question_answer

A compound microscope uses an objective lens of focal length 4cm and eyepiece lens of focal length 10cm. An object is placed at 6cm from the objective lens.

Calculate the magnifying power of the compound microscope. If final image is formed at the least distance of distinct vision. Also, calculate the length of microscope.

Answer:

Given that, \[{{f}_{e}}=10\,\,cm,\,\,{{f}_{o}}=4\,\,cm,\,\,u=-\,6\,\,cm\] Use lens formula for objective lens, \[\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\]             \[\Rightarrow \]               \[\frac{1}{v}-\frac{1}{(-\,6)}=\frac{1}{4}\]             \[\Rightarrow \]               \[\frac{1}{v}=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\]             \[\Rightarrow \]               \[v=12\,\,cm\] Magnifying power, \[m=\frac{-{{v}_{0}}}{{{u}_{0}}}\,\,\left( 1+\frac{D}{{{f}_{e}}} \right)\] Adjusting \[\therefore \]      \[m=\frac{-12}{6}\,\,\left( 1+\frac{25}{10} \right)=\frac{-\,2\times 35}{10}=-7\] Negative sign shows that the image is inverted. Length of the microscope is given by, \[L=\left| {{v}_{0}} \right|+\left| {{u}_{e}} \right|\] For eyepiece of the microscope, \[\frac{1}{{{f}_{e}}}=\frac{1}{{{v}_{e}}}-\frac{1}{{{u}_{e}}}=\frac{1}{{{u}_{e}}}=\frac{1}{{{v}_{e}}}-\frac{1}{{{f}_{e}}}=-\frac{1}{25}-\frac{1}{10}\] We have \[{{v}_{e}}=D=-\,25\,\,cm\] \[\Rightarrow \]   \[{{u}_{e}}=\frac{-\,50}{7}\,\,cm=-7.14\] \[\therefore \]    \[L=12+7.14=19.14\text{ }cm\]