A) 9 cm and 18 cm
B) 6 cm and 12 cm
C) 3 cm and 6 cm
D) 4.5 cm and 9 cm
Correct Answer: D
Solution :
: Let \[{{R}_{1}}=R\] \[\therefore \] \[{{R}_{2}}=2R\] According to lens makers formula \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] Here, \[f=6cm,\,\mu =1.5\] \[{{R}_{1}}=R,\,\,{{R}_{2}}=-2R\] \[\therefore \]\[\frac{1}{6}=(1.5-1)\left( \frac{1}{R}-\frac{1}{-2R} \right)=0.5\left( \frac{1}{R}+\frac{1}{2R} \right)\] \[\frac{1}{6}=0.5\left( \frac{3}{2R} \right)\] or \[\frac{1}{6}=\frac{1.5}{2R}\] \[R=\frac{1.5\times 6}{2}=4.5cm\] \[\therefore \] \[{{R}_{1}}=4.5cm,\,{{R}_{2}}=9cm\]
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