A block of transparent polymer with a refractive index of 1.25 is placed over a small mark… — JAMB UTME

A block of transparent polymer with a refractive index of $1.25$ is placed over a small mark on a table. When viewed vertically from directly above, the mark appears to be shifted upward by $3.0\text{ cm}$ . What is the real thickness of the polymer block in centimeters?

Answer: 15 cm

Answer

The real thickness of the polymer block is

15.0\text{ cm}

Refraction at the interface produces an upward shift

s = d\left(1 - \frac{1}{n}\right)

. Substituting

s = 3.0\text{ cm}

and

n = 1.25

gives

3.0 = d(1 - 0.80) = 0.20d

, which solves to a real thickness

d = 15.0\text{ cm}

Step-by-Step Solution

Relate apparent depth

d'

to real depth

d

using the refractive index.

d' = \frac{d}{n}

Light refracting at the surface of a denser medium causes the object to appear at a shallower depth.

Express the apparent displacement (upward shift)

s

in terms of

d

and

n

s = d - d' = d\left(1 - \frac{1}{n}\right)

The upward shift is the vertical distance between the true position and the apparent position.

Substitute

s = 3.0\text{ cm}

and

n = 1.25

into the equation and solve for

d

3.0 = d\left(1 - \frac{1}{1.25}\right) = d(1 - 0.80) = 0.20 d \implies d = 15.0\text{ cm}

Algebraic evaluation yields the exact real thickness of the polymer block.

Key Concept

Apparent depth and upward displacement due to refraction

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