Question

A coin is at the bottom of water column of depth 4cm in a vessel.An immiscible liquid of density less than that of water is poured in the same vessel.To an observer in the air above the surface of liquid , the image of coin is at exact interface of water and liquid. If the refractive indices of water and liquid are$\frac{4}{3}$ and $\frac{5}{4}$ respectively, the depth of liquid is

• A. 12cm
• B. 15cm
• C. 18cm
• D. 2cm

Answer 1

Answer: (B)

15cm

Suppose X is the height of liquid
First Image I will from due to refraction of ray from water to liquid.
$I{I}^1=\frac{4}{{\mu }_{wl}}$

$Now,IwillbehavelikeanobjectforI{I}^{nd}refraction$

$soS{I}^1=\frac{II{}^1+S{I}^1}{{\mu }_L}$

$X=\frac{\frac{4}{\mu WL}+X}{{\mu }_L}({\mu }_{WL}=\frac{\frac{4}{3}}{\frac{5}{4}}=\frac{16}{15})$

$\frac{5x}{4}=\frac{15}{4}+x$

$x=15cm$