Question

In the case of Ramsdens eyepiece, L is length of eyepiece, $f_f$ is the focal length of field lens and $f_e$ is the effective focal length of eyepiece. The ratio ( L : $L_f:L_e$ ) is :

• A. 8:9:12
• B. 9:8:12
• C. 8:12:9
• D. 9:12:8

Answer 1

Answer: (C)

8:12:9

For Ramsden's eyepiece we have

$L=\frac{2f_f}{3}$

$L_f=f_f$

$L_e=\frac{3f_f}{4}$

So, $L:L_f:L_e=8:12:9$

Hence the correct answer is option (C) 8:12:9