A) 1.6
B) 1.66
C) 2.6
D) 2.66
E) 1.4
Correct Answer: D
Solution :
\[\frac{1}{f}(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] For planoconvex lens \[{{R}_{1}}=\infty ,{{R}_{2}}=-R=-1.5\,cm,\mu =1.4\] \[\therefore \]\[\frac{1}{f}=(1.4-1)\left( 0+\frac{1}{15} \right)\]or\[\frac{1}{f}=0.4\times \frac{1}{15}\] Therefore, power of the lens in diopter \[P=\frac{100}{f}=\frac{40}{15}=2.66D\]
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