Reflection and Refraction of Light: Quick Revision

A brief note on Reflection and Refraction of Light. Read it before a test. Try each question at the end before you open its answer.

Mirrors

• Law of reflection: $i = r$, with both angles from the normal. The incident ray, the reflected ray and the normal lie in one plane.
• Plane mirror: a virtual, upright, same-size image, as far behind the mirror as the object is in front. A mirror that turns by $\theta$ turns the reflected ray by $2\theta$.
• Spherical mirror: $f = \dfrac{R}{2}$ for rays close to the axis.
• New Cartesian signs: light travels left to right. A real object has $u < 0$. A concave mirror has $f < 0$ and a convex mirror has $f > 0$.
• Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$. Magnification: $m = -\dfrac{v}{u}$.
• Real is positive (A-Level, IB): $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$ with $u$ and $v$ positive for real objects and images.

Refraction

• Refractive index: $n = \dfrac{c}{v}$. The frequency stays the same. The wavelength is $\dfrac{\lambda_0}{n}$.
• Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$.
• Apparent depth (looking straight down): real depth divided by $n$.
• Slab: the emergent ray is parallel to the incident ray, shifted by $d = \dfrac{t\sin(i - r)}{\cos r}$.
• Critical angle: $\sin C = \dfrac{n_2}{n_1}$, for light going from the denser to the rarer material.
• Total internal reflection: the angle of incidence is larger than $C$. Uses: diamond, optical fibre with $\sin\theta_a = \sqrt{n_1^2 - n_2^2}$, mirage, prism.

A spherical surface

• Formula: $\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}$, with $R > 0$ when the centre of curvature is on the side of the outgoing light.
• Magnification: $m = \dfrac{n_1 v}{n_2 u}$.
• Flat surface: $v = \dfrac{n_2}{n_1}u$.
• Ball lens: the focus is at $\dfrac{nR}{2(n - 1)}$ from the centre.

A worked example

A concave mirror has $f = -20\ \text{cm}$. A pen $4.0\ \text{cm}$ tall stands $50\ \text{cm}$ in front of it. Find the image distance and the height of the image.

1. Distance. $v = \dfrac{uf}{u - f} = \dfrac{(-50\ \text{cm})(-20\ \text{cm})}{-50\ \text{cm} + 20\ \text{cm}} = -33.3\ \text{cm}$.
2. Magnification. $m = -\dfrac{v}{u} = -\dfrac{-33.3\ \text{cm}}{-50\ \text{cm}} = -0.667$.
3. Height. $h' = mh = -0.667 \times 4.0\ \text{cm} = -2.7\ \text{cm}$. The image is real, inverted and smaller.

Practise with new numbers

Check yourself

1. A convex mirror has $f = +15\ \text{cm}$. An object is $45\ \text{cm}$ in front of it. Where is the image?

2. A ray in air meets a pool of water at $70°$ to the normal. Can it be totally reflected?

3. A coin lies $1.33\ \text{m}$ under water ($n = 1.33$). How deep does it seem, looking straight down?

Sources

• NCERT Physics Part II (Class 12), Chapter 9, the sections on mirrors, refraction and total internal reflection.
• OpenStax University Physics Volume 3, Chapters 1 and 2, the sections on reflection, refraction and mirrors.
• OpenStax College Physics 2e, Chapter 25, the sections on reflection, refraction and total internal reflection.