A coin lies at the bottom of a vessel filled with a liquid to a depth of 14.0\text{ cm}. I… — JAMB UTME

A coin lies at the bottom of a vessel filled with a liquid to a depth of $14.0\text{ cm}$ . If the refractive index of the liquid relative to air is $1.40$ , calculate the apparent upward displacement of the coin in centimeters when viewed vertically from directly above.

Answer: 4 cm

Answer

The apparent upward displacement of the coin is

4.0\text{ cm}

Refraction at the liquid-air boundary makes an object at real depth

h = 14.0\text{ cm}

appear at an apparent depth

h' = \frac{h}{n} = \frac{14.0}{1.40} = 10.0\text{ cm}

. The apparent upward displacement is the difference between real depth and apparent depth:

d = 14.0\text{ cm} - 10.0\text{ cm} = 4.0\text{ cm}

Step-by-Step Solution

Identify the given values and formula for refractive index in terms of depth.

Real depth

h = 14.0\text{ cm}

, refractive index

n = 1.40

. Formula:

n = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{h}{h'}

Light rays bending away from the normal upon leaving the denser liquid cause the coin to appear closer to the surface.

Calculate the apparent depth (

h'

h' = \frac{14.0\text{ cm}}{1.40} = 10.0\text{ cm}

Rearranging the refractive index formula gives

h' = \frac{h}{n}

Calculate the apparent upward displacement (

d

d = h - h' = 14.0\text{ cm} - 10.0\text{ cm} = 4.0\text{ cm}

The displacement is the distance between the actual position at the bottom and the virtual image position.

Key Concept

Real depth, apparent depth, and apparent displacement

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