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Compound Microscope - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • Compound Microscope is considered one of the most asked concept.

  • 9 Questions around this concept.

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A compound microscope has an eyepiece of focal length \mathrm{10\ cm} and an objective of focal length \mathrm{4\ cm}. Calculate the magnification, if an object is kept at a distance of \mathrm{5\ cm} from the objective, so that the final image is formed at the least distance of distinct vision \mathrm{20\ cm}:

A compound microscope has an eyepiece of focal length\mathrm{10\ cm} and an objective of focal length \mathrm{4\ cm}. Calculate the magnification, if an object is kept at a distance of \mathrm{5\ cm} from the objective so that the final image is formed at the least distance of distinct vision \mathrm{25\ cm}:

In a compound microscope, the focal lengths of two lenses are \mathrm{1.5\ cm} and \mathrm{6.25\ cm}. If an object is placed at \mathrm{2\ cm} from objective and the final image is formed at \mathrm{25\ cm} from the eye lens, the distance between the two lenses is:

Concepts Covered - 1

Compound Microscope

Compound Microscope: 

A compound microscope is of two converging lenses called objective and eye lens. It is used to view magnified images of small objects on a glass slide. It can achieve higher levels of magnification than stereo or other low power microscopes and reduce chromatic aberration.

       

f_{\text {eyelens }}>f_{\text {objective }} \text{and} \ (diameter) _{\text {eyelens }}>( diameter) _{objective }
 Intermediate image is real and enlarged. The final image is magnified, virtual and inverted.

Here in the diagram 

 u_o = Distance of object from objective (o), 

v_o= Distance of image (A′B′) formed by objective from objective, 

u_e= Distance of  A′B′ from eye lens, 

v_e= Distance of final image from eye lens, 

f_o=  Focal length of objective, 

f_e= Focal length of eye lens.

Case 1: Final image is formed at D :  Magnification m_{D}=\frac{v_{o}}{u_{o}}\left(1+\frac{D}{f_{e}}\right) and length of the microscope tube (distance between two lenses) is L_{D}=v_{e}+u_{e} .
Generally, object is placed very near to the principal focus of the objective hence u_{o} \equiv f_{o} . The eye piece is also of small focal length and the image formed by the objective is also very near the eyepiece.

So v_{o} \equiv L_{D},  the length of the tube.
Hence, we can write

m_{D}=\frac{L}{f_{o}}\left(1+\frac{D}{f_{e}}\right).
Case 2:  Final image is formed at \infty: Magnification
m_{\infty}=\frac{v_{0}}{u_{0}} . \frac{D}{f_{e}}$ and length of tube $L_{\infty}=v_{0}+f_{e}
In terms of length    m_{\infty}=\frac{\left(L_{\infty}-f_{o}-f_{e}\right) D}{f_{o} f_{e}}.

  •  For large magnification of the compound microscope, both f_{o} \ \text{and} \ f_{e} should be small.
  •  If the length of the tube of microscope increases, then its magnifying power increases.
  • The magnifying power of the compound microscope may be expressed asM=m_{o} \times m_{e} \ \text{where } m_{0}  is  the magnification of  the  objective and  m_{e} is magnifying the  of eyepiece.

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Compound Microscope

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Compound Microscope

Concepts of Physics Part-1

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