If in a compound microscope $m_1$ and $m_2$ are the linear magnifications of the objective lens and the eye lens respectively,then the magnifying power of the compound microscope will be:

  • A
    $m_1 - m_2$
  • B
    $\sqrt{m_1 + m_2}$
  • C
    $(m_1 + m_2) / 2$
  • D
    $m_1 \times m_2$

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What is the tube length of a compound microscope?

$A$ compound microscope is used to enlarge an object kept at a distance $0.03\,m$ from its objective,which consists of several convex lenses in contact and has a focal length of $0.02\,m$. If a lens of focal length $0.1\,m$ is removed from the objective,then by what distance must the eyepiece of the microscope be moved to refocus the image? (in $cm$)

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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 . . . . . . .

$A$ 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 $9\,cm$) held close to the eye. What is the angular magnification (magnifying power) of the lens?

The magnifying power of a compound microscope increases when

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