Understanding Real Depth to Apparent Depth Ratio

When viewing an object through a liquid in a beaker with refractive index 1.5 relative to air, it appears shallower due to refraction—this is the real depth to apparent depth ratio. This fundamental optics concept, where the ratio equals the refractive index (1.5), is crucial for exams like CSIR NET physics sections. Master the formula:

μ = real depth / apparent depth

or equivalently, μ = apparent depth / real depth depending on the medium configuration.

Why Apparent Depth Differs from Real Depth

Light rays bend at the air–liquid interface, making objects in denser media (like the liquid with μ = 1.5) appear closer to the surface. The true depth of the object is denoted by h, whereas the apparent depth observed from air is h′ = h / μ.

For normal viewing in a beaker, this ratio remains constant regardless of depth, giving real depth / apparent depth = 1.5.

Step-by-Step Derivation and Formula

1. Consider an object at real depth h emitting rays that refract at the air–liquid interface.
2. By Snell’s law, sin i = 1.5 sin r.
3. For paraxial rays, we approximate tan i / tan r = 1.5.
4. From geometry, tan i = x / h′ and tan r = x / h.
5. Therefore, h / h′ = 1.5.

This proves that the real depth to apparent depth ratio equals the refractive index μ = 1.5.

Practical Calculation Example

If an object is 10 cm deep in the beaker, then its apparent depth is:

h′ = 10 / 1.5 ≈ 6.67 cm

Thus, the ratio h / h′ = 10 / 6.67 = 1.5.

This ratio holds for any depth when viewed from above the liquid in air.

Exam Tips for Refractive Index Problems

• Always assume normal incidence unless otherwise specified.
• For air-to-liquid transition, the ratio μ = 1.5.
• Do not confuse apparent shift given by h(1 − 1/μ) with the ratio of depths.