In Ramsden eyepiece, the two planoconvex lenses each of focal length f are separated by a distance 12 cm. The equivalent focal length (in cm ) of the eyepiece is

Question

In Ramsden eyepiece, the two planoconvex lenses each of focal length f are separated by a distance 12 cm. The equivalent focal length (in cm ) of the eyepiece is (A) 10.5 (B) 12.0 (C) 13.5 (D) 15.5

Tardigrade Admin · Accepted Answer

d=(2 f/3) or f=(3 d/2)=(3 × 12/2)=18 cm Equivalent focal length is f'=(f1 f2/f1+f2)+(f/4) =(18 × 18/18+18)+(18/4) =9+4.5 =13.5 cm

Q. In Ramsden eyepiece, the two planoconvex lenses each of focal length $f$ are separated by a distance $12 c m$ . The equivalent focal length (in $c m$ ) of the eyepiece is

10.5

12.0

13.5

15.5

$d = \frac{2 f}{3}$
or $f = \frac{3 d}{2} = \frac{3 \times 12}{2} = 18 c m$
Equivalent focal length is
$f^{'} = \frac{f _{1} f _{2}}{f _{1} + f _{2}} + \frac{f}{4}$
$= \frac{18 \times 18}{18 + 18} + \frac{18}{4}$
$= 9 + 4.5$
$= 13.5 c m$

Q. In Ramsden eyepiece, the two planoconvex lenses each of focal length f are separated by a distance 12cm. The equivalent focal length (in cm ) of the eyepiece is