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Light - Images Formed by Plane Mirrors

Grade 7CBSE

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

🔑Concepts

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A plane mirror is a flat, smooth reflecting surface that forms a specific type of image through the process of regular reflection.

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The image formed is always Virtual, meaning it cannot be obtained on a screen because the light rays do not actually meet but only appear to diverge from a point behind the mirror.

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The image is Erect, which means it is upright and oriented in the same way as the object.

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The size of the image is exactly equal to the size of the object: himage=hobjecth_{image} = h_{object}.

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The distance of the image from the mirror is equal to the distance of the object from the mirror: v=uv = u.

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Lateral Inversion: This is a characteristic property where the left side of the object appears as the right side of the image, and the right side of the object appears as the left side of the image.

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Laws of Reflection: These govern how light behaves on the mirror. 1. The angle of incidence is equal to the angle of reflection: ∠i=∠r\angle i = \angle r. 2. The incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane.

📐Formulae

∠i=∠r\angle i = \angle r

u=vu = v

Total distance between object and image=u+v=2u\text{Total distance between object and image} = u + v = 2u

💡Examples

Problem 1:

David is observing his image in a plane mirror. The distance between the mirror and his image is 4 m4\text{ m}. If he moves 1 m1\text{ m} towards the mirror, then find the distance between David and his image.

Solution:

6 m6\text{ m}

Explanation:

Initially, image distance v=4 mv = 4\text{ m}, so object distance u=4 mu = 4\text{ m}. When David moves 1 m1\text{ m} towards the mirror, the new object distance becomes u′=4 m−1 m=3 mu' = 4\text{ m} - 1\text{ m} = 3\text{ m}. Since in a plane mirror u=vu = v, the new image distance v′v' is also 3 m3\text{ m}. The total distance between David and his image is u′+v′=3 m+3 m=6 mu' + v' = 3\text{ m} + 3\text{ m} = 6\text{ m}.

Problem 2:

A ray of light strikes a plane mirror such that the angle between the incident ray and the mirror surface is 30∘30^\circ. Calculate the angle of reflection.

Solution:

60∘60^\circ

Explanation:

The angle between the incident ray and the mirror is 30∘30^\circ. Since the normal is perpendicular (90∘90^\circ) to the mirror, the angle of incidence is ∠i=90∘−30∘=60∘\angle i = 90^\circ - 30^\circ = 60^\circ. According to the law of reflection, ∠i=∠r\angle i = \angle r, therefore the angle of reflection ∠r=60∘\angle r = 60^\circ.

Problem 3:

If the letter 'P' is placed in front of a plane mirror, how will its image appear?

Solution:

It will appear as 'q'.

Explanation:

This happens due to Lateral Inversion. The left side of the letter 'P' (the vertical line) remains on the same side relative to the mirror's plane, but the loop on the right side of the 'P' appears on the left side in the image.

Images Formed by Plane Mirrors - Revision Notes & Key Formulas | CBSE Class 7 Science