Optics — Practice Questions with Answers

Refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium: $n = c/v$ — a measure of how much a medium slows down and bends light.

Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$, where $\theta_1$ and $\theta_2$ are the angles of incidence and refraction respectively, relative to the normal.

The critical angle is the specific angle of incidence (in the denser medium) at which the angle of refraction becomes exactly 90° — for angles of incidence beyond this, total internal reflection occurs instead of refraction.

Using Snell's law with $\theta_2=90°$: $n_1\sin\theta_c = n_2\sin90° = n_2$

$\sin\theta_c = \dfrac{n_2}{n_1}$ (where $n_1 > n_2$, denser to rarer medium)

The lens maker's formula: $\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$

Here, $f$ is the focal length, $n$ is the refractive index of the lens material relative to the surrounding medium, and $R_1$, $R_2$ are the radii of curvature of the lens's two surfaces (with appropriate sign convention). This formula shows that a lens's focal length depends both on the material it's made from and its specific curved geometry.

Dispersion is the splitting of white (composite) light into its constituent colours (wavelengths) as it passes through a prism.

This occurs because the refractive index of the prism's material is slightly different for different wavelengths of light (a property called dispersion) — since violet light has a shorter wavelength and is refracted (bent) more than red light, each individual colour emerges from the prism at a slightly different angle, spreading out into the visible spectrum.

Myopia (nearsightedness): Distant objects appear blurred, since the eye focuses their image in front of the retina — corrected using a diverging (concave) lens, which shifts the focus back onto the retina.

Hypermetropia (farsightedness): Nearby objects appear blurred, since the eye focuses their image behind the retina (or the eye cannot accommodate sufficiently) — corrected using a converging (convex) lens, which brings the focus forward onto the retina.

• The two light sources must be coherent — maintaining a constant phase relationship with each other over time.
• The sources should ideally have the same frequency/wavelength, and comparable amplitude, for clear, high-contrast fringes.
• The sources should be narrow and close together (relative to the screen distance) to produce a fringe pattern of measurable, observable width.

Interference results from the superposition of light waves from two (or more) distinct, coherent sources — producing a pattern of alternating bright and dark fringes of generally similar/equal width and intensity.

Diffraction results from the bending/spreading of light waves around obstacles or through narrow openings, and technically involves the superposition of wavelets from different parts of the same wavefront — the resulting pattern typically has a bright, wide central maximum, with progressively fainter secondary maxima on either side, unlike the generally uniform interference pattern.

Polarisation refers to restricting the oscillation of a light wave's electric field vector to a single specific plane (rather than oscillating in all possible perpendicular directions, as in ordinary unpolarised light).

Polarisation is possible only for transverse waves (where oscillation is perpendicular to propagation direction) — since longitudinal waves oscillate along their direction of travel, they have no "orientation" perpendicular to propagation that could be selectively restricted. The experimentally observed ability to polarise light (e.g., using polaroid filters) is therefore direct evidence confirming that light is fundamentally a transverse wave.

Sunlight, containing all visible wavelengths, interacts with the tiny molecules present in Earth's atmosphere — this interaction causes Rayleigh scattering, where scattering intensity is strongly wavelength-dependent, being much more effective for shorter wavelengths (blue light) than longer wavelengths (red light).

Since blue light is scattered far more strongly and in many more directions across the sky compared to other colours, an observer looking at any part of the daytime sky (away from the Sun's direct position) predominantly receives this strongly scattered blue light, which is precisely why the sky appears blue rather than another colour.

An optical fibre consists of a central "core" of high refractive index glass, surrounded by an outer "cladding" of slightly lower refractive index material.

Light entering the fibre at an appropriate angle strikes the core-cladding boundary at an angle exceeding the critical angle for that interface, undergoing total internal reflection — this reflected light continues bouncing repeatedly off the core-cladding boundary, effectively trapped and guided along the length of the fibre with minimal loss, allowing efficient, long-distance transmission of light-encoded signals (widely used in telecommunications).

Using sign convention (concave mirror, real object): $u=-30\text{ cm}$, $f=-10\text{ cm}$

$\dfrac{1}{f} = \dfrac{1}{v}+\dfrac{1}{u} \Rightarrow \dfrac{1}{v} = \dfrac{1}{f}-\dfrac{1}{u} = \dfrac{1}{-10}-\dfrac{1}{-30}$

$\dfrac{1}{v} = -\dfrac{1}{10}+\dfrac{1}{30} = -\dfrac{3}{30}+\dfrac{1}{30} = -\dfrac{2}{30} = -\dfrac{1}{15}$

$v = \mathbf{-15\text{ cm}}$ — the image forms 15 cm in front of the mirror, and is real, inverted, and diminished (since $|v|<|u|$).

$n_1\sin\theta_1 = n_2\sin\theta_2$

$1\times\sin30° = 1.5\times\sin\theta_2$

$0.5 = 1.5\sin\theta_2 \Rightarrow \sin\theta_2 = \dfrac{0.5}{1.5} = 0.333$

$\theta_2 = \sin^{-1}(0.333) \approx \mathbf{19.5°}$

Using sign convention: $u=-30\text{ cm}$, $f=+20\text{ cm}$

$\dfrac{1}{f} = \dfrac{1}{v}-\dfrac{1}{u} \Rightarrow \dfrac{1}{v} = \dfrac{1}{f}+\dfrac{1}{u} = \dfrac{1}{20}+\dfrac{1}{-30}$

$\dfrac{1}{v} = \dfrac{3}{60}-\dfrac{2}{60} = \dfrac{1}{60}$

$v = 60\text{ cm}$

Magnification: $m = \dfrac{v}{u} = \dfrac{60}{-30} = \mathbf{-2}$ — image is real, inverted, and magnified 2 times.

$\sin\theta_c = \dfrac{n_{air}}{n_{water}} = \dfrac{1}{1.33}$

$\sin\theta_c \approx 0.7519$

$\theta_c = \sin^{-1}(0.7519) \approx \mathbf{48.75°}$

For lenses in contact: $\dfrac{1}{f} = \dfrac{1}{f_1}+\dfrac{1}{f_2} = \dfrac{1}{10}+\dfrac{1}{-20}$

$\dfrac{1}{f} = \dfrac{2}{20}-\dfrac{1}{20} = \dfrac{1}{20}$

$f = 20\text{ cm} = 0.2\text{ m}$

Power: $P = \dfrac{1}{f(\text{in metres})} = \dfrac{1}{0.2} = \mathbf{+5\text{ D}}$

$\beta = \dfrac{\lambda D}{d} \Rightarrow \lambda = \dfrac{\beta d}{D}$

$\lambda = \dfrac{1.2\times10^{-3}\times0.5\times10^{-3}}{1}$

$\lambda = 0.6\times10^{-6}\text{ m} = 600\times10^{-9}\text{ m}$

$\lambda = \mathbf{600\text{ nm}}$

The corrective lens must form a virtual image of a distant (infinity) object at the person's actual far point (2 m), so the eye can then focus on this virtual image.

$u=-\infty$, $v=-2\text{ m}$ (virtual image, same side as object, using sign convention)

$\dfrac{1}{f} = \dfrac{1}{v}-\dfrac{1}{u} = \dfrac{1}{-2}-\dfrac{1}{-\infty} = -\dfrac{1}{2}-0 = -\dfrac{1}{2}$

$f = -2\text{ m}$; Power $P = \dfrac{1}{f} = \mathbf{-0.5\text{ D}}$ (a diverging/concave lens, as expected for myopia correction)

$n = \dfrac{\sin\left(\dfrac{A+D_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$

$n = \dfrac{\sin\left(\dfrac{60+30}{2}\right)}{\sin\left(\dfrac{60}{2}\right)} = \dfrac{\sin(45°)}{\sin(30°)}$

$n = \dfrac{0.707}{0.5} = \mathbf{1.414}$ (approximately $\sqrt{2}$)

For a real, inverted image, magnification is negative: $m = -3$ (given magnitude 3, real image so negative sign), $u=-20\text{ cm}$

$m = -\dfrac{v}{u} \Rightarrow -3 = -\dfrac{v}{-20} \Rightarrow -3 = \dfrac{v}{20} \Rightarrow v = -60\text{ cm}$

Using mirror formula: $\dfrac{1}{f} = \dfrac{1}{v}+\dfrac{1}{u} = \dfrac{1}{-60}+\dfrac{1}{-20} = -\dfrac{1}{60}-\dfrac{3}{60} = -\dfrac{4}{60} = -\dfrac{1}{15}$

$f = \mathbf{-15\text{ cm}}$ (concave mirror, as expected)

Using the thin-prism approximation for deviation: $D = (n-1)A$

$D_{red} = (1.51-1)\times60° = 0.51\times60° = 30.6°$

$D_{violet} = (1.53-1)\times60° = 0.53\times60° = 31.8°$

Angular dispersion: $D_{violet} - D_{red} = 31.8° - 30.6° = \mathbf{1.2°}$