A) 7.5 cm
B) 6.75 cm
C) 4.5 cm
D) 1.5 cm
Correct Answer: D
Solution :
Here: Actual depth of liquid h= 6 cm Refractive index of the liquid \[=\frac{4}{3}\] Using the relation \[\mu =\frac{actual\,depth\,(h)}{apparent\,depth\,(x)}\] or \[x=h\times \frac{3}{4}=6\times \frac{3}{4}=4.5\,cm\] Hence, the coin will appear at a depth of \[=6-4.5=1.5\,cm\]You need to login to perform this action.
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