A) 3/2
B) 4/3
C) 5/3
D) 2
Correct Answer: C
Solution :
| [c] When the lens in air, we have |
| \[{{P}_{a}}=\frac{1}{{{f}_{a}}}=\frac{{{\mu }_{g}}-{{\mu }_{a}}}{{{\mu }_{a}}}\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\] |
| When the lens is in liquid, we have |
| \[{{P}_{l}}=\frac{1}{{{f}_{I}}}=\frac{{{\mu }_{g}}-{{\mu }_{a}}}{{{\mu }_{l}}}\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\] |
| Here, \[{{P}_{a}}=5,{{P}_{l}}=-1,{{\mu }_{a}}=1,{{\mu }_{g}}=1.5\] |
| On solving, we get, \[{{\mu }_{l}}=\frac{5}{3}\] |
You need to login to perform this action.
You will be redirected in
3 sec
