[JEE Main 10-4-2019 Afternoon]
A) \[\frac{{{b}^{2}}c}{a}\]
B) \[\frac{{{b}^{2}}}{ac}\]
C) \[\frac{a}{c}\]
D) \[\frac{b}{c}\]
Correct Answer: D
Solution :
\[\frac{1}{\text{v}}-\frac{1}{u}=\frac{1}{f}\] \[1-\frac{\text{v}}{u}=\frac{\text{v}}{f}\] \[1-m=\frac{\text{v}}{f}\] \[m=1-\frac{\text{v}}{f}\] At\[\text{v}=a,{{m}_{1}}=1-\frac{a}{f}\] At\[\text{v}=a+b,{{m}_{2}}=1-\frac{a+b}{f}\] \[{{m}_{2}}-{{m}_{1}}=c=\left[ 1-\frac{a+b}{f} \right]-\left[ 1-\frac{a}{f} \right]\] \[c=\frac{b}{f}\] \[f=\frac{b}{c}\]
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