$A$ compound microscope produces a magnification of $24$. The focal length of the eyepiece is $5 \ cm$. The final image is formed at the least distance of distinct vision. The magnification produced by the objective is
- A$4$
- B$5$
- C$6$
- D$7$
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If the focal length of the objective lens and the eye lens are $4 \, mm$ and $25 \, mm$ respectively in a compound microscope,and the length of the tube is $16 \, cm$,find its magnifying power for the relaxed eye position.
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View SolutionThe 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
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 . . . . . . .
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?
DifficultTS EAMCET 2007
View SolutionIn a compound microscope,the objective of focal length ${f_o}$ and the eyepiece of focal length ${f_e}$ are placed at a distance $L$ such that $L$ equals:
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