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Compound Microscope - Practice Questions & MCQ

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

Quick Facts

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

  • 10 Questions around this concept.

Solve by difficulty

When a beam of white light is allowed to pass through convex lens parallel to principal axis, the different colours of light converge at different point on the principle axis after refraction. This is called:

A compound microscope has an eye piece 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 eye piece 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 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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