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The separation L between the objective (f=0.5cm) and the eyepiece (f=5cm) of a compound microscope is 7cm. The angular magnification produced by this microscope when eye is least strained is:

Option: 1

-5


Option: 2

-10


Option: 3

-15


Option: 4

-20


Answers (1)

best_answer

As we learn

Compound Microscope -

m= -\frac{v_{o}}{u_{o}}\cdot \frac{D}{u_{e}}
 

- wherein

v_{o}\, and \, u_{o} is the distance from the objective.

u_{e} Distance from the eyepiece.

Maximum magnification m= -\frac{v_{o}}{u_{0}}\left ( 1+\frac{D}{f_{e}} \right )

 

 Since the eye is least strained hence the final image will form at infinity. In such a case image formed by the object should form at the focus of the eyepiece.

v_{0}=7cm-5cm=2cm                    f_{0}=0.5cm

\frac{1}{u}=\frac{1}{v_{0}}-\frac{1}{f_{0}}=\frac{1}{2}-\frac{1}{0.5}=\frac{-3}{2}

or u=\frac{-2}{3}cm

Angular magnification m=\frac{v_{0}}{u_{0}}\cdot \frac{D}{f_{e}}=\frac{-25}{5} =-15cm

                  

 

Posted by

Deependra Verma

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