Solve this.
36. When a small lamp is held 1.5 m above the surface of water in a tank, its image formed by reflection as the surface appears to coincide with image of the bottom of the tank ($\mu$ of water = 4/3). The depth of tank is
(a) 2 m
(b) 1.5 m
(c) 1 m
(d) 4 m

Dear Student,
The equation used for this problem is as follows,

Real depth(d₁)/Apparent depth(d₂) = n₂/n₁ , n₂ is the refractive index of the denser medium (water) and n₁ is refractive index of the rarer medium (air). 

Here the d₂ is given as 1.5 m and for the case when the object in denser medium is viewed from the rarer medium
d₁>d₂.  So here d₁>1.5 and can be estimated as.

$d1 =\frac{n2\times d2}{n1}$ , where n₂= 4/3 and n₁=1  substituting the values, we have
The real depth of tank, d₁ = 2 m.

Hence the solution is option (A).

I hope I solved your problem thoroughly. 
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