A vessel of depth 2d cm is half filled with a liquid of refractive index -1 and the upper half with a liquid of refractive index -2- The apparent depth of the vessel seen perpendicularly is

Μ=Real depth/apparent depth h'=d_1/μ_1+d_2/μ_2=d (1/μ_1+1/μ_2 ) ...

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Byju's Answer

Standard XIII

Physics

Apparent Depth

A vessel of d...

Question

A vessel of depth 2d cm 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

d (\frac{μ_{1} μ_{2}}{μ_{1} + μ_{2}})

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d (\frac{1}{μ_{1}} + \frac{1}{μ_{2}})

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2 d (\frac{1}{μ_{1}} + \frac{1}{μ_{2}})

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2 d (\frac{1}{μ_{1} μ_{2}})

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Solution

The correct option is B $d (\frac{1}{μ_{1}} + \frac{1}{μ_{2}})$
$μ = \frac{R e a l d e p t h}{a p p a r e n t d e p t h} h^{'} = \frac{d_{1}}{μ_{1}} + \frac{d_{2}}{μ_{2}} = d (\frac{1}{μ_{1}} + \frac{1}{μ_{2}})$

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A vessel of depth 2d 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

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The correct option is B $d (\frac{1}{μ_{1}} + \frac{1}{μ_{2}})$
$μ = \frac{R e a l d e p t h}{a p p a r e n t d e p t h} h^{'} = \frac{d_{1}}{μ_{1}} + \frac{d_{2}}{μ_{2}} = d (\frac{1}{μ_{1}} + \frac{1}{μ_{2}})$