A layer of benzene $(μ = 1.50) 6 c m$ deep floats on water ( $μ = \frac{4}{3}) 4 c m$ deep. When viewed vertically through air, the apparent distance of the bottom of the vessel below the free surface of the benzene is

10 c m

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5 c m

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7 c m

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3.5 c m

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Solution

The correct option is B $7 c m$
Image of AB will be situated at a distance OI deep to refraction from water to benzene
So OI $= h (\frac{μ_{1}}{μ_{w}} - 1)$
$= 4 (\frac{1.5}{1.33} - 1)$
$= 0.5 c m$
New, Image of oI will be formed deep to refraction of ray from benzene to Air, So $= \frac{O I + A C + C E}{μ_{6}} = 6.93 c m$
$= 7 c m$

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A layer of benzene (μ=1.50)6cm deep floats on water ( μ=43)4cm deep. When viewed vertically through air, the apparent distance of the bottom of the vessel below the free surface of the benzene is

The correct option is B 7cmImage of AB will be situated at a distance OI deep to refraction from water to benzeneSo OI =h(μ1μw−1)=4(1.51.33−1)=0.5 cmNew, Image of oI will be formed deep to refraction of ray from benzene to Air, So=OI+AC+CEμ6=6.93 cm=7 cm

A layer of benzene $(μ = 1.50) 6 c m$ deep floats on water ( $μ = \frac{4}{3}) 4 c m$ deep. When viewed vertically through air, the apparent distance of the bottom of the vessel below the free surface of the benzene is