Question

Explain reflection by plane mirror with appropriate diagram.​

Answer 1

Answer:

Each small portion of the extended object facing the mirror acts like a point source. The incident rays AN and AQ are drawn from points A of the object. The corresponding reflected rays NA and QR are drawn applying the laws of reflection.

Answer 2

Answer:

In the above figure, an extended object AO of height h is kept in front of the plane mirror MM' at a distance u.

In the above figure, an extended object AO of height h is kept in front of the plane mirror MM' at a distance u.Each small portion of the extended object facing the mirror acts like a point source.

In the above figure, an extended object AO of height h is kept in front of the plane mirror MM' at a distance u.Each small portion of the extended object facing the mirror acts like a point source.The incident rays AN and AQ are drawn from points A of the object.

In the above figure, an extended object AO of height h is kept in front of the plane mirror MM' at a distance u.Each small portion of the extended object facing the mirror acts like a point source.The incident rays AN and AQ are drawn from points A of the object.The corresponding reflected rays NA and QR are drawn applying the laws of reflection. As the reflected rays NA and QR are divergent rays, they cannot meet in front of the mirror, but they intersect at 'A', when extended behind the mirror as shown in figure. Thus, 'A' is the virtual image of the real object A. Similarly, all point sources between A and O will form corresponding images between A' and I.

In the above figure, an extended object AO of height h is kept in front of the plane mirror MM' at a distance u.Each small portion of the extended object facing the mirror acts like a point source.The incident rays AN and AQ are drawn from points A of the object.The corresponding reflected rays NA and QR are drawn applying the laws of reflection. As the reflected rays NA and QR are divergent rays, they cannot meet in front of the mirror, but they intersect at 'A', when extended behind the mirror as shown in figure. Thus, 'A' is the virtual image of the real object A. Similarly, all point sources between A and O will form corresponding images between A' and I.Thus, we see that a plane mirror forms a virtual, erect and inverted image at the same distance behind the mirror as that of an object kept in front of the mirror.