Question

A compound microscope has a magnifing
power of 100 when the image is formed
at infinity. The objective has a tocal length
of 0.5cm and the tubelength is 6.5cm..
Then the focal length of the eyepiece is?

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Answer 1

Answer:

${\bf{\green{For}}}$${\bf{\red{the}}}$${\bf{\blue{given}}}$${\bf{\pink{compound }}}$${\bf{\green{microscope, }}}$${\bf{\orange{Focal length=0.5cm}}}$,${\bf{\blue{tube length =6.5cm}}}$,

${\bf{\red{magnifying power }}}$ ${\bf{\green{=100}}}$

${\bf{\pink{since, }}}$${\bf{\blue{the}}}$${\bf{\red{image }}}$${\bf{\blue{is}}}$ ${\bf{\pink{formed }}}$${\bf{\green{at}}}$${\bf{\orange{infinity }}}$${\bf{\blue{, the}}}$${\bf{\red{real}}}$${\bf{\green{image}}}$${\bf{\pink{produced}}}$${\bf{\blue{by}}}$ ${\bf{\green{objective }}}$${\bf{\red{lens}}}$${\bf{\blue{should }}}$${\bf{\pink{lie}}}$${\bf{\green{on}}}$${\bf{\orange{the}}}$${\bf{\blue{focus }}}$${\bf{\red{of}}}$${\bf{\green{the}}}$${\bf{\pink{eye}}}$${\bf{\blue{piece.}}}$

So,

${\bf{\red{=>Vo+fe=6.5cm.. EQ1}}}$

${\bf{\blue{magnifying power}}}$ ${\bf{\pink{=Vo/uo×D/fe}}}$${\bf{\green{(for}}}$${\bf{\orange{normal}}}$${\bf{\blue{adjustment) }}}$

${\bf{\red{=>m=-[1-vo/fo]D/fe }}}$${\bf{\green{(since;}}}$${\bf{\pink{vo/uo=1-vo/fo) }}}$

${\bf{\blue{=>100=-[1-vo/0.5]25/fe}}}$ ${\bf{\green{(since;D=25cm)}}}$

${\bf{\red{=>100fe=(-1+vo/0.5)25}}}$

${\bf{\blue{=>4fe=(-1+vo/0.5)}}}$

${\bf{\pink{=>2fe=-0.5+vo}}}$

${\bf{\green{=>2fe-vo=-0.5...EQ2}}}$

${\bf{\orange{adding EQ1 and EQ2}}}$

${\bf{\blue{=>vo+fe+2fe-vo=6.5-0.5}}}$

${\bf{\red{=>3fe=6}}}$

${\bf{\green{=>fe=6/3}}}$

${\bf{\pink{=>fe=2cm}}}$

${\bf{\blue{sub,}}}$${\bf{\red{fe=2cm}}}$${\bf{\blue{in}}}$${\bf{\pink{EQ1}}}$

${\bf{\green{then,}}}$

${\bf{\orange{=>vo+2cm=6.5}}}$

${\bf{\blue{=>vo=6.5-2cm}}}$

${\bf{\red{=>Vo=4.5cm}}}$

${\bf{\green{so}}}$${\bf{\pink{the}}}$${\bf{\blue{focal}}}$ ${\bf{\green{length }}}$${\bf{\red{of}}}$${\bf{\blue{eye}}}$${\bf{\pink{piece }}}$${\bf{\green{fe}}}$${\bf{\orange{=}}}$${\bf{\blue{2cm}}}$

Answer 2

Avnakka nene draw chesa.........