A coin kept inside water [µ= 4/3 ] when viewed from air in a vertical direction appears to be raised by 3.0 mm. Find the depth of the coin in water.

Given: μ = 4/3, apparent rise of the coin = 3.0 mm

Let the apparent depth be a and the real depth be d.

μ = real depth ÷ apparent depth d = (4/3) a … (1)

The apparent rise is how much shallower the coin looks: rise = d − a = 3.0 mm … (2)

Substituting (1) into (2): (4/3)a − a = 3 (4a − 3a) ÷ 3 = 3 a ÷ 3 = 3 a = 9 mm

Then from (1): d = (4/3) × 9 = 12 mm

Depth of the coin in water = 12 mm

The 3.0 mm is the RISE, not the depth — it is the difference between where the coin really is and where it appears to be. Treating it as the apparent depth and multiplying by 4/3 gives 4 mm, which is the standard wrong answer.

Set up two equations, one for the refractive index and one for the rise, and solve them together.

There is a shortcut worth knowing: rise = d(1 − 1/μ). Here 3 = d(1 − 3/4) = d/4, so d = 12 mm in a single line. Check it back: real 12 mm, apparent 9 mm, rise 3 mm ✓.