An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification '6', gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to __________ cm.

Given, magnification, M = 6<br/><br/>Since we know that magnifying power of a simple microscope is given by<br/><br/>$M = 1 + {D \over {{f_0}}}$<br/><br/>where, D = least distance of distinct vision = 25 cm<br/><br/>and f₀ = focal length of objective lens.<br/><br/>$$ \Rightarrow 6 = 1 + {D \over {{f_0}}} \Rightarrow 6 = 1 + {{25} \over {{f_0}}} \Rightarrow 5 = {{25} \over {{f_0}}} \Rightarrow {f_0} = 5$$ cm<br/><br/>For compound microscope, magnifying power is given by<br/><br/>$M = {{I\,.\,D} \over {{f_0}{f_e}}} = 2$M_{simple microscope}<br/><br/>where, f₀ = f_e are the focal lengths of the objective lens and eye piece respectively<br/><br/>and l = length of the given tube = 0.6 m<br/><br/>$\Rightarrow 12 = {{60 \times 25} \over {5\,.\,{f_e}}}$ [$\because$ magnification is doubled]<br/><br/>$\Rightarrow$ f_e = 25 cm<br/><br/>This is the required focal length of eyepiece.