Question

Explain apparent depth by refraction

Answer 1

Apparent depth is the perceived shallower depth of an object submerged in a denser medium (like water) when viewed from a rarer medium (like air), caused by the refraction of light.

Answer 2

Explanation of apparent depth by refraction:

When an object is placed in a denser medium (e.g., water) and viewed from a rarer medium (e.g., air), it appears to be at a shallower depth than its actual depth. This phenomenon is known as apparent depth and occurs due to the refraction of light as it travels from the denser to the rarer medium.

Setup and Ray Tracing:

Consider a point object 'O' placed at the bottom of a transparent denser medium of refractive index 'n' (or μ). The interface separating this medium from the rarer medium (air, refractive index ≈ 1) is denoted by PQ. The real depth of the object is AO.

1. Normal Ray: A ray of light OA travels perpendicularly from the object 'O' to the surface PQ. Since it strikes the surface normally, it passes undeviated into the rarer medium.
2. Oblique Ray: Another ray OB emerges from 'O' and strikes the surface PQ at point B at an angle of incidence 'i' with the normal N'BN.
3. Refraction: As light travels from the denser medium (n) to the rarer medium (1), it bends away from the normal. The refracted ray BC makes an angle of refraction 'r' with the normal.
4. Image Formation: When an observer's eye views these rays, the undeviated ray OA and the refracted ray BC appear to originate from a point 'I'. This point 'I' is the virtual image of the object 'O', and the distance AI is the apparent depth.

Mathematical Derivation:

Let's consider the angles:

• Angle of incidence, $i = \angle OBN'$ (where N'B is the normal at B). Since AO is parallel to N'B, we have $i = \angle AOB$ (alternate interior angles).
• Angle of refraction, $r = \angle CBN$ (where BN is the normal at B). When the refracted ray BC is extended backward, it appears to come from I. Thus, $r = \angle AIB$ (corresponding angles, as IA is parallel to BN).

Now, consider the right-angled triangles $\triangle ABO$ and $\triangle ABI$:

In $\triangle ABO$:

$$\sin i = \frac{AB}{OB}$$

In $\triangle ABI$:

$$\sin r = \frac{AB}{IB}$$

According to Snell's Law, for light traveling from a denser medium (refractive index 'n') to a rarer medium (refractive index '1'):

$$n \sin i = 1 \sin r$$ $$n = \frac{\sin r}{\sin i}$$

Substitute the expressions for $\sin i$ and $\sin r$:

$$n = \frac{\frac{AB}{IB}}{\frac{AB}{OB}}$$ $$n = \frac{OB}{IB}$$

For paraxial rays (when the eye is positioned directly above the object, and thus the angles $i$ and $r$ are very small), point B is very close to point A.

Therefore, we can approximate:

$OB \approx OA$ (Real Depth)

$IB \approx IA$ (Apparent Depth)

Substituting these approximations into the equation for 'n':

$$n = \frac{OA}{IA}$$ $$n = \frac{\text{Real Depth}}{\text{Apparent Depth}}$$

This relation shows that the refractive index of the denser medium is the ratio of the real depth to the apparent depth. Since $n > 1$ for a denser medium relative to air, it implies that the apparent depth ($IA$) is always less than the real depth ($OA$).

Solution:

Apparent depth is the perceived shallower depth of an object submerged in a denser medium (like water) when viewed from a rarer medium (like air), caused by the refraction of light. Light rays from the object bend away from the normal as they cross the interface from denser to rarer medium. The eye traces these refracted rays back to a virtual image that is closer to the surface than the actual object.

The relationship between the real depth ($OA$), apparent depth ($IA$), and the refractive index ($n$) of the denser medium with respect to the rarer medium is derived using Snell's Law and paraxial approximation:

1. Consider a point object 'O' at real depth $OA$ in a medium of refractive index $n$.
2. A ray $OA$ travels normally and undeviated to the surface.
3. Another ray $OB$ strikes the surface at point $B$ with angle of incidence $i = \angle AOB$.
4. It refracts away from the normal with angle of refraction $r = \angle AIB$.
5. Using trigonometry, $\sin i = \frac{AB}{OB}$ and $\sin r = \frac{AB}{IB}$.
6. By Snell's Law, $n \sin i = 1 \sin r \implies n = \frac{\sin r}{\sin i}$.
7. Substituting the trigonometric values, $n = \frac{OB}{IB}$.
8. For paraxial rays (small angles), $OB \approx OA$ and $IB \approx IA$.
9. Therefore, the refractive index $n = \frac{\text{Real Depth}}{\text{Apparent Depth}}$.