Question

A compound microscope is 20 cm long and the power of its eye-piece is 20 D. If the objective is bi-convex (radius of curvature of either surface 2 cm) and made of a material of = 1.5, the magnifying power of the microscope is

• A. 54
• B. 60
• C. 13.5
• D. 41.5

Answer 1

Answer: (D)

: For objective lens, $\frac{1}{f_0}=(\mu -1)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=(1.5-1)\times \frac{2}{R}=\frac{0.5\times 2}{2}=\frac{1}{2}$ Or $f_0=2cm$ ...(i) $f_c=\frac{100}{D}=\frac{100}{20}=5cm$ For eyepiece, $\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$ Final image is at 25 cm from it. $\therefore$ $\frac{-1}{25}-\frac{1}{u_e}=\frac{1}{5}$ or $u_e=-\frac{25}{6}cm=$ Image is on the object (virtual) side of lens. $\therefore$ Tube length $=u_e+v_0$ or $v_0=20-\frac{25}{6}=\frac{95}{6}cm$ ...(iii) For objective lens, $\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}$ Or $\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}=\frac{6}{95}-\frac{1}{2}=\frac{12-95}{190}=\frac{-83}{190}$ Or $u_0=\frac{-190}{83}cm$ $\therefore$ Magnifying power of microscope $=\frac{v_0}{u_0}\left(1+\frac{D}{f_e}\right)$ Here final image is formed at least distance of distinct vision (25 cm) from eyepiece. The final image is virtual and magnified.
$M=\frac{v_0}{u_0}\left(1+\frac{D}{f_e}\right)$ $M=\frac{95/6}{190/83}\left(1+\frac{25}{5}\right)=\frac{95}{6}\times \frac{83}{190}(1+5)$ $=\frac{83\times 6}{6\times 2}=41.5$ Magnifying power = 41.5