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Waves - Light

Grade 12IGCSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Reflection: The law of reflection states that the angle of incidence θi\theta_i is equal to the angle of reflection θr\theta_r relative to the normal line.

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Refraction: The change in direction of light as it passes from one medium to another due to a change in speed. The refractive index nn is defined as the ratio of the speed of light in a vacuum cc to the speed of light in the medium vv.

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Snell's Law: Relates the angles of incidence and refraction to the refractive indices of the two media: n1sin⁡(θ1)=n2sin⁡(θ2)n_1 \sin(\theta_1) = n_2 \sin(\theta_2).

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Critical Angle (cc): The angle of incidence in a denser medium for which the angle of refraction is 90∘90^\circ. Total Internal Reflection (TIR) occurs when the angle of incidence exceeds the critical angle.

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Dispersion: The separation of white light into its constituent colors (spectrum) when passing through a prism, because different wavelengths (colors) have slightly different refractive indices.

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Converging (Convex) Lenses: These bring parallel rays of light to a principal focus FF. The distance from the center of the lens to the focus is the focal length ff.

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Real vs Virtual Images: Real images are formed where rays actually converge and can be projected onto a screen; virtual images are formed where rays appear to diverge from a point.

📐Formulae

n=sin⁡isin⁡rn = \frac{\sin i}{\sin r}

n=cvn = \frac{c}{v}

sin⁡c=1n\sin c = \frac{1}{n}

n1sin⁡θ1=n2sin⁡θ2n_1 \sin \theta_1 = n_2 \sin \theta_2

1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

m=hiho=vum = \frac{h_i}{h_o} = \frac{v}{u}

💡Examples

Problem 1:

A ray of light travels from air into a glass block with a refractive index of 1.521.52. If the angle of incidence is 45∘45^\circ, calculate the angle of refraction.

Solution:

Using Snell's Law: n1sin⁡θ1=n2sin⁡θ2n_1 \sin \theta_1 = n_2 \sin \theta_2. Given n1=1.0n_1 = 1.0 (air), θ1=45∘\theta_1 = 45^\circ, and n2=1.52n_2 = 1.52. 1.0×sin⁡(45∘)=1.52×sin⁡(r)1.0 \times \sin(45^\circ) = 1.52 \times \sin(r) sin⁡r=0.70711.52≈0.4652\sin r = \frac{0.7071}{1.52} \approx 0.4652 r=arcsin⁡(0.4652)≈27.7∘r = \arcsin(0.4652) \approx 27.7^\circ

Explanation:

Light slows down as it enters the denser glass medium, causing the ray to bend toward the normal, resulting in an angle of refraction smaller than the angle of incidence.

Problem 2:

Calculate the critical angle cc for a diamond-air boundary if the refractive index of diamond is 2.422.42.

Solution:

Using the formula sin⁡c=1n\sin c = \frac{1}{n}: sin⁡c=12.42≈0.4132\sin c = \frac{1}{2.42} \approx 0.4132 c=arcsin⁡(0.4132)≈24.4∘c = \arcsin(0.4132) \approx 24.4^\circ

Explanation:

The critical angle is the threshold above which light is entirely reflected back into the diamond. A lower critical angle means light is more likely to be trapped and reflected, contributing to the 'sparkle' of a diamond.

Problem 3:

An object is placed 12 cm12\text{ cm} from a converging lens with a focal length of 8 cm8\text{ cm}. Determine the image distance vv and its nature.

Solution:

Using the lens formula: 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}. Given f=8 cmf = 8\text{ cm} and u=12 cmu = 12\text{ cm}. 18=112+1v\frac{1}{8} = \frac{1}{12} + \frac{1}{v} 1v=18−112=3−224=124\frac{1}{v} = \frac{1}{8} - \frac{1}{12} = \frac{3-2}{24} = \frac{1}{24} v=24 cmv = 24\text{ cm} Since vv is positive, the image is real.

Explanation:

Because the object is placed beyond the focal point (u>fu > f), the lens forms a real, inverted image on the opposite side of the lens. The magnification would be m=2412=2m = \frac{24}{12} = 2.

Light - Revision Notes & Key Formulas | IGCSE Grade 12 Physics