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Physics - Light (Reflection, Refraction, and Color)

Grade 8IGCSE

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.

Light (Reflection, Refraction, and Color) Revision - Grade 8 Science IGCSE