Question

A compound microscope has an objective of focal length 2 cm and eye piece of focal length 5 cm.The distance between two lenses is 25 cm. If final image is at 25 cm from the eye-piece. Then the magnifying power of the microscope.

• A. 56.5
• B. 65.2
• C. 25.6
• D. 35.2

Answer 1

Answer: (A)

56.5

$f_0=2cm$

$f_e=5cm$

$L=25$

$v_e=25cm$

$M_D=\frac{v_0}{u_0}(1+\frac{D}{f_e})$
$L=v_0+\frac{D{f}_e}{D+f_e}$

$25=v_0+\frac{25\times \text{/}5}{\text{/}{30}_6}$

$v_0=\frac{125}{6}$

$\frac{2\times 6}{2\times 125}-\frac{1}{u}=\frac{125}{2\times 125}$

$-\frac{1}{u}=\frac{125}{250}-\frac{12}{250}$

$u_0=\frac{250}{113}$

$M_D=\frac{125\times 113}{6\times 250}[1+\frac{\text{/}{25}^5}{\text{/}a5}]=56.5$