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  • (a) Draw a labelled ray diagram of compound microscope, when final image forms at the least distance of distinct vision. (b) Why is its objective of short focal length and of short aperture, compared to its eyepiece? Explain. (c) Th

(a) Draw a labelled ray diagram of compound microscope, when final image forms at the least distance of distinct vision.

(b) Why is its objective of short focal length and of short aperture, compared to its eyepiece? Explain.

(c) The focal length of the objective is 4 cm while that of eyepiece is 10 cm. The object is placed at a distance of 6 cm from the objective lens.

(i) Calculate the magnifying power of the compound microscope, if its final image is formed at the near point.

(ii) Also, calculate length of the compound microscope.

 

 

 
 

Answers (1)

(a)

(b) The focal length of the objective and its aperture should be short,

to have high magnifying power and high resolution.

\because m=m_{0}\times m_{e}

(c) (i) To calculate the magnifying power of a compound microscope.

  We have given,

u_0=6

f _{0}=4

f _{e}=10

Thus, for the objective lens ;

 u_0=-6

Then,

 \frac{1}{\nu _{0}}=\frac{1}{4}+\frac{1}{-6}

\frac{1}{\nu _{0}}=\frac{1}{4}+\frac{1}{-6}

\frac{1}{\nu _{0}}=\frac{1}{4}+\frac{1}{-6}

\nu _{0}=12\; cm

Since for eyepiece (\nu _{0})=-d=-25\; cm

\frac{1}{u _{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}

u _{e}=-7.14\; cm    

Hence,

 m=\frac{\vartheta _{0}}{u_{0}}\left ( 1+\frac{D}{f_{e}} \right )  (formula of magnifying power) m=\frac{12}{6}\left ( 1+\frac{25}{10} \right ) =7

 m=7 it is the magnifying power of the compound microscope.

(ii) The length of the microscope can be determined as;-

        L=|\nu _{0}|+\left | u _{e} \right |

        L=12+7.14

        L=19.14\; cm

Posted by

Safeer PP

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