A vessel contains water up to a height of 20 cm and above it an oil up to another 20 cm. The refractive indices of the water and the oil are 1.33 and 1.30 respectively. Find the apparent depth of the vessel when viewed from above.

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Solution

Shifted due to water,

$D t_{w} = 1 - (\frac{1}{μ}) d$

$= [1 - (\frac{1}{1.33})] 20$

$= \frac{1}{4} (20) = 5 cm$

Shifted to oil,

$D t_{0} = [1 - (\frac{1}{1.33})] 20$

= 4.6 cm

Total shift, Kt = 5 + 4.6 = 9.6 cm

Apparent depth = 40 - (9.6)

= 30.4 cm between the surface.

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A vessel contains water up to a height of 20 cm and above it an oil up to another 20 cm. The refractive indices of the water and the oil are 1.33 and 1.30 respectively. Find the apparent depth of the vessel when viewed from above.

Shifted due to water, Dtw=1−(1μ)d =[1−(11.33)]20 =14(20)=5 cm Shifted to oil, Dt0=[1−(11.33)]20 = 4.6 cm Total shift, Kt = 5 + 4.6 = 9.6 cm Apparent depth = 40 - (9.6) = 30.4 cm between the surface.

Shifted due to water,

$D t_{w} = 1 - (\frac{1}{μ}) d$

$= [1 - (\frac{1}{1.33})] 20$

$= \frac{1}{4} (20) = 5 cm$

Shifted to oil,

$D t_{0} = [1 - (\frac{1}{1.33})] 20$

= 4.6 cm

Total shift, Kt = 5 + 4.6 = 9.6 cm

Apparent depth = 40 - (9.6)

= 30.4 cm between the surface.