Question

An object O is locked at the bottom of a tank containing two immiscible liquids and is seen vertically from above. The lower and upper liquids are of depth h₁&h₂ respectively and of refractive indices μ₁& μ₂ respectively. Location of the position of the image of the object o from the surface 

(a) (h₁/μ₁)+ (h₂/μ₂) 

(b) (h₁/μ₂) + (h₂/μ₁) 

(c) (h₁+h₂)/μ₁+μ₂) 

(d) none

Answer 1

Correct Option :- (a) (h₁/μ₁)+ (h₂/μ₂) 

Explanation :-

As shown in the figure refraction at the 1st Surface μ₁ sin i = μ₂ sin r 

For small angle sin θ= tan θ = θ 

μ₁ tan i = μ₂ tan r 

μ₁ [AB/OB]= μ₂ [AB/BI1] 

BI₁ =(μ₂/μ₁)h₁                                     ..............(1) 

For refraction at the second surface μ₂ sin r = 1 sin i₂ μ₂ tan r = tan i₂ 

μ₂ [CD/DI₁]= [CD/DI₂] 

DI₂ = DI₁/μ₂                                         ............(2) 

DI₁ = DB +BI₁ = h₂ + ( μ₂/μ₁) h₁         ............(3) 

Substituting (3) in (2) 

DI₂ = [h₂+(μ₂/μ₁)h₁]/μ₂ = (h₂/μ₂)+( h₁+μ₁) 

DI₂ =(h₂/μ₂)+( h₁+μ₁)