
Magnifying power of a compound -microscope increases when,
A. The focal length of the objective lens is increased and that of the eyepiece is decreased.
B. The focal length of the eyepiece is increased and that of the objective lens is decreased.
C. Focal lengths of both objective and eyepiece are increased.
D. Focal length of both objective and eyepiece are decreased.
Answer
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Hint: Recall the image formation by the compound microscope. Using the formula for magnification of the compound microscope, determine on which factors the magnification of compound microscope depends.
Complete step by step answer:
We know the formula for magnification of compound microscope,
\[m = - \dfrac{l}{{{f_o}}}\dfrac{D}{{{f_e}}}\]
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.
We can write the proportionality of the terms in the above formula as,
\[m \propto \dfrac{1}{{{f_o}}}\] and, \[m \propto \dfrac{1}{{{f_e}}}\]
From the above formula, we can conclude that the magnifying power of the compound microscope increases when the focal lengths of both objective and eyepiece lenses decrease.
So, the correct answer is option (D).
Additional information: A compound microscope is constructed using two convex lenses: eyepiece lens and objective lens. The eyepiece lens is also called an ocular lens since it is closer to the eye. The objective lens is close to the object being examined.
The overall magnification of the compound microscope is due to individual magnification of eyepiece and objective lens. The image of the object formed by the objective lens is inverted and it behaves as the object for eyepiece lenses. The image formed by the eyepiece lens is virtual.
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 then 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.
Complete step by step answer:
We know the formula for magnification of compound microscope,
\[m = - \dfrac{l}{{{f_o}}}\dfrac{D}{{{f_e}}}\]
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.
We can write the proportionality of the terms in the above formula as,
\[m \propto \dfrac{1}{{{f_o}}}\] and, \[m \propto \dfrac{1}{{{f_e}}}\]
From the above formula, we can conclude that the magnifying power of the compound microscope increases when the focal lengths of both objective and eyepiece lenses decrease.
So, the correct answer is option (D).
Additional information: A compound microscope is constructed using two convex lenses: eyepiece lens and objective lens. The eyepiece lens is also called an ocular lens since it is closer to the eye. The objective lens is close to the object being examined.
The overall magnification of the compound microscope is due to individual magnification of eyepiece and objective lens. The image of the object formed by the objective lens is inverted and it behaves as the object for eyepiece lenses. The image formed by the eyepiece lens is virtual.
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 then 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.
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