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  4. A layer of benzene (μ=1.5) 6 cm deep floats on water (μ=(4/3)), 4 cm deep. When viewed vertically through air, the apparent distance of the bottom of the vessel below the free surface of benzene is

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Q. A layer of benzene $(\mu=1.5) 6 \,cm$ deep floats on water $\left(\mu=\frac{4}{3}\right), 4\, cm$ deep. When viewed vertically through air, the apparent distance of the bottom of the vessel below the free surface of benzene is

Ray Optics and Optical Instruments

Solution:

Apparent depth $=\frac{t_{b}}{\mu_{b}}+\frac{t_{w}}{\mu_{w}}$
$=\frac{6 cm }{1.5}+\frac{4 cm }{4 / 3}=(4+3) cm =7 cm$