A) 62.2cm
B) 5.82cm
C) 58.2cm
D) 6.22cm
Correct Answer: C
Solution :
[c] : When the lens is in air \[\frac{1}{{{f}_{a}}}=({{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[\frac{1}{20}=(1.5-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] When lens is in water, \[\frac{1}{{{f}_{w}}}=\left( \frac{{{\mu }_{g}}}{{{\mu }_{w}}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] or\[\frac{1}{{{f}_{w}}}=\left( \frac{1.5-1.33}{1.33} \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] or\[\frac{{{f}_{w}}}{20}=(1.5-1)\left( \frac{1.33}{1.5-1.33} \right)\] or\[{{f}_{w}}=20\times 0.5\times \frac{1.33}{0.17}=78.2cm\] \[=78.2cm\] The change in focal length \[=78.2-20=58.2cm\]
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