A) convex of focal length 70 cm
B) concave of focal length 70 cm
C) concave of focal length 66.6 cm
D) convex of focal length 66.6 cm
E) concave of focal length 72.5 cm
Correct Answer: C
Solution :
The focal length \[\frac{1}{{{f}_{1}}}=(\mu -1)\frac{1}{-R}\] \[\frac{1}{{{f}_{1}}}=(1.5-1)\frac{1}{-25}\] \[{{f}_{1}}=-50\,cm\] The focal length \[\frac{1}{{{f}_{2}}}=(1.5-1)\times \frac{1}{-20}\] \[\frac{1}{{{f}_{2}}}=0.5\times \frac{1}{-20}m\] \[{{f}_{2}}=-\,40\,cm\] The focal length of bi-convex lens \[\frac{1}{{{f}_{3}}}={{(}_{1}}{{n}_{2}}-1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\] \[\frac{1}{{{f}_{3}}}=\left( \frac{4}{3}-1 \right)\left( \frac{1}{20}+\frac{1}{25} \right)\] \[\frac{1}{{{f}_{3}}}=\frac{1}{3}\times \left( \frac{5+4}{100} \right)\] \[\frac{1}{{{f}_{3}}}=\frac{1}{3}\times \frac{9}{100}\] \[\frac{1}{{{f}_{3}}}=\frac{3}{100}\] \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}+\frac{1}{{{f}_{3}}}\] \[=-\frac{1}{50}-\frac{1}{40}+\frac{3}{100}\] \[\frac{1}{F}=\frac{-200}{3}\] \[F=-\,66.6\text{ }cm\] Negative sign represent the concave lens.You need to login to perform this action.
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