The total magnification produced by a compound microscope is $24$ when the final image is formed at the least distance of distinct vision. If the focal length of the eyepiece is $5 \ cm$, the magnification produced by the objective is

  • A
    $4$
  • B
    $4.8$
  • C
    $120$
  • D
    $6$

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$A$ microscope has an objective of focal length $1 \ cm$ and an eye-piece of focal length $6 \ cm$. If the tube length is $30 \ cm$ and the image is formed at the least distance of distinct vision,what is the magnification produced by the microscope? Take $D = 25 \ cm$.

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In a microscope, the objective has a focal length $f_0 = 2 \ cm$ and the eyepiece has a focal length $f_e = 4 \ cm$. The tube length is $32 \ cm$. The magnification produced by this microscope for normal adjustment is . . . . . . .

Obtain the equation of magnification for a compound microscope.

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