A) 0. 1 m and 1.0 m
B) 1.0 m and 0.1 m
C) 0.6 m and 0.5 m
D) 0.5 m and 0.6 m
Correct Answer: B
Solution :
Given: magnifying power =10 The length of the astronomical telescope \[=1.1\text{ m}\] Length of telescope\[={{f}_{o}}+{{f}_{e}}\] \[1.1={{f}_{o}}+{{f}_{e}}\] ?(i) Magnifying power of telescope \[m=\frac{{{f}_{o}}}{{{f}_{e}}}\]hence , \[\frac{{{f}_{o}}}{{{f}_{e}}}=10\] ?(ii) or \[{{f}_{o}}=10{{f}_{e}}\] Now, putting\[{{f}_{o}}=10{{f}_{e}}\]in Eq. (i) \[10{{f}_{e}}+{{f}_{e}}=1.1\] \[11{{f}_{e}}=1.1\Rightarrow {{f}_{e}}=\frac{1.1}{11}=0.1\,m\] So, \[{{f}_{o}}=1m\]
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