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Light Energy - Reflection of Light

Grade 7ICSE

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

🔑Concepts

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Light Energy: A form of energy that enables us to see objects. It travels in a straight line, a property known as the Rectilinear Propagation of Light.

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Reflection of Light: The phenomenon of bouncing back of light into the same medium after striking a polished surface like a mirror.

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Terminology: The ray striking the surface is the Incident Ray, the ray bouncing back is the Reflected Ray, and the perpendicular drawn at the point of incidence is the Normal.

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Laws of Reflection: 1. The angle of incidence ∠i\angle i is always equal to the angle of reflection ∠r\angle r. 2. The incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane.

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Types of Reflection: Regular Reflection occurs from smooth surfaces (like a plane mirror) and forms clear images. Irregular (Diffuse) Reflection occurs from rough surfaces (like a wall) where light scatters in different directions.

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Characteristics of Image in a Plane Mirror: The image is virtual (cannot be caught on a screen), erect (upright), the same size as the object, and undergoes Lateral Inversion (left side appears right and vice versa).

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Object and Image Distance: In a plane mirror, the distance of the object from the mirror uu is exactly equal to the distance of the image from the mirror vv.

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Periscope: An optical device used to see over obstacles, consisting of a tube with two plane mirrors fixed parallel to each other and inclined at an angle of 45∘45^\circ to the horizontal.

📐Formulae

∠i=∠r\angle i = \angle r

u=vu = v

Total Distance=u+v=2u\text{Total Distance} = u + v = 2u

c≈3×108 m/sc \approx 3 \times 10^8 \text{ m/s}

💡Examples

Problem 1:

A ray of light strikes a plane mirror such that the angle between the incident ray and the reflected ray is 70∘70^\circ. Calculate the angle of incidence and the angle of reflection.

Solution:

Given that the total angle between the incident ray and reflected ray is 70∘70^\circ. According to the Law of Reflection, ∠i=∠r\angle i = \angle r. Therefore, ∠i+∠r=70∘\angle i + \angle r = 70^\circ. Substituting ∠r\angle r with ∠i\angle i: 2×∠i=70∘  ⟹  ∠i=35∘2 \times \angle i = 70^\circ \implies \angle i = 35^\circ. Thus, ∠i=35∘\angle i = 35^\circ and ∠r=35∘\angle r = 35^\circ.

Explanation:

Since the angle of incidence and reflection are equal, the total angle formed by the two rays is exactly double the angle of incidence.

Problem 2:

An object is placed at a distance of 15 cm15 \text{ cm} in front of a plane mirror. If the object is moved 5 cm5 \text{ cm} towards the mirror, what is the new distance between the object and its image?

Solution:

Initial object distance u1=15 cmu_1 = 15 \text{ cm}. After moving 5 cm5 \text{ cm} closer, the new object distance is u2=15 cm−5 cm=10 cmu_2 = 15 \text{ cm} - 5 \text{ cm} = 10 \text{ cm}. Since u=vu = v for a plane mirror, the image distance v2=10 cmv_2 = 10 \text{ cm}. The total distance between the object and the image is u2+v2=10 cm+10 cm=20 cmu_2 + v_2 = 10 \text{ cm} + 10 \text{ cm} = 20 \text{ cm}.

Explanation:

The distance between an object and its image in a plane mirror is always twice the distance between the object and the mirror surface.

Problem 3:

A ray of light is incident on a plane mirror such that it makes an angle of 30∘30^\circ with the mirror surface (glancing angle). Find the angle of reflection.

Solution:

The angle with the mirror surface is the glancing angle θ=30∘\theta = 30^\circ. The Normal is at 90∘90^\circ to the surface. Therefore, the angle of incidence ∠i=90∘−30∘=60∘\angle i = 90^\circ - 30^\circ = 60^\circ. According to the laws of reflection, ∠r=∠i=60∘\angle r = \angle i = 60^\circ.

Explanation:

The angle of incidence is measured between the incident ray and the normal, not the mirror surface.