What is the magnifying power of a compound microscope whose eyepiece has a focal length of 10 cm and the magnifying power of objective is 4? (Use least distance of distinct vision as 25 cm)

Concept-Magnifying power (m) –The magnifying power of a compound microscope is defined as the ratio of the ANGLES subtended by the image and the object at the eye when both are at the least distance of distinct vision from the eye.\(Magnifying\;power = \frac{{Angle\;subtended\;by\;the\;image\;at\;the\;least\;distance\;of\;distant\;vision}}{{Angle\;subtended\;by\;the\;object\;at\;the\;least\;distance\;of\;distant\;vision}}\)∴ \(m = \frac{\beta }{\alpha }\)The magnifying power of the compound microscope is \(m = \;\frac{L}{{{u_o}}}\left( {1 + \frac{D}{{{f_e}}}} \right)\)Or, \(m=\ \frac{L}{{{f}_{o}}}\left( 1+\frac{D}{{{f}_{E}}} \right)\Rightarrow m={{m}_{0}}\times {{m}_{e}}\)Wherevo = image distance from the objective lens,uo = object distance from the objective lens,D = least distance of distinct vision from the eye,fe = focal length of the eyepieceL is the distance between the two tubes. mo = magnification power of the eyepiece (L/f0)me = magnification power of the eyepiece \(\left( {1 + \frac{D}{{{f_e}}}} \right)\)Calculation:Given that,fe = 10 cmmo =L/f0 =4D = 25 cmThe magnifying power of the compound microscope is \(m =m_0\left( {1 + \frac{D}{{{f_e}}}} \right)\)\(\therefore m=4\times (1+\frac{25}{10})=14\)Hence option 2 is correct among all