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Waves - Light (Reflection, refraction and thin converging lenses)

Grade 11IGCSEPhysics

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. In a plane mirror, the image is virtual, upright, the same size as the object, and laterally inverted.

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Refraction: The bending of light as it passes from one optical medium to another due to a change in speed. When light enters a denser medium (e.g., air to glass), it bends towards the normal.

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Refractive Index (nn): A ratio of the speed of light in a vacuum cc to the speed of light in a medium vv. It is also defined by Snell's Law using the ratio of the sine of the angle of incidence to the sine of the angle of refraction.

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Total Internal Reflection (TIR): Occurs when light travels from an optically denser medium to a less dense medium and the angle of incidence is greater than the critical angle cc. If the angle of incidence equals the critical angle, the refracted ray travels along the boundary (r=90∘r = 90^\circ).

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Thin Converging Lenses: These lenses are thicker in the middle and converge parallel rays of light to a point called the principal focus FF. The distance from the optical center to the principal focus is the focal length ff.

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Image Formation: Real images are formed when rays actually meet and can be projected on a screen. Virtual images are formed when rays only appear to originate from a point and cannot be projected on a screen. For a converging lens, if the object distance u>fu > f, the image is real; if u<fu < f, the image is virtual (magnifying glass).

📐Formulae

n=sin⁡isin⁡rn = \frac{\sin i}{\sin r}

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

sin⁡c=1n\sin c = \frac{1}{n}

Linear Magnification=Image heightObject height=vu\text{Linear Magnification} = \frac{\text{Image height}}{\text{Object height}} = \frac{v}{u}

💡Examples

Problem 1:

A ray of light in air strikes a glass block (n=1.52n = 1.52) at an angle of incidence of 40∘40^\circ. Calculate the angle of refraction rr inside the glass.

Solution:

Using Snell's Law: n=sin⁡isin⁡r  ⟹  1.52=sin⁡40∘sin⁡rn = \frac{\sin i}{\sin r} \implies 1.52 = \frac{\sin 40^\circ}{\sin r}. Rearranging for rr: sin⁡r=sin⁡40∘1.52≈0.64281.52≈0.4229\sin r = \frac{\sin 40^\circ}{1.52} \approx \frac{0.6428}{1.52} \approx 0.4229. Therefore, r=arcsin⁡(0.4229)≈25∘r = \arcsin(0.4229) \approx 25^\circ.

Explanation:

Apply the refractive index formula n=sin⁡isin⁡rn = \frac{\sin i}{\sin r} where ii is the angle in air and rr is the angle in the denser medium.

Problem 2:

Determine the critical angle cc for a water-air boundary if the refractive index of water is 1.331.33.

Solution:

Using the formula sin⁡c=1n\sin c = \frac{1}{n}, we have sin⁡c=11.33≈0.7519\sin c = \frac{1}{1.33} \approx 0.7519. Thus, c=arcsin⁡(0.7519)≈48.8∘c = \arcsin(0.7519) \approx 48.8^\circ.

Explanation:

The critical angle is the angle of incidence in the denser medium that results in an angle of refraction of 90∘90^\circ in the rarer medium.

Problem 3:

An object is placed 12 cm12\text{ cm} away from a converging lens with a focal length f=8 cmf = 8\text{ cm}. Describe the nature of the image.

Solution:

Since the object distance u=12 cmu = 12\text{ cm} is between ff (8 cm8\text{ cm}) and 2f2f (16 cm16\text{ cm}), the image formed will be real, inverted, and magnified.

Explanation:

For a thin converging lens, when f<u<2ff < u < 2f, the rays converge on the opposite side of the lens beyond 2f2f, creating an enlarged real image.

Light (Reflection, refraction and thin converging lenses) Revision - Grade 11 Physics IGCSE