In a compound microscope,the focal length of | vedclass

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(A) Given: $f_{o} = 1.2 \, cm$,$f_{e} = 3.0 \, cm$,$u_{o} = -1.25 \, cm$.For the objective lens,using the lens formula $\frac{1}{f_{o}} = \frac{1}{v_{

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$200$

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(A) Given: $f_{o} = 1.2 \, cm$,$f_{e} = 3.0 \, cm$,$u_{o} = -1.25 \, cm$.For the objective lens,using the lens formula $\frac{1}{f_{o}} = \frac{1}{v_{o}} - \frac{1}{u_{o}}$:$\frac{1}{1.2} = \frac{1}{v_{o}} - \frac{1}{-1.25}$$\frac{1}{v_{o}} = \frac{1}{1.2} - \frac{1}{1.25} = \frac{1.25 - 1.2}{1.5} = \frac{0.05}{1.5} = \frac{1}{30}$So,$v_{o} = 30 \, cm$.The magnifying power for the final image at infinity is given by $M = -\frac{v_{o}}{u_{o}} 	imes \frac{D}{f_{e}}$,where $D = 25 \, cm$ is the least distance of distinct vision.$M = -\frac{30}{-1.25} 	imes \frac{25}{3.0}$$M = 24 	imes 8.333 = 200$.Thus,the magnifying power is $200$.

In a compound microscope,the focal length of the objective lens is $1.2 \, cm$ and the focal length of the eyepiece is $3.0 \, cm$. When the object is kept at $1.25 \, cm$ in front of the objective,the final image is formed at infinity. The magnifying power of the compound microscope is:

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

The intermediate image formed by the objective of a compound microscope is

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