Ray Optics And Optical Instruments Ques 66
66. A astronomical telescope has objective and eyepiece of focal lengths $40$ $ cm$ and $4 $ $cm$ respectively. To view an object $200 $ $cm$ away from the objective, the lenses must be separated by a distance :
[2016]
(a) $37.3 $ $cm$
(b) $46.0 $ $cm$
(c) $50.0 $ $cm$
(d) $54.0 $ $cm$
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Answer:
Correct Answer: 66.(d)
Solution:
- (d) Given: Focal length of objective, $f_0=40 $ $cm$
Focal length of eye - piece $f_e=4 $ $cm$
image distance, $v_0=200 $ $cm$
Using lens formula for objective lens
$\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0} \Rightarrow \frac{1}{v_0}=\frac{1}{f_0}+\frac{1}{u_0}$
$\Rightarrow \frac{1}{v_0}=\frac{1}{40}+\frac{1}{-200}=\frac{+5-1}{200}$
$\Rightarrow v_0=50 $ $cm$
Tube length $\ell=|v_0|+f_e=50+4=54 $ $cm$.
In a telescope, if field and eye lenses are interchanged magnification will change from $(\frac{f_o}{f_e})$ to $(\frac{f_e}{f_o})$ i.e., it will change from $m$ to $\frac{1}{m}$ i.e., will become $(\frac{1}{m^{2}})$ times of its initial value.
BETA
