of magnification on various factors: Equation (8) indicates that magnification produced by the compound microscope depends on the following factors:
- Length of the tube (L)
- Focal length of the objective (f0)
- Focal length of the eyepiece (fe)
However, the length of the tube cannot be increased indiscriminately as it will make handling the instrument difficult. For optimum use and advantage the length of the tube is maintained within 20-25 cm.
Again, if the magnification of the image is very large, its brightness decreases considerably. So to obtain sufficiently bright and magnified Image the object is to be Illuminated with a separate bright light.
To obtain the image free from the defects like spherical abberation, chromatic abberation etc. more than one combination of lenses are used for objective and eyepiece.
Uses: In laboratories compound microscopes are widely used for different examinations; e.g., to examine blood in medical science, to study different cells for Botany and Zoology etc.
Numerical Examples
Example 1.
The focal lengths of objective and eyepiece of a compound microscope are 0.5 cm and 1.5 cm respectively. If the least distance of distinct vision is 25 cm and magnification is 500 then what is the distance between objective and eyepiece?
Solution:
Here f0 = 0.5 cm, fe = 1.5 cm, D = 25 cm
Let L = distance between eyepiece and objective.
we know that magnification of compound microscope,
m = \(\frac{L D}{f_o f_e}\) or, 500 = \(\frac{L \times 25}{0.5 \times 1.5}\) or, L = 15 cm
Example 2.
The focal length of the two lenses of a compound microscope are 0.5 cm and 1 cm respectively. An object is placed at a distance of 1 cm from the objective. If the final image of the object is formed at a distance of 25 cm from the eye, what is the distance between the two lenses and the magnifying power of the microscope?
Solution:
In the compound microscope the focal length of the eyepiece is greater than that of the objective. So f0 = 0.5 cm and fe = 1 cm. For the objective u = -1 cm.
If v be the image distance, then according to the equation of the lens,
\(\frac{1}{v}\) + \(\frac{1}{1}\) = \(\frac{1}{0.5}\) or \(\frac{1}{v}\) = -1 + 2 = 1 or v = 1 cm
So, the image is formed on the other side of the objective at a distance of 1 cm.
For the eyepiece, ve = – 25 cm (the image is virtual and formed at near point) and fe = 1 cm
We have from the equation of the lens,
\(-\frac{1}{25}\) – \(\frac{1}{u_e}\) = \(\frac{1}{1}\) or, \(\frac{1}{u_e}\) = \(\frac{1}{-25}\) – 1 = –\(\frac{26}{25}\)
or, ue = –\(\frac{25}{26}\) = -0.96 cm
∴ |ue| = 0.96 cm
∴ The distance between the two lenses
= |v| + |ue| = 1 + 0.96 = 1.96 cm
The magnifying power of the microscope,
m = \(\frac{v}{u}\)(1 + \(\frac{D}{f_e}\)) = \(\frac{1}{1}\)(1 + \(\frac{25}{1}\)) = 26
Example 3.
The focal length of the objective and the eyepiece are 1 cm and 4 cm respectively. The distance between them is 14.5 cm. If an object of height 1 mm is placed at a distance of 1.1 cm from the objective what will be the position and the size of the image seen through the microscope?
Solution:
Here f0 = 1 cm and fe = 4 cm.
For the objective, u = -1.1 cm. If the image of the object is formed at a distance v, we have, from the equation of the lens,
\(\frac{1}{v}\) + \(\frac{1}{1.1}\) = \(\frac{1}{1}\) or, \(\frac{1}{v}\) + \(\frac{10}{11}\) = 1 or, v = 11 cm
So, the image formed by the objective is at a distance 11 cm on the other side of the objective and this image is real. This image acts as an object to the eyepiece.
∴ m1 = \(\frac{v}{u}\) = \(\frac{11}{1.1}\) = 10
Now, the object distance relative to the eyepiece
= -(14.5 – 11) = -3.5 cm
If the image is formed at a distance of V from the eyepiece we have from the equation of the lens,
\(\frac{1}{V}\) + \(\frac{1}{3.5}\) = \(\frac{1}{4}\) or, \(\frac{1}{V}\) = –\(\frac{2}{7}\) + \(\frac{1}{4}\) = –\(\frac{1}{28}\)
or, V = -28 cm
This image is virtual and will be formed at a distance of 28 cm in front of the eyepiece.
∴ m2 = \(\frac{28}{3.5}\) = 8
∴ Magnification of the final image,
m = m1 × m2 = 10 × 8 = 80
∴ The size of the final image = 80 × 1 = 80 mm = 8 cm
Example 4.
The focal length of the objective and the eyepiece of a compound microscope are 1 cm and 5 cm respectively and the distance between the centres of the lenses is 15 cm. If the final image is formed at the least distance of distinct vision, what is the magnifying power of the microscope?
Solution:
For the eyepiece, v = least distance of distinct vision = -25 cm, fe = 5 cm, object distance = u.
According to the equation of the lens we have,
\(\frac{1}{-25}\) – \(\frac{1}{u}\) = \(\frac{1}{5}\) or, \(\frac{1}{u}\) = -(\(\frac{1}{25}\) + \(\frac{1}{5}\)) = –\(\frac{6}{25}\)
or, u = –\(\frac{25}{6}\) cm
So, the image formed by the objective is formed at a distance of \(\frac{25}{6}\) cm in front of the eyepiece.
Therefore, this image is formed behind the objective at a distance (15 – \(\frac{25}{6}\)) or, \(\frac{65}{6}\) cm i.e., the image distance of the object relative to the objective, v = \(\frac{65}{6}\) cm.
Therefore, total magnification of the compound microscope,
m = (\(\frac{v}{f_o}\) – 1)(1 + \(\frac{D}{f_e}\)) = \(\left(\frac{\frac{65}{6}}{1}-1\right)\)(1 + \(\frac{25}{5}\)) = \(\frac{59}{6}\) × 6 = 59
Example 5.
The focal lengths of the objective and the eyepiece of a compound microscope are 1 cm and 2 cm respectively and the distance between them is 12 cm. If the least distance of distinct vision of the observer is 25 cm, where should a small object be placed to see it?
Solution:
For the eyepiece, fe = 2 cm, v = 25 cm. So, from the equation of the lens we have,
\(\frac{1}{-25}\) – \(\frac{1}{u}\) = \(\frac{1}{2}\) or, \(\frac{1}{u}\) = -(\(\frac{1}{25}\) + \(\frac{1}{2}\)) = –\(\frac{27}{50}\)
or u = –\(\frac{50}{27}\) cm
So, the image formed by the objective is formed in front of the eyepiece at a distance \(\frac{50}{27}\) cm. Hence, the image is formed behind the objective at a distance (12 – \(\frac{50}{27}\)) or, \(\frac{274}{27}\) cm i.e., the image distance of the object relative to the objective, v = \(\frac{274}{27}\) cm.
The focal Length of the objective, f0 = 1 cm from the equation of the lens we have,
\(\frac{\frac{1}{274}}{27}\) – \(\frac{1}{u}\) = \(\frac{1}{1}\) or, \(\frac{27}{274}\) – \(\frac{1}{u}\) = 1
or, –\(\frac{1}{u}\) = 1 – \(\frac{27}{274}\) = \(\frac{247}{274}\) or, u = –\(\frac{274}{247}\) = -1.11 cm
∴ The required object distance = 1.11 cm.
Example 6.
An object is placed at a distance of 5 cm from the objective of a compound microscope. If the final image is formed at the least distance of distinct vision and coincides with the object then calculate the focal lengths of the objective and the eyepiece. Given that the least distance of distinct vision = 25 cm and the magnifying power of the instrument = 15.
Solution:
For the objective,
image distance = v, object distance = u
For the eyepiece, image distance = v1, object distance = u1
∴ According to the problem,
u = 5 cm, v1 = 25 cm
Since the object and the final image coincide, so the distance between the objective and the eyepiece, L = 25 – 5 = 20 cm.
∴ v + u1 = 20 …………. (1)
Again, total magnification of the instrument = magnification by the objective × magnification by the eyepiece
∴ 15 = \(\frac{v}{u}\) × \(\frac{v_1}{u_1}\) or, 15 = \(\frac{v}{5}\) × \(\frac{25}{u_1}\)
or, \(\frac{v}{u_1}\) = 3 ……… (2)
Solving equations (1) and (2) we have,
v = 15 cm and u1 = 5 cm
For the objective,
u = -5 cm and v = +15 cm [∵ image is real]
According to the equation of lens we have,
\(\frac{1}{15}\) – \(\frac{1}{-5}\) = \(\frac{1}{f_o}\) or, f0 = 3.75 cm
For the eyepiece,
u1 = -5 cm and v1 = -25 cm [as image is virtual]
According to the equation of lens we have,
\(\frac{1}{-25}\) – \(\frac{1}{-5}\) = \(\frac{1}{f_e}\) or fe = 6.25 cm
So, focal length of the objective and eyepiece are 3.75 cm and 6.25 cm respectively.