Question

A compound microscope is 20 cm long and the power of its eye-piece is 20 D. If the objective is bi-convex (radius of curvature of either surface 2 cm) and made of a material of = 1.5, the magnifying power of the microscope is

• A. 54
• B. 60
• C. 13.5
• D. 41.5

Answer 1

Answer: (D)

: For objective lens, $ \frac{1}{{{f}_{0}}}=(\mu -1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)=(1.5-1)\times \frac{2}{R}=\frac{0.5\times 2}{2}=\frac{1}{2} $ Or $ {{f}_{0}}=2\,cm $ ...(i) $ {{f}_{c}}=\frac{100}{D}=\frac{100}{20}=5\,cm $ For eyepiece, $ \frac{1}{{{v}_{e}}}-\frac{1}{{{u}_{e}}}=\frac{1}{{{f}_{e}}} $ Final image is at 25 cm from it. $ \therefore $ $ \frac{-1}{25}-\frac{1}{{{u}_{e}}}=\frac{1}{5} $ or $ {{u}_{e}}=-\frac{25}{6}cm= $ Image is on the object (virtual) side of lens. $ \therefore $ Tube length $ ={{u}_{e}}+{{v}_{0}} $ or $ {{v}_{0}}=20-\frac{25}{6}=\frac{95}{6}cm $ ...(iii) For objective lens, $ \frac{1}{{{v}_{0}}}-\frac{1}{{{u}_{0}}}=\frac{1}{{{f}_{0}}} $ Or $ \frac{1}{{{u}_{0}}}=\frac{1}{{{v}_{0}}}-\frac{1}{{{f}_{0}}}=\frac{6}{95}-\frac{1}{2}=\frac{12-95}{190}=\frac{-83}{190} $ Or $ {{u}_{0}}=\frac{-190}{83}cm $ $ \therefore $ Magnifying power of microscope $ =\frac{{{v}_{0}}}{{{u}_{0}}}\left( 1+\frac{D}{{{f}_{e}}} \right) $ Here final image is formed at least distance of distinct vision (25 cm) from eyepiece. The final image is virtual and magnified.
$ M=\frac{{{v}_{0}}}{{{u}_{0}}}\left( 1+\frac{D}{{{f}_{e}}} \right) $ $ M=\frac{95/6}{190/83}\left( 1+\frac{25}{5} \right)=\frac{95}{6}\times \frac{83}{190}(1+5) $ $ =\frac{83\times 6}{6\times 2}=41.5 $ Magnifying power = 41.5