Question

An object o is kept at a depth H below the surface of a liquid of refractive index μ. What is apparent depth when the angle of vision at the surface is θ. 

(a) h=(H/μ)[(μ cos θ)/(√μ²-sin²θ)]³

(b) h=(H/μ)[(μ sin θ)/(√μ²-sin²θ)]³ 

(c) h=(μ/H)[(μ cos θ)/(√μ²-sin²θ)]³ 

(d) none  

Answer 1

Correct Option :-  (d) none 

Explanation :-

Here it must be noted that the object when viewed from an angle θ not only shifts upwards by H-h but also laterally (a-b) with respect to its real position. If h is the apparent depth

a= H tan i, b = h tan θ 

a-b = H tan i – h tan θ 

a, b , H , h are independent of small variation in θ & i 

0 = H sec² i di – h sec² θ dθ 

di/dθ = [h sec²θ/H sec²i] ⇒ h =(H sec² i/sec²θ)(di/dθ) = H(cos²θ/cos²i)(di/dθ) 

(1) But as at the surface Snell’s Law 

μ sin i = 1 sin θ 

μ cos i di = cos θ dθ 

(di/dθ) = cos θ/μ cos i ...(2) 

Substituting (2) in (1) we get 

h= H (cos²θ/cos²i)(di/dθ) = H (cos²θ/cos²i) (cos θ/ μ cos i) 

h = H[cos³θ/μ cos³i] (3) 

μ sin i = sin θ 

sin i = (sinθ)/μ 

√1-cos² i = (sinθ)/μ 

1-cos² i =(sin²θ)/μ 

1-(sin θ)/μ² = cos² i 

(μ²- sin²θ)/μ²= cos² i 

√(μ²- sin²θ)/μ = cos i (4) 

Substituting cos i in (3) we get 

h= H (cos³θ/μ cos³i) =(H/μ)[(μ³ cos3θ)/(μ2- sin2θ)^3/2] 

h = H(μ cos θ)³/μ(μ²- sin²θ)^3/2