A) 8.8 cm
B) 11.7 cm
C) 6.7 cm
D) 13.4 cm
Correct Answer: A
Solution :
Light incident from particle P will be reflected at mirror \[u=-5cm,\,f=-\frac{R}{2}=-20cm\] \[\frac{1}{\text{v}}+\frac{1}{u}=\frac{1}{f}\] \[\] This image will act as object for light getting refracted at water surface So, object distance\[d=5+\frac{20}{3}=\frac{25}{3}cm\]below water surface. After refraction, final image is at\[d'=d\left( \frac{{{\mu }_{2}}}{{{\mu }_{1}}} \right)\] \[=\left( \frac{35}{3} \right)\left( \frac{1}{4/3} \right)\] \[=\frac{35}{4}=8.75cm\] \[\approx 8.8cm\]You need to login to perform this action.
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