Question

In a compound microscope the focal length of an objective lens is $1.2cm$ and focal length of eyepiece is $3cm$. When the object is kept at $1.25cm$ in front of the objective, the final image is formed at infinity. Magnifying power of the compound microscope should be:
A. 400
B. 200
C. 150
D. 100

Answer 1

Compound microscope is a system of lenses which produces enlarged images of the object. It follows lens law $\dfrac{1}{u_{o}}+\dfrac{1}{v_{o}}=\dfrac{1}{f_{o}}$ and magnification is given by $m=\dfrac{v_{o}}{u_{o}}(\dfrac{D}{f_{e}})$, where $D=25cm$ is the distance till which the relaxed eyes can see.

Formula used:
$\dfrac{1}{u_{o}}+\dfrac{1}{v_{o}}=\dfrac{1}{f_{o}}$
$m=\dfrac{v_{o}}{u_{o}}(\dfrac{D}{f_{e}})$

Complete step-by-step answer:
Given that,
Focal length of eyepiece=$f_{e}=3cm$,
Focal length of objective=$f_{o}=1.2cm$,
Object distance=$u_{o}=1.25cm$
Image distance=$v_{e}=\infty$
From lens formula, $\dfrac{1}{u_{o}}+\dfrac{1}{v_{o}}=\dfrac{1}{f_{o}}$
$\dfrac{1}{f_{o}}-\dfrac{1}{u_{o}}=\dfrac{1}{v_{o}}$
$\dfrac{1}{1.2}-\dfrac{1}{1.25}=\dfrac{1}{v_{o}}$
${v_{o}}=30cm$
Magnification of microscope $m=\dfrac{v_{o}}{u_{o}}(\dfrac{D}{f_{e}})$, where $D=25cm$ is the distance till which the relaxed eyes can see .
Substituting, we get
$m=\dfrac{30}{1.25}(\dfrac{25}{3})=200$,
Hence, the magnifying power of the compound microscope is 200. So the correct option is B.

Additional Information:
Microscopes are used to magnify objects, the compound microscope has two convex lenses, namely the objective lens and the eyepiece. Both contribute to the final magnification. The objective lens produces a magnification value of about $5\times \; to \; 100\times$ the original size. The object is focused by adjusting the two lenses to produce an enlarged image.
The object produces an image with certain magnification by the objective lens with $f_{o}$ , this image now acts a virtual object to the eyepiece with $f_{e}$ which in turn produces a virtual image, which is magnified further. The final image is generally inverted, and can be seen through the normal eyes.
Lens formula is the relationship between the distance of an object $u$, distance of image $v$ and the focal length of the lens $f$. This law can be used for both concave and convex lenses with appropriate sign conventions. The thickness of the lens is neglected.
Lens formula: $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
And magnification equation is given by $M=\dfrac{Height\; of \;image}{Height\; of \;object}=-\dfrac{distance\; of\; image}{distance\; of\; object}$ if$M=+$ then the image is magnified and if $M=-$ then image is diminished.
But here since two lenses are used the magnification is given by $m=\dfrac{v_{o}}{u_{o}}(\dfrac{D}{f_{e}})$.

Note: The compound microscope has two convex lenses, namely the objective lens and the eyepiece. Both contribute to the final magnification. The object produces an image with certain magnification by the objective lens with $f_{o}$ , this image now acts a virtual object to the eyepiece with $f_{e}$ which in turn produces a virtual image, which is magnified further.