Optics • Topic 2 of 3

Ray Optics: Refraction, Lenses & Instruments

Refraction is the bending of light as it changes speed passing between media. The refractive index is $n = \dfrac{c}{v}$, and Snell's law relates the angles: $n_1 \sin i = n_2 \sin r$. Light bends towards the normal entering a denser medium and away entering a rarer one.

From a denser to a rarer medium, if the angle of incidence exceeds the critical angle $i_c$, all light is reflected — total internal reflection (TIR), with $\sin i_c = \dfrac{1}{n}$. TIR underlies optical fibres, diamond brilliance and reflecting prisms.

Refraction at a spherical surface gives $\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}$. Applying this at both surfaces of a thin lens yields the lens maker's formula $\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$. The thin lens formula $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ locates the image, with magnification $m = \dfrac{v}{u}$. A converging (convex) lens has $f > 0$, a diverging (concave) lens $f < 0$. The power is $P = \dfrac{1}{f\,(\text{m})}$ in dioptres, and lenses in contact add: $P = P_1 + P_2$.

A prism deviates a ray by $\delta$, which is minimum ($\delta_m$) when the ray passes symmetrically. Then $n = \dfrac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$, where $A$ is the refracting angle. Since $n$ varies with wavelength (dispersion), white light splits into a spectrum — violet deviating most, red least.

Optical instruments enlarge the visual angle. A simple microscope gives $M = 1 + \dfrac{D}{f}$ ($D = 25\,\text{cm}$). A compound microscope uses an objective and eyepiece, $M \approx \dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$. An astronomical telescope in normal adjustment has $M = \dfrac{f_o}{f_e}$ and tube length $L = f_o + f_e$; a large objective and small eyepiece focal length give high magnification and a bright image.

Converging (convex) lens: real, inverted image of an object beyond 2F2FFF2FObjectImage
1
Worked Example
Light passes from air into glass of refractive index $1.5$ at an angle of incidence of $30^\circ$. Find the angle of refraction.
Solution
  1. Snell's law: $n_1 \sin i = n_2 \sin r$, with $n_1 = 1$, $n_2 = 1.5$, $i = 30^\circ$.
  2. $1 \times \sin 30^\circ = 1.5 \times \sin r \Rightarrow \sin r = \dfrac{0.5}{1.5} = 0.333$.
  3. $r = \sin^{-1}(0.333) \approx 19.5^\circ$.

Answer: The angle of refraction is about $19.5^\circ$ (light bends towards the normal in the denser medium).

2
Worked Example
The refractive index of water is $1.33$. Calculate the critical angle for a water-air interface.
Solution
  1. For total internal reflection at a denser-to-rarer (water-air) boundary, $\sin i_c = \dfrac{1}{n}$.
  2. $\sin i_c = \dfrac{1}{1.33} = 0.752$.
  3. $i_c = \sin^{-1}(0.752) \approx 48.8^\circ$.

Answer: The critical angle is about $48.8^\circ$.

3
Worked Example
A biconvex lens has both radii of curvature equal to $20\,\text{cm}$ and is made of glass of refractive index $1.5$. Find its focal length.
Solution
  1. Lens maker's formula: $\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$.
  2. For a biconvex lens $R_1 = +20\,\text{cm}$, $R_2 = -20\,\text{cm}$.
  3. $\dfrac{1}{f} = (1.5 - 1)\left(\dfrac{1}{20} - \dfrac{1}{-20}\right) = 0.5 \times \dfrac{2}{20} = \dfrac{1}{20}$.
  4. $f = +20\,\text{cm}$.

Answer: The focal length is $20\,\text{cm}$ (converging lens).

4
Worked Example
An object is placed $30\,\text{cm}$ from a convex lens of focal length $20\,\text{cm}$. Find the image distance and magnification.
Solution
  1. $u = -30\,\text{cm}$, $f = +20\,\text{cm}$. Lens formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$.
  2. $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{20} + \dfrac{1}{-30} = \dfrac{3 - 2}{60} = \dfrac{1}{60}$.
  3. $v = +60\,\text{cm}$.
  4. $m = \dfrac{v}{u} = \dfrac{60}{-30} = -2$.

Answer: Image is $60\,\text{cm}$ from the lens, real, inverted and magnified two times.

5
Worked Example
The angle of a prism is $60^\circ$ and the angle of minimum deviation is $40^\circ$. Find the refractive index of the prism.
Solution
  1. Use $n = \dfrac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$ with $A = 60^\circ$, $\delta_m = 40^\circ$.
  2. $\dfrac{A + \delta_m}{2} = \dfrac{60 + 40}{2} = 50^\circ$; $\dfrac{A}{2} = 30^\circ$.
  3. $n = \dfrac{\sin 50^\circ}{\sin 30^\circ} = \dfrac{0.766}{0.5}$.
  4. $n = 1.532$.

Answer: The refractive index of the prism is about $1.53$.

6
Worked Example
An astronomical telescope has an objective of focal length $100\,\text{cm}$ and an eyepiece of focal length $5\,\text{cm}$. Find the magnifying power and tube length in normal adjustment.
Solution
  1. In normal adjustment $M = \dfrac{f_o}{f_e} = \dfrac{100}{5}$.
  2. $M = 20$.
  3. Tube length $L = f_o + f_e = 100 + 5$.
  4. $L = 105\,\text{cm}$.

Answer: Magnifying power $= 20$; tube length $= 105\,\text{cm}$.

Key Points

  • Snell's law: $n_1 \sin i = n_2 \sin r$; refractive index $n = c/v$.
  • Total internal reflection occurs beyond the critical angle, where $\sin i_c = \dfrac{1}{n}$.
  • Lens maker's formula: $\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$; lens formula $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$; power $P = 1/f$ (m) in dioptres.
  • At minimum deviation $n = \dfrac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$.
  • Telescope (normal adjustment) $M = \dfrac{f_o}{f_e}$; simple microscope $M = 1 + \dfrac{D}{f}$.
Tap an option to check your answer0 / 4
Q1.Light bends towards the normal when it passes from:
Explanation: Entering a denser medium the speed decreases, so the ray bends towards the normal.
Q2.The power of a convex lens of focal length $50\,\text{cm}$ is:
Explanation: $P = 1/f(\text{m}) = 1/0.5 = +2\,\text{D}$; positive for a converging lens.
Q3.Total internal reflection can occur when light travels from:
Explanation: TIR needs a denser-to-rarer path and an incidence angle greater than the critical angle.
Q4.In an astronomical telescope in normal adjustment, the magnifying power is:
Explanation: Normal adjustment (final image at infinity) gives $M = f_o/f_e$.