Q.Focal power of the eye lens of a compound microscope is 6 dioptre. The microscope is to be used for maximum magnifying power (angular magnification) of at least 12.5. The packing instructions demand that length of the microscope should be 25 cm. Determine minimum focal power of the objective. How much will its radius of curvature be if it is a biconvex lens of n = 1.5.

[!TLDR] With fe=1/6 m and Me_max=2.5, achieving overall M=12.5 needs mo=5; the L=25cm constraint then fixes vo=15cm, giving fo=2.5cm (Po=40 D) and, for a biconvex n=1.5 lens, R=2.5cm. [!A …

Eye lens power 6 D gives fe=1/6 m=100/6 cm=16.667 cm. Maximum eye-lens magnifying power, Me_max=1+D/fe=1+25/16.667=1+1.5=2.5. For overall M at least 12.5, the objective's linear magnification must satisfy mo x Me_max >= 12.5, so mo >= 12.5/2.5=5; for the MINIMUM objective power (the boundary case), take mo=5 exactly. For Me_max, the intermediate image sits at the eye lens's near-point-conjugate distance: using 1/fe=1/ue+1/D (image at D), ue=Dfe/(D+fe)=(25 x 16.667)/(25+16.667) about 10 cm. Given microscope length L=25 cm=vo+ue, vo=25-10=15 cm. Then mo=vo/uo=5 gives uo=vo/5=15/5=3 cm. Applying the thin lens formula to the objective (uo=-3 cm, vo=+15 cm): 1/fo=1/vo-1/uo=1/15-1/(-3)=1/15+1/3=1/15+5/15=6/15=0.4, so fo=2.5 cm, i …

Compute the eye lens's maximum angular magnification (Me_max=1+D/fe), use the required overall M=12.5 to find the minimum needed objective linear magnification mo, use the fixed total length L to find vo (via the eye lens's own near-point object distance ue), then apply the thin lens formul …

Forgetting that Me in the product M=mo x Me should use its MAXIMUM value (since the final image is explicitly at its biggest/nearest, D) rather than …