Question

The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece along is 5. The microscope is focused on a certain object. The distance between the objective and eyepiece is observed to be 14cm. Then the focal lengths of the objective and eyepiece are respectively, if final image formed at near point

(D$=$20 cm)

• A. 2cm, 5cm
• B. 5 cm, 10 cm
• C. 3 cm, 9 cm
• D. 4 cm, 12 cm

Answer 1

The correct option is A 2cm, 5cm

Using the lens formula and the magnification formula for the objective and eyepiece independently,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

and m = $\frac{v}{u}$

Now, we have 2 relations,
$v_o+u_e=14$

and $m_o{m}_e=20$

Solving them simultaneously,

$f_o=2$ $f_e=5$