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Q. Focal length of objective of a compound microscope is $4 \,mm$ and the image is formed at a distance of $224 \,mm$ from it. If angular magnification is $550$ , then focal length (in cm) of eyepiece for normal adjustment is

Ray Optics and Optical Instruments

Solution:

$m_{0}=\frac{4-224}{4}=-55$
$m=m_{0} \times m_{e} $
$\Rightarrow-550=-55 \times m_{e} $
$\Rightarrow m_{e}=10$
$\frac{D}{f_{e}}=10 $
$\Rightarrow f_{e}=\frac{D}{10}=2.5\, cm$