In a compound microscope,let $u_0$ and $v_0$ be the object distance and image distance respectively. The objective of focal length $f_0$ magnifies a tiny object into a real,inverted image. The linear magnification of the objective is

  • A
    $\frac{f_0+u_0}{u_0 f_0}$
  • B
    $\frac{f_0}{f_0+u_0}$
  • C
    $\frac{u_0}{u_0 f_0+1}$
  • D
    $\frac{u_0 f_0}{f_0+u_0}$

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Similar Questions

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: For a simple microscope,the angular size of the object equals the angular size of the image.
Reason $R$: Magnification is achieved as the small object can be kept much closer to the eye than $25\, cm$ and hence it subtends a large angle.
In the light of the above statements,choose the most appropriate answer from the options given below:

$A$ simple microscope is used to see an object first in blue light and then in red light. Due to the change from blue to red light, what is the effect on its magnifying power?

In a compound microscope,the objective lens and eyepiece have focal lengths of $0.95 \ cm$ and $5 \ cm$ respectively,and are kept at a distance of $20 \ cm$. The final image is formed at a distance of $25 \ cm$ from the eyepiece. Calculate the magnifying power.

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In a Ramsden eyepiece, two plano-convex lenses, each of focal length $f$, are separated by a distance of $d = 2f/3$. If the separation distance is $12 \ cm$, what is the equivalent focal length (in $cm$) of the eyepiece?

$A$ card sheet divided into squares each of size $1 \, mm^{2}$ is being viewed at a distance of $9 \, cm$ through a magnifying glass (a converging lens of focal length $10 \, cm$) held close to the eye.
$(a)$ What is the magnification produced by the lens? How much is the area of each square in the virtual image?
$(b)$ What is the angular magnification (magnifying power) of the lens?
$(c)$ Is the magnification in $(a)$ equal to the magnifying power in $(b)$? Explain.

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