Advertisements
Advertisements
Question
Complete the following diagrams shown in the below figure by drawing the reflected ray for each incident ray.

Advertisements
Solution

A light ray when produced backwards passes through principal focus as shown in the problem figure. We draw the normal through the centre of curvature at the point of incidence and draw the reflected ray at an angle equal to the angle of incidence thus following the laws of reflection. The reflected ray is parallel to the principal axis. The other ray is passing through the centre of curvature as shown in the problem figure. This ray retraces its incident path because it strikes the mirror normally i.e. 90 degrees. These two reflected rays when produced backwards coincide at a point where the image is formed. The image, A'B' is virtual, erect, and diminished in size.
APPEARS IN
RELATED QUESTIONS
Give one use each of a concave and a convex mirror.
A ray of light is incident normally on a plane mirror. What will be the
angle of reflection?
What is the name of the phenomenon in which the right side of an object appears to be the left side of the image in a plane mirror?
Name the spherical mirror which can produce a virtual and diminished image of an object.
The diagram below in Figure, shows a convex mirror. C is its centre of curvature and F is its focus. (i) Draw two rays from A and hence locate the position of image of object OA. Label the image IB. (ii) State three characteristics of the image.
State whether the image in part (a) is real or virtual?
Numerical problem.
The radius of curvature of a spherical mirror is 25 cm. Find its focal length.
The image formed in a plane mirror is always inverted.
