Question

When an object is placed 6cm from the objective lense of a compound microscope the final image is two-third from the object and it is at the LDDV, calculate the focal length of the two lenses assuming that the magnification power of the instrument is 14. calculate also the separation of the two lenses

Answer 1

${v_e\,is\,the\,least\,distance\,of\,distinct\,vision\,which\,is\,always\,24 cm.}\\ {L\,is\,24-6\,which\,is\,equal\,to\,18cm}\\ {u_o = 6cm}\\{v_e = 24cm}\\{L= 18cm}\\{M=14}\\ {f_o=focal\,length\,of\,objective\,lense}\\{f_e=focal\,length\,of\,eyepiece\,lense}\\ {L_D=v_o+\frac{Df_e}{D+f_e}}\\ {M_D=\frac{v_o}{6}{(1+\frac{25}{f_e})}}\\ {v_o=\frac{84fe}{f_e+25}}\\ {v_o=\frac{84*7}{32}=18.34}\\ {v_o=18.34}\\ {\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}}\\ {\frac{1}{f_o}=\frac{8}{147}-\frac{1}{6}}\\ {f_o=8.91cm}\\ {18=\frac{84f_e}{f_e+24}+\frac{24f_e}{f_e+24}}\\ {18=\frac{108f_e}{f_e+24}}\\ {18f_e+432=108f_e}\\ {90f_e=432} {f_e=\frac{432}{90}}\\ {f_e=4.8cm}\\ {Distance\,between\,lenses={u_e+v_o}}\\ {\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}}\\ {\frac{-1}{u_e}=\frac{90}{432}-\frac{1}{24}}\\ {u_e=6cm}\\ {Distance\,between\,lenses=6+18.34=24.34cm}$