A small object lies at the bottom of a transparent vessel containing two immiscible liquid… — JAMB UTME

A small object lies at the bottom of a transparent vessel containing two immiscible liquid layers, $A$ and $B$ . Layer $A$ (top) has a real thickness of $7.0\text{ cm}$ and a refractive index of $1.40$ . Layer $B$ (bottom) has a real thickness of $8.0\text{ cm}$ and a refractive index of $1.60$ . Calculate the apparent displacement of the object, in centimeters, when viewed vertically from directly above.

Answer: 5 cm

Answer

The apparent displacement of the object is

5.0\text{ cm}

For multiple refractive layers viewed normally, the total apparent depth is the sum of individual layer apparent depths:

\frac{7.0}{1.40} + \frac{8.0}{1.60} = 5.0 + 5.0 = 10.0\text{ cm}

. Subtracting this total apparent depth from the total real depth of

15.0\text{ cm}

gives an apparent vertical shift (displacement) of

5.0\text{ cm}

Step-by-Step Solution

Calculate the apparent depth of the top liquid layer (Layer A)

Apparent depth of Layer A =

5.0\text{ cm}

Apparent depth for a single medium is obtained by dividing real depth by its refractive index:

\frac{7.0}{1.40} = 5.0\text{ cm}

Calculate the apparent depth of the bottom liquid layer (Layer B)

Apparent depth of Layer B =

5.0\text{ cm}

Apparent depth for Layer B is

\frac{8.0}{1.60} = 5.0\text{ cm}

Calculate total real depth and total apparent depth

Total real depth =

15.0\text{ cm}

; Total apparent depth =

10.0\text{ cm}

Depths in composite media are additive.

Calculate vertical apparent displacement

Apparent displacement =

5.0\text{ cm}

Displacement is the difference between total real depth and total apparent depth:

15.0\text{ cm} - 10.0\text{ cm} = 5.0\text{ cm}

Key Concept

Refraction through composite media and vertical apparent displacement

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