A vessel of depth 2 d cm is half filled with a liquid of refractive index μ1 and the upper half with a liquid of refractive index μ2. The apparent depth of the vessel seen perpendicularly is

Question

A vessel of depth 2 d cm is half filled with a liquid of refractive index μ1 and the upper half with a liquid of refractive index μ2. The apparent depth of the vessel seen perpendicularly is (A) ((μ1 μ2/μ1+μ2)) d (B) ((1/μ1)+(1/μ2)) d (C) ((1/μ1)+(1/μ2)) 2 d (D) ((1/μ1 μ2)) d

Tardigrade Admin · Accepted Answer

Apparent depth =( text real depth / text refractive index ) ⇒ depth seen =(d/μ1)+(d/μ2) =d((1/μ1)+(1/μ2))

Q. A vessel of depth $2 d c m$ is half filled with a liquid of refractive index $μ_{1}$ and the upper half with a liquid of refractive index $μ_{2}$ . The apparent depth of the vessel seen perpendicularly is

$(\frac{μ _{1} μ _{2}}{μ _{1} + μ _{2}}) d$

$(\frac{1}{μ _{1}} + \frac{1}{μ _{2}}) d$

$(\frac{1}{μ _{1}} + \frac{1}{μ _{2}}) 2 d$

$(\frac{1}{μ _{1} μ _{2}}) d$

Apparent depth $= \frac{real depth}{refractive index}$
$\Rightarrow$ depth seen $= \frac{d}{μ _{1}} + \frac{d}{μ _{2}}$
$= d (\frac{1}{μ _{1}} + \frac{1}{μ _{2}})$

Q. A vessel of depth 2dcm is half filled with a liquid of refractive index μ1​ and the upper half with a liquid of refractive index μ2​. The apparent depth of the vessel seen perpendicularly is

(μ1​+μ2​μ1​μ2​​)d

(μ1​1​+μ2​1​)d

(μ1​1​+μ2​1​)2d

(μ1​μ2​1​)d