Laws of Refraction & Snell's Law - Tenth Grade Science
Light travels rectilinearly at a characteristic constant velocity through any uniform, homogeneous optical medium. However, when an oblique ray of light passes across the boundary separating two transparent media of different optical densities—such as from air into water, or from air into optical crown glass—it undergoes an abrupt change in velocity that causes the ray to bend. This optical phenomenon is termed refraction. Refraction governs everyday optical effects, from the apparent bending of a pencil immersed in water to the shimmering of desert mirages and the focusing power of the human eye. In this chapter, you will investigate the wave mechanics of refraction, analyze the two fundamental Laws of Refraction, explore Willebrord Snell's law ($\sin i / \sin r = \text{constant}$), trace the lateral displacement of light passing through a rectangular glass slab, and evaluate optical phenomena resulting from boundary refraction.
The Physical Cause and Wave Mechanics of Refraction
Refraction is the deviation in the direction of propagation of a wave of light as it crosses the interface between two transparent optical media possessing different propagation speeds.
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| REFRACTION ACROSS AN OPTICAL BOUNDARY |
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| |
| Incident Ray Normal (NN') |
| \ | |
| \ | OPTICALLY RARER MEDIUM |
| \ Angle of | (Higher speed of light, v1)|
| \ Incidence (i) | e.g., AIR |
| \ | |
| \ | |
| ========================*=====================*============================= |
| Optical Boundary \ | OPTICALLY DENSER MEDIUM |
| \ Angle of | (Lower speed of light, v2) |
| \ Refraction (r) | e.g., GLASS / WATER |
| \ | |
| v | |
| Refracted Ray v |
| |
| Wave bends TOWARD the normal when entering an optically denser medium (v2 < v1). |
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The Velocity Differential Mechanism
The physical cause of refraction is the difference in the speed of light across different optical substances: In a vacuum, light propagates at its maximum speed: $c \approx 3.0 \times 10^8\text{ m/s}$. In water, the speed of light drops to approximately $2.25 \times 10^8\text{ m/s}$. In crown glass, the speed of light drops to approximately $2.0 \times 10^8\text{ m/s}$.
When a wavefront of light strikes an optical interface obliquely, one side of the wavefront enters the new medium and changes speed before the other side. This difference in speed across the wavefront pivots the trajectory of the wave, bending the light beam.
During refraction across any medium boundary: The speed ($v$) of the light wave changes. The wavelength ($\lambda$) of the light wave changes proportionally ($v \propto \lambda$). The frequency ($\nu$) of the light wave remains constant, because frequency is determined by the oscillating source that emits the electromagnetic wave.
The Two Rules of Directional Bending
Rule 1 (Rarer to Denser): When a ray of light passes from an optically rarer medium (higher light speed, e.g., air) into an optically denser medium (lower light speed, e.g., glass or water), the refracted ray bends toward the normal line ($\angle i > \angle r$).
Rule 2 (Denser to Rarer): When a ray of light passes from an optically denser medium into an optically rarer medium, the refracted ray bends away from the normal line ($\angle i < \angle r$).
If a light ray strikes the interface normally ($\angle i = 0^\circ$), it passes straight into the second medium without changing direction ($\angle r = 0^\circ$), although its speed and wavelength change.
The Laws of Refraction and Snell's Formulation
The propagation of refracted light is governed by two empirical laws of refraction.
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| THE TWO LAWS OF REFRACTION |
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| |
| FIRST LAW OF REFRACTION: |
| The incident ray, the refracted ray, and the normal to the interface at the |
| point of incidence all lie in the same geometric plane. |
| |
| SECOND LAW OF REFRACTION (SNELL'S LAW): |
| The ratio of the sine of the angle of incidence to the sine of the angle of |
| refraction is a constant for a given pair of media and for light of a given color|
| |
| sin(i) |
| -------- = constant = n_21 |
| sin(r) |
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Snell's Law of Refraction
Formulated by Dutch astronomer Willebrord Snell in 1621, the Second Law of Refraction states:
For a given pair of transparent optical media and for light of a given wavelength (color), the ratio of the sine of the angle of incidence ($\sin i$) to the sine of the angle of refraction ($\sin r$) is a constant:
$$\frac{\sin i}{\sin r} = \text{constant} = n_{21} = \frac{n_2}{n_1}$$
This constant ratio is termed the relative refractive index of the second medium with respect to the first medium ($n_{21}$).
Written in symmetric form:
$$n_1 \sin i = n_2 \sin r$$
where $n_1$ is the absolute refractive index of the incident medium and $n_2$ is the absolute refractive index of the refracting medium.
Snell's Law demonstrates that the degree of bending depends not on the angles themselves, but on the trigonometric sines of those angles. If the angle of incidence is doubled, the angle of refraction does not double; the sines of the angles scale proportionally.
Refraction Through a Rectangular Glass Slab: Lateral Displacement
When a ray of light traverses a parallel-sided rectangular glass slab, it undergoes two sequential refractions at opposite parallel surfaces.
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| REFRACTION THROUGH A RECTANGULAR GLASS SLAB |
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| |
| Incident Ray A |
| \ |
| \ Angle i1 |
| --------------------*--------------------------------------------- |
| Air (n1) \ Angle r1 |
| \ |
| Glass Slab (n2) \ Refracted Ray inside glass |
| \ |
| \ Angle i2 = r1 (Alternate angles) |
| --------------------------*--------------------------------------- |
| Air (n1) \ |
| \ Angle of Emergence (e = i1) |
| \ |
| v Emergent Ray CD |
| | <--- Lateral Displacement (d) |
| - - - - Original Incident Path Extended |
| |
| Result: The Emergent Ray is PARALLEL to the original Incident Ray direction, |
| but is shifted sideways by a perpendicular distance d (Lateral Shift). |
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Derivation: Emergent Ray Parallelism ($\angle i = \angle e$)
Consider an incident ray $AB$ striking the top surface of a rectangular glass slab of thickness $t$ at an angle of incidence $i_1$: At the first air-to-glass interface: Light enters an optically denser medium and bends toward the normal, with an angle of refraction $r_1$. According to Snell's Law:
$$\frac{\sin i_1}{\sin r_1} = n_{ga} = \frac{n_g}{n_a} \quad \text{--- (Equation 1)}$$
At the second glass-to-air interface: The ray travels through the glass and strikes the bottom parallel face at point $C$. The normal at $C$ is parallel to the normal at $B$. Therefore, the angle of incidence at the bottom surface ($i_2$) is an alternate interior angle equal to the angle of refraction at the top surface:
$$i_2 = r_1$$
The ray exits the glass into air, bending away from the normal with an angle of emergence ($e$). Applying Snell's Law at the second interface:
$$\frac{\sin i_2}{\sin e} = \frac{\sin r_1}{\sin e} = n_{ag} = \frac{n_a}{n_g} = \frac{1}{n_{ga}} \quad \text{--- (Equation 2)}$$
Multiplying Equation 1 and Equation 2:
$$\left(\frac{\sin i_1}{\sin r_1}\right) \times \left(\frac{\sin r_1}{\sin e}\right) = n_{ga} \times \frac{1}{n_{ga}} = 1 \implies \frac{\sin i_1}{\sin e} = 1 \implies \sin i_1 = \sin e$$
Therefore:
$$\angle i_1 = \angle e$$
The angle of emergence ($\angle e$) is equal to the angle of incidence ($\angle i_1$). Because the emergent ray emerges into the same optical medium at the same angle to parallel normals, the emergent ray is parallel to the path of the original incident ray.
Lateral Displacement ($d$)
Although the emergent ray is parallel to the incident ray, it is displaced sideways. The perpendicular distance between the original incident ray path and the emergent ray is termed lateral displacement ($d$).
The magnitude of lateral displacement depends on three factors: Thickness of the Glass Slab ($t$): Lateral displacement is directly proportional to slab thickness; thicker slabs produce larger lateral shifts. Refractive Index of the Glass ($n$): Denser optical media with higher refractive indices bend light more sharply, increasing lateral displacement. Angle of Incidence ($i$): Larger angles of incidence increase lateral displacement.
Everyday Optical Phenomena Caused by Refraction
Refraction produces observable optical effects in daily life:
Apparent Depth of Submerged Objects: When an object (such as a coin or pebble) rests at the bottom of a pool of water, light rays radiating from the object bend away from the normal upon exiting the water into air. Tracing these diverging rays back into the water makes them appear to originate from an apparent depth shallower than the true physical depth:
$$\text{Apparent Depth} = \frac{\text{Real Depth}}{n_{\text{water}}}$$
Because water has a refractive index of $n \approx 1.33$ ($4/3$), a pool of water appears roughly three-quarters of its actual physical depth.
Apparent Bending of a Submerged Stick: A straight stick or pencil placed obliquely in a beaker of water appears bent at the water surface. Light rays from the submerged portion refract away from the normal upon entering air, displacing the apparent position of the submerged portion upward relative to the portion in air.
Chapter Exercises
Exercise 1: A ray of light traveling in air strikes the flat surface of a glass block at an angle of incidence of $60^\circ$. If the refractive index of the glass is $1.50$, calculate the angle of refraction inside the glass block using Snell's Law, and state whether the ray bends toward or away from the normal.
Exercise 2: Prove that when a ray of light undergoes refraction through a parallel-sided rectangular glass slab surrounded by air on both sides, the emergent ray is strictly parallel to the original incident ray, showing all algebraic steps involving Snell's Law.
Exercise 3: Explain why a coin resting at the bottom of a glass beaker filled with water appears elevated when viewed from above, and derive the mathematical relationship connecting real depth, apparent depth, and the refractive index of water.
Solutions and Explanations
Solution 1: The incident medium is air ($n_1 = 1.00$), and the refracting medium is glass ($n_2 = 1.50$). The angle of incidence is $i = 60^\circ$. Using Snell's Law: $n_1 \sin i = n_2 \sin r \implies 1.00 \times \sin(60^\circ) = 1.50 \times \sin(r)$. Since $\sin(60^\circ) = \sqrt{3}/2 \approx 0.8660$: $0.8660 = 1.50 \times \sin(r) \implies \sin(r) = 0.8660 / 1.50 \approx 0.5773$. Taking the inverse sine: $r = \arcsin(0.5773) \approx 35.26^\circ$. The angle of refraction inside the glass block is approximately $35.3^\circ$. Because the light travels from an optically rarer medium (air) into an optically denser medium (glass), the speed of light decreases, and the refracted ray bends toward the normal ($\angle r = 35.3^\circ < \angle i = 60.0^\circ$).
Solution 2: Consider a light ray $AB$ striking the top face of a rectangular glass slab with parallel surfaces at angle of incidence $i$. At the first air-to-glass boundary, the ray refracts at angle $r_1$. According to Snell's Law: $\sin(i) / \sin(r_1) = n_g / n_a$, where $n_a$ is the refractive index of air and $n_g$ is the refractive index of glass. Inside the slab, the ray travels to the bottom face, striking it at angle $i_2$. Because the top and bottom faces are parallel, the normals erected at the two surfaces are parallel lines; therefore, the angle of incidence at the bottom surface is an alternate interior angle to the first angle of refraction: $i_2 = r_1$. At the bottom glass-to-air boundary, the ray emerges into air at angle of emergence $e$. Applying Snell's Law at the second boundary: $\sin(i_2) / \sin(e) = \sin(r_1) / \sin(e) = n_a / n_g$. Multiplying the two boundary equations: $[\sin(i) / \sin(r_1)] \times [\sin(r_1) / \sin(e)] = [n_g / n_a] \times [n_a / n_g] = 1$. The term $\sin(r_1)$ cancels out: $\sin(i) / \sin(e) = 1 \implies \sin(i) = \sin(e) \implies i = e$. Because the angle of incidence at the first interface equals the angle of emergence at the second interface relative to parallel normals, the emergent ray exits in a direction parallel to the original incident ray.
Solution 3: A coin resting at the bottom of a beaker filled with water appears elevated due to the refraction of light rays at the water-air interface. Light rays radiate outward in all directions from point $O$ on the submerged coin. A ray traveling perpendicular to the surface ($\angle i = 0^\circ$) passes into air without deviation. An adjacent ray striking the surface obliquely at a small angle of incidence $i$ passes from an optically denser medium (water, $n$) into an optically rarer medium (air, $n \approx 1.0$), bending away from the normal at an angle of refraction $r > i$. An observer's eye collects these diverging rays. Tracing the refracted rays backward in straight lines, they appear to intersect at a virtual image point $I$ located higher than the actual coin $O$. For small angles of incidence where $\sin i \approx \tan i$ and $\sin r \approx \tan r$: Let $d_{\text{real}}$ be the real depth of the water ($OA$), and $d_{\text{apparent}}$ be the apparent depth ($IA$). From the right triangles formed with the surface normal: $\tan i = AB / OA = AB / d_{\text{real}}$, and $\tan r = AB / IA = AB / d_{\text{apparent}}$. Applying Snell's Law: $n \sin i = 1 \sin r \implies n \tan i \approx \tan r \implies n (AB / d_{\text{real}}) = AB / d_{\text{apparent}}$. Canceling $AB$ yields: $n / d_{\text{real}} = 1 / d_{\text{apparent}}$, which rearranges to $d_{\text{apparent}} = d_{\text{real}} / n$. Because the refractive index of water is $n \approx 1.33$ ($4/3$), the apparent depth is roughly $3/4$ of the real depth, making the bottom of the beaker appear elevated.