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The magnifying power of an astronomical telescope for relaxed vision is 16 . On adjusting, the distance between the objective and eye lens is $34 \mathrm{~cm}$. Then the focal length of objective and eye lens will be respectively
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The correct answer is:
$32 \mathrm{~cm}, 2 \mathrm{~cm}$
Correct option is 2. $20 \mathrm{~cm}, 14 \mathrm{~cm}$
In this case $|m|=\frac{f_0}{f_e}=16$....(i)
and length of telescope $=f_0+f_e=34 \ldots$...(ii)
Solving (i) and (ii), we get $f_{\mathrm{k}}=2 \mathrm{~cm}, \mathrm{Fo}_0=32 \mathrm{~cm}$
In this case $|m|=\frac{f_0}{f_e}=16$....(i)
and length of telescope $=f_0+f_e=34 \ldots$...(ii)
Solving (i) and (ii), we get $f_{\mathrm{k}}=2 \mathrm{~cm}, \mathrm{Fo}_0=32 \mathrm{~cm}$
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