Refraction at Plane Surfaces
Comprehensive NEB syllabus guide with full mathematical derivations, print-ready ray diagrams, interactive simulation, and step-by-step solved numerical problems.
Refraction of Light & The Laws of Refraction
Refraction is the phenomenon of bending of a ray of light when it passes obliquely from one transparent optical medium into another transparent medium of different optical density. This bending occurs due to a change in the phase velocity of light as it transitions between media.
Fundamental Laws of Refraction
- First Law: The incident ray, the refracted ray, and the normal to the refracting interface at the point of incidence all lie in the same plane.
- Second Law (Snell's Law): For light of a given color and for a given pair of media, the ratio of the sine of the angle of incidence ($i$) to the sine of the angle of refraction ($r$) is constant. $$\frac{\sin i}{\sin r} = {}_1\mu_2 = \frac{\mu_2}{\mu_1}$$
Figure 15.1: Ray Geometry for Refraction at Interface
Vector Form of Snell's Law: If $\hat{i}$ is the unit vector along the incident ray, $\hat{r}$ is the unit vector along the refracted ray, and $\hat{n}$ is the unit vector normal to the interface pointing into Medium 1, Snell's law can be expressed in vector form as: $$\mu_1 (\hat{i} \times \hat{n}) = \mu_2 (\hat{r} \times \hat{n})$$
Refractive Index of a Medium
The Absolute Refractive Index ($\mu$) of a medium is defined as the ratio of the speed of light in vacuum ($c$) to the speed of light in that medium ($v$).
Since frequency ($f$) of light remains unchanged during refraction (determined solely by the source), $c = f \lambda_0$ and $v = f \lambda$, where $\lambda_0$ is the wavelength in vacuum and $\lambda$ is the wavelength in the medium. $$\mu = \frac{f \lambda_0}{f \lambda} = \frac{\lambda_0}{\lambda}$$
Relative Refractive Index (${}_1\mu_2$)
The refractive index of Medium 2 with respect to Medium 1: $${}_1\mu_2 = \frac{\mu_2}{\mu_1} = \frac{v_1}{v_2} = \frac{\lambda_1}{\lambda_2}$$
Cauchy's Dispersion Formula
Refractive index depends on wavelength $\lambda$: $$\mu(\lambda) = A + \frac{B}{\lambda^2} + \frac{C}{\lambda^4} + \dots$$ Violet light ($\lambda_V$ shorter) experiences higher $\mu$ than Red light ($\lambda_R$ longer).
Principle of Reversibility & Refraction Through Compound Slabs
Principle of Reversibility of Light: If the path of a ray of light is reversed after undergoing any number of refractions or reflections, it retraces its entire path backwards.
Mathematical Proof:
For light going from Medium 1 to 2: $\frac{\sin i}{\sin r} = {}_1\mu_2$.
If reversed from Medium 2 to 1: $\frac{\sin r}{\sin i} = {}_2\mu_1$.
Multiplying both equations yields:
$${}_1\mu_2 \times {}_2\mu_1 = 1 \implies {}_1\mu_2 = \frac{1}{{}_2\mu_1}$$
Refraction Through Multiple Parallel Media (Compound Slab)
Consider a composite block containing Air (a), Water (w), and Glass (g) placed in parallel contact. A ray entering from Air at angle $i_1$ refracts into Water at $r_1$, then into Glass at $r_2$, and finally re-emerges into Air at angle $i_1$.
Applying Snell's Law at each interface sequentially:
1. Air to Water interface: ${}_a\mu_w = \frac{\sin i_1}{\sin r_1}$
2. Water to Glass interface: ${}_w\mu_g = \frac{\sin r_1}{\sin r_2}$
3. Glass to Air interface: ${}_g\mu_a = \frac{\sin r_2}{\sin i_1}$
Multiplying all three equations:
$${}_a\mu_w \times {}_w\mu_g \times {}_g\mu_a = \frac{\sin i_1}{\sin r_1} \times \frac{\sin r_1}{\sin r_2} \times \frac{\sin r_2}{\sin i_1} = 1$$
Hence, the relative refractive index of glass w.r.t water is given by:
$${}_w\mu_g = \frac{1}{{}_a\mu_w \times {}_g\mu_a} = \frac{{}_a\mu_g}{{}_a\mu_w}$$
Real Depth, Apparent Depth & Normal Displacement
When an object placed in an optically denser medium is observed from a rarer medium, it appears to be raised due to refraction at the interface.
Step-by-Step Mathematical Derivation
Consider a point object $O$ located at real depth $OA = d$ below the surface of a denser medium of refractive index $\mu$. A ray $OA$ incident normally on the interface passes straight without deviation along $AA'$. Another ray $OB$ incident at an angle $i$ to the normal refracts along $BC$ into air at an angle of refraction $r$.
1. Apply Snell's Law at point B:
$$\mu \sin i = 1 \cdot \sin r \implies \mu = \frac{\sin r}{\sin i}$$2. Trigonometric relations from right triangles:
In $\Delta OAB$: $\sin i = \frac{AB}{OB}$In $\Delta IAB$: $\sin r = \frac{AB}{IB}$
Substituting these into Snell's law gives: $$\mu = \frac{AB / IB}{AB / OB} = \frac{OB}{IB}$$
3. Paraxial approximation for near-normal viewing:
For rays viewed nearly normally (small angles $i$ and $r$), point $B$ lies very close to point $A$. Thus, $OB \approx OA$ and $IB \approx IA$. $$\mu = \frac{OA}{IA} = \frac{\text{Real Depth } (d)}{\text{Apparent Depth } (d')}$$ $$\text{Apparent Depth } (d') = \frac{d}{\mu}$$Normal Shift / Displacement ($s$):
The distance through which the object appears raised: $$s = \text{Real Depth} - \text{Apparent Depth} = d - d' = d - \frac{d}{\mu}$$ $$s = d \left( 1 - \frac{1}{\mu} \right)$$Lateral Shift (Lateral Displacement) Derivation
When a ray of light passes through a parallel-sided glass slab, the emergent ray is parallel to the incident ray but is shifted laterally by a perpendicular distance known as the Lateral Shift ($x$).
Complete Mathematical Proof
Let $t$ be the thickness of the glass slab and $i$ be the angle of incidence at the upper surface. Let $r$ be the angle of refraction inside the slab. The ray suffers a deviation of $(i - r)$ at the first surface.
Step 1: Express lateral shift $x$ in right-angled triangle $\Delta BCE$:
$$\sin(i - r) = \frac{CE}{BC} = \frac{x}{BC} \implies x = BC \sin(i - r) \quad \text{--- (Equation 1)}$$Step 2: Relate hypotenuse $BC$ to thickness $t$ in $\Delta BCF$:
In right triangle $\Delta BCF$ (where $CF \perp BF$, $CF = t$): $$\cos r = \frac{BF}{BC} = \frac{t}{BC} \implies BC = \frac{t}{\cos r} \quad \text{--- (Equation 2)}$$Step 3: Substitute Equation 2 into Equation 1:
$$x = t \frac{\sin(i - r)}{\cos r}$$Alternative Expression in terms of $\mu$ and $i$:
Expanding $\sin(i - r) = \sin i \cos r - \cos i \sin r$: $$x = t \frac{\sin i \cos r - \cos i \sin r}{\cos r} = t \sin i \left( 1 - \frac{\cos i \sin r}{\sin i \cos r} \right)$$ Using Snell's law ($\frac{\sin i}{\sin r} = \mu \implies \frac{\sin r}{\sin i} = \frac{1}{\mu}$) and $\tan r = \frac{\sin r}{\sqrt{1 - \sin^2 r}} = \frac{\sin i}{\sqrt{\mu^2 - \sin^2 i}}$: $$x = t \sin i \left( 1 - \frac{\cos i}{\sqrt{\mu^2 - \sin^2 i}} \right)$$Factors Affecting Lateral Shift ($x$):
- Thickness of slab ($t$): Directly proportional ($x \propto t$). Thicker slab produces greater shift.
- Angle of incidence ($i$): Increases with increase in angle of incidence $i$.
- Refractive Index ($\mu$): Higher $\mu$ causes smaller angle $r$, increasing $(i - r)$ and thus $x$.
- Wavelength ($\lambda$): Lateral shift is larger for shorter wavelength (violet) than longer wavelength (red).
Total Internal Reflection (TIR) & Critical Angle
When light travels from an optically denser medium to an optically rarer medium, the refracted ray bends away from the normal ($r > i$). As $i$ increases, $r$ reaches $90^\circ$. If $i$ exceeds this threshold, refraction ceases and light is entirely reflected back into the denser medium.
Critical Angle ($\theta_c$ or $i_c$) Definition
The critical angle is defined as the angle of incidence in the optically denser medium for which the angle of refraction in the optically rarer medium is $90^\circ$.
Mathematical Relation for Critical Angle
Applying Snell's law at the interface for critical incidence ($i = \theta_c$ and $r = 90^\circ$): $$\mu_1 \sin \theta_c = \mu_2 \sin 90^\circ$$ For denser medium ($\mu_1 = \mu$) and air/rarer medium ($\mu_2 = 1$): $$\mu \sin \theta_c = 1 \cdot 1 \implies \sin \theta_c = \frac{1}{\mu}$$ $$\theta_c = \sin^{-1}\left( \frac{1}{\mu} \right)$$
Two Essential Conditions for Total Internal Reflection:
- Light must travel from an optically denser medium towards an optically rarer medium.
- The angle of incidence ($i$) in the denser medium must be strictly greater than the critical angle ($\theta_c$) for the given pair of media ($i > \theta_c$).
Applications of Total Internal Reflection
1. Mirage Formation
On hot summer days, air near the ground gets intensely heated and becomes rarer, while upper atmospheric layers remain cooler and denser. Light rays from distant trees bend away from the normal progressively as they travel downward through air layers of decreasing refractive index. When angle $i > \theta_c$, TIR occurs, bending rays upward to the observer's eyes. The observer perceives an inverted image, creating the illusion of a water pool.
2. Sparkling of Diamond
Diamond has an exceptionally high refractive index ($\mu \approx 2.42$), resulting in a tiny critical angle ($\theta_c \approx 24.4^\circ$). Diamond cutters shape facets so light entering the crystal hits internal surfaces at angles greater than $24.4^\circ$, undergoing multiple internal reflections before exiting concentrated through specific faces, producing brilliant sparkle.
3. Totally Reflecting Glass Prisms (Porro Prisms)
Right-angled isosceles prisms ($45^\circ - 90^\circ - 45^\circ$) made of crown glass ($\mu \approx 1.5, \theta_c \approx 42^\circ$) utilize TIR to deviate or invert rays without loss of intensity.
Working Principle & Structure of Optical Fiber
An optical fiber is a hair-thin flexible glass or plastic wave-guide that transmits light signals over long distances via continuous total internal reflections with extremely low loss of signal power.
Derivation of Acceptance Angle ($\theta_a$) & Numerical Aperture ($NA$)
Consider a light ray entering the fiber core from air ($\mu_0 = 1$) at an angle of incidence $i = \theta_a$. It refracts into the core ($\mu_1$) at angle $r$. The ray hits the core-cladding interface at angle $\phi = (90^\circ - r)$.
1. Condition for TIR at Core-Cladding Boundary:
For TIR to occur, $\phi \ge \theta_c \implies \sin\phi \ge \sin\theta_c = \frac{\mu_2}{\mu_1}$.Since $\sin\phi = \sin(90^\circ - r) = \cos r$, we get: $$\cos r \ge \frac{\mu_2}{\mu_1}$$
2. Apply Snell's Law at the air-core entrance:
$$\mu_0 \sin\theta_a = \mu_1 \sin r = \mu_1 \sqrt{1 - \cos^2 r}$$ Substituting $\cos r = \frac{\mu_2}{\mu_1}$ into the limit case: $$\sin\theta_a = \mu_1 \sqrt{1 - \left(\frac{\mu_2}{\mu_1}\right)^2} = \sqrt{\mu_1^2 - \mu_2^2}$$Numerical Aperture ($NA$):
$$NA = \sin\theta_a = \sqrt{\mu_1^2 - \mu_2^2}$$Acceptance Angle ($\theta_a$):
$$\theta_a = \sin^{-1}\left( \sqrt{\mu_1^2 - \mu_2^2} \right)$$Medical Application (Endoscopy)
Flexible fiber optical bundles transmit light down interior body organs (stomach, lungs) and bring back real-time diagnostic images without invasive surgery.
Telecommunications
High-frequency optical pulses transmit terabits of data per second across continental and subsea internet networks with negligible electromagnetic interference.
Interactive Refraction & TIR Ray Simulator
Real-Time Data
Critical Angle ($\theta_c$): N/A
Refraction Angle ($r$): 28.1°
Status: Standard Refraction
Solved NEB Standard Numerical Problems
Q1: A glass vessel contains water ($\mu_w = 4/3$) up to a height of $12\text{ cm}$ and oil ($\mu_o = 1.4$) floated above it to a depth of $7\text{ cm}$. Find the apparent depth of the vessel's bottom when viewed normally from air.
Step-by-Step NEB Solution:
Given: Depth of water layer ($d_1$) = $12\text{ cm}$, Refractive index of water ($\mu_1$) = $4/3$
Depth of oil layer ($d_2$) = $7\text{ cm}$, Refractive index of oil ($\mu_2$) = $1.4$
Formula: Total Apparent Depth $d' = \frac{d_1}{\mu_1} + \frac{d_2}{\mu_2}$
Calculation: $$d' = \frac{12}{4/3} + \frac{7}{1.4} = \left(12 \times \frac{3}{4}\right) + 5 = 9\text{ cm} + 5\text{ cm} = 14\text{ cm}$$
Result: The apparent depth of the bottom is $14\text{ cm}$ (Apparent shift $s = 19 - 14 = 5\text{ cm}$).
Q2: A ray of light strikes a parallel glass slab of thickness $10\text{ cm}$ at an angle of incidence $60^\circ$. If the refractive index of glass is $1.5$, calculate the lateral shift of the emergent ray.
Step-by-Step NEB Solution:
Given: Thickness ($t$) = $10\text{ cm}$, Angle of incidence ($i$) = $60^\circ$, Refractive index ($\mu$) = $1.5$
1. Find angle of refraction $r$: $$\sin r = \frac{\sin i}{\mu} = \frac{\sin 60^\circ}{1.5} = \frac{0.8660}{1.5} = 0.5773 \implies r \approx 35.26^\circ$$
2. Calculate $(i - r) = 60^\circ - 35.26^\circ = 24.74^\circ$ and $\cos r = \cos(35.26^\circ) \approx 0.8165$.
3. Apply lateral shift formula: $$x = t \frac{\sin(i - r)}{\cos r} = 10 \times \frac{\sin 24.74^\circ}{0.8165} = 10 \times \frac{0.4185}{0.8165} \approx 5.13\text{ cm}$$
Result: The lateral displacement of the ray is $5.13\text{ cm}$.
Q3: An optical fiber has a core of refractive index $1.55$ and cladding of refractive index $1.48$. Compute (a) the critical angle at the core-cladding boundary, (b) Numerical Aperture, and (c) Acceptance angle in air.
Step-by-Step NEB Solution:
Given: Core $\mu_1 = 1.55$, Cladding $\mu_2 = 1.48$
(a) Critical Angle $\theta_c = \sin^{-1}\left(\frac{\mu_2}{\mu_1}\right) = \sin^{-1}\left(\frac{1.48}{1.55}\right) = \sin^{-1}(0.9548) \approx 72.71^\circ$
(b) Numerical Aperture $NA = \sqrt{\mu_1^2 - \mu_2^2} = \sqrt{(1.55)^2 - (1.48)^2} = \sqrt{2.4025 - 2.1904} = \sqrt{0.2121} \approx 0.4605$
(c) Acceptance Angle $\theta_a = \sin^{-1}(NA) = \sin^{-1}(0.4605) \approx 27.42^\circ$
Result: Critical Angle = $72.71^\circ$, $NA = 0.461$, Acceptance Angle = $27.42^\circ$.