krit.club logo

Light - Rectilinear Propagation and Reflection

Grade 7CBSE

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

🔑Concepts

•

Light travels along a straight line in a medium; this property is known as the Rectilinear Propagation of Light.

•

The phenomenon of bouncing back of light rays after striking a polished or shiny surface is called the Reflection of Light.

•

A plane mirror forms an image that is virtual, erect, and of the same size as the object.

•

The image formed by a plane mirror is at the same distance behind the mirror as the object is in front of it, represented as u=vu = v.

•

Lateral Inversion is the property of a plane mirror where the left side of the object appears as the right side of the image and vice versa.

•

The Angle of Incidence ∠i\angle i is the angle between the incident ray and the normal.

•

The Angle of Reflection ∠r\angle r is the angle between the reflected ray and the normal.

•

Laws of Reflection: 1) The angle of incidence is always 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

Object Distance (u)=Image Distance (v)\text{Object Distance } (u) = \text{Image Distance } (v)

Total distance between Object and Image=u+v=2u\text{Total distance between Object and Image} = u + v = 2u

💡Examples

Problem 1:

An incident ray makes an angle of 35∘35^\circ with the surface of a plane mirror. Find the angle of reflection.

Solution:

The normal is perpendicular (90∘90^\circ) to the mirror surface. Therefore, the angle of incidence ∠i=90∘−35∘=55∘\angle i = 90^\circ - 35^\circ = 55^\circ. According to the law of reflection, ∠r=∠i=55∘\angle r = \angle i = 55^\circ.

Explanation:

Since the given angle is with the surface, we subtract it from the normal (90∘90^\circ) to find the angle of incidence, then apply ∠i=∠r\angle i = \angle r.

Problem 2:

David is standing 5 m5\text{ m} away from a plane mirror. He moves 2 m2\text{ m} towards the mirror. What is the new distance between David and his image?

Solution:

Initial distance u=5 mu = 5\text{ m}. New object distance unew=5 m−2 m=3 mu_{new} = 5\text{ m} - 2\text{ m} = 3\text{ m}. Since u=vu = v in a plane mirror, the image distance vnew=3 mv_{new} = 3\text{ m}. Total distance =unew+vnew=3 m+3 m=6 m= u_{new} + v_{new} = 3\text{ m} + 3\text{ m} = 6\text{ m}.

Explanation:

In a plane mirror, the image is as far behind the mirror as the object is in front of it. The total distance is the sum of the object-to-mirror and mirror-to-image distances.