Question

Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope have short focal lengths? Explain.

Answer 1

Hint: Recall the formula for magnifying power of the compound microscope when the image is placed at infinity. Then we can define the magnifying power of the compound microscope using the formula you derived. Using the formula for magnification of the compound microscope, determine on which factors the magnification of compound microscope depends.

Formula used:
\[M = - \dfrac{l}{{{f_o}}}\left( {1 + \dfrac{D}{{{f_e}}}} \right)\]
Here, l is the length of the tube, \[{f_o}\] is the focal length of the objective lens, D is the distance of distinct vision and \[{f_e}\] is the focal length of the eyepiece lens.

Complete step by step answer:
The magnifying power of a compound microscope is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the eye. We can define the magnifying power of the compound microscope when the final image is formed at infinity as the product of magnification produced by the objective lens and the magnification produced by the eyepiece lens. We know the formula for magnifying power of the compound microscope is,
\[M = - \dfrac{l}{{{f_o}}}\left( {1 + \dfrac{D}{{{f_e}}}} \right)\]…… (1)
Here, l is the length of the tube, \[{f_o}\] is the focal length of the objective lens, D is the distance of distinct vision and \[{f_e}\] is the focal length of the eyepiece lens.
When the image forms at infinity, we can write the magnification of eyepiece as,
\[{m_e} = \dfrac{D}{{{f_e}}}\]
Therefore, we can write equation (1) as,
\[M = - \dfrac{l}{{{f_o}}} \times \dfrac{D}{{{f_e}}}\] …… (2)
But we know that the term, \[ - \dfrac{l}{{{f_o}}}\] is the magnification of objective lens and the term \[\dfrac{D}{{{f_e}}}\] is the magnification of eyepiece lens. Therefore, we can write the above equation as,
\[M = {m_o} \times {m_e}\]
So, the magnification of the compound microscope when the image is formed at infinity is defined as the product of magnification produced by the objective lens and the magnification produced by the eyepiece lens.
From equation (2), we have,
\[M \propto \dfrac{l}{{{f_o}}}\,\,{\text{and }}M \propto \dfrac{D}{{{f_e}}}\]
From the above equation, we have, the magnification of the compound microscope is inversely proportional to the focal length of the objective lens and focal length of the eyepiece lens.

So, the magnification will be greater for the microscope with short focal lengths of objective and eyepiece lens.

Note:The image formed by the eyepiece lens is virtual and enlarged. It forms far from the eye because the eye cannot focus the object at a distance less than the distance of distinct vision. We see the magnification is proportional to the length of the tube of compound microscope and distance of distinct vision. We should note that we cannot increase the length of the tube. Also, the distance of distinct vision is the constant value and its value is 25 cm.