A) 15 cm
B) 20 cm
C) 30 cm
D) 10 cm
Correct Answer: C
Solution :
\[A{{C}^{2}}=A{{E}^{2}}+C{{E}^{2}}\]or \[{{R}^{2}}={{a}^{2}}+{{(R-t)}^{2}}={{a}^{2}}+{{R}^{2}}-2Rt+{{t}^{2}}\] or \[2Rt\left\{ 1-\frac{t}{2R} \right\}={{a}^{2}}\] since \[t<\,<R,\]the term \[t/2R\]can be neglected. 2R \[t={{a}^{2}}\] \[\therefore \]\[R=\frac{{{a}^{2}}}{2t}=15\,cm.\] \[\mu =\frac{c}{v}=1.5\]\[\therefore \]\[\frac{1}{f}=(\mu -1)\times \frac{1}{R}\] \[=(1.5-1)\times \frac{1}{15}=\frac{1}{30}\] Hence, the correction option is (c).You need to login to perform this action.
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