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Physics - Light (Reflection, Refraction, and Color)

Grade 8Cambridge (IGCSE)

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

🔑Concepts

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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 occurs when light changes speed as it moves from one medium to another, causing it to bend. Light bends towards the normal when entering a denser medium (vv decreases) and away from the normal when entering a less dense medium (vv increases).

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The refractive index nn is a measure of how much a medium slows down light, 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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Total Internal Reflection (TIR) occurs when light travels from a denser to a less dense medium at an angle of incidence greater than the critical angle θc\theta_c.

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Dispersion is the splitting of white light into its constituent colors (the spectrum) through a prism because different frequencies of light refract by different amounts; Violet light refracts the most, while Red light refracts the least.

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The three primary colors of light are Red, Green, and Blue. They combine to form white light. Secondary colors are Cyan (G+BG+B), Magenta (R+BR+B), and Yellow (R+GR+G).

📐Formulae

θi=θr\theta_i = \theta_r

n=sin⁡isin⁡rn = \frac{\sin i}{\sin r}

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

sin⁡c=1n\sin c = \frac{1}{n}

💡Examples

Problem 1:

A ray of light travels from air into a glass block with a refractive index of n=1.5n = 1.5. If the angle of incidence ii is 30∘30^{\circ}, calculate the angle of refraction rr.

Solution:

1.5=sin⁡30∘sin⁡r1.5 = \frac{\sin 30^{\circ}}{\sin r}
sin⁡r=sin⁡30∘1.5=0.51.5≈0.333\sin r = \frac{\sin 30^{\circ}}{1.5} = \frac{0.5}{1.5} \approx 0.333
r=arcsin⁡(0.333)≈19.47∘r = \arcsin(0.333) \approx 19.47^{\circ}

Explanation:

We use Snell's Law to find the angle of refraction. Since light is entering a denser medium (1.5>1.01.5 > 1.0), the ray bends 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 which has a refractive index of n=2.42n = 2.42.

Solution:

sin⁡c=1n=12.42≈0.413\sin c = \frac{1}{n} = \frac{1}{2.42} \approx 0.413
c=arcsin⁡(0.413)≈24.4∘c = \arcsin(0.413) \approx 24.4^{\circ}

Explanation:

The critical angle is the angle of incidence that results in an angle of refraction of 90∘90^{\circ}. For materials with a high refractive index like diamond, the critical angle is small, making total internal reflection very likely and giving the diamond its 'sparkle'.

Problem 3:

A red filter is placed in front of a white light source, and the resulting beam is shone onto a blue object. What color does the object appear?

Solution:

The object appears Black.

Explanation:

A red filter only allows red light to pass through. When this red light hits a blue object, the blue object absorbs the red light (as it only reflects blue light). Since no light is reflected back to the eye, the object appears black.