10. LIGHT – REFLECTION AND REFRACTION
Sign Convention for Reflection by Spherical Mirrors
New
Cartesian Sign Convention:
In this convention, the pole (P) of the mirror is taken
as the origin. The principal axis of the mirror is taken as the x-axis
(X’X) of the coordinate system.
The New Cartesian Sign
Convention for spherical mirrors
The
conventions are as follows:
a) The object is
always placed to the left of the mirror. i.e., light from the object
falls on the mirror from the left-hand side.
b) All distances parallel to
the principal axis are measured from the pole of the mirror.
c) All the distances
measured to the right of the origin (along + x-axis) are taken as
positive while those measured to the left of the origin (along
– x-axis) are taken as negative.
d) Distances measured
perpendicular to and above the principal axis (along + y-axis) are taken
as positive.
e) Distances measured
perpendicular to and below the principal axis (along – y-axis) are taken
as negative.
Sign conventions
are applied to obtain the mirror formula and solve related numerical
problems.
Mirror Formula and Magnification
In a spherical mirror,
the distance of the object from its pole is called the object distance (u).
The distance of the image
from the pole of the mirror is called the image distance (v).
The distance of the principal focus from the pole is called the focal length (f).
This formula is valid in all situations for all spherical mirrors for all positions of the object.
Magnification (m)
It is the enlargement of the image formed
by a spherical mirror, relative to the size of the object.
It is the ratio of the height of the image (h′) to
the height of the object (h).
Magnification is also related to the object distance (u)
and image distance (v). It can be expressed as:
The height of the
object is taken to be positive as the object is placed above the
principal axis.
The height of the image
is taken as positive for virtual images and negative for real
images.
A negative sign in
the value of the magnification indicates that the image is real. A positive
sign indicates that the image is virtual.
Problem: A convex mirror used for
rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is
located at 5.00 m from this mirror, find the position, nature and size of the
image.
Solution
Radius
of curvature, R = + 3.00 m
Object-distance, u =
– 5.00 m
Image-distance, v= ?
Height of the image, h′=
?
Focal length, f = R/2 = + 300 m/2
The
image is 1.15 m at the back of the mirror.
The
image is virtual, erect & smaller by a factor of 0.23.
Problem: An object, 4.0 cm in
size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0
cm. At what distance from the mirror should a screen be placed in order to
obtain a sharp image? Find the nature and the size of the image.
Solution
Object-size,
h = + 4.0 cm
Object-distance,
u = – 25.0 cm
Focal
length, f = –15.0 cm
Image-distance, v= ?
Image-size,
h′= ?
v = – 37.5 cm
The
screen should be placed at 37.5 cm in front of the mirror. The image is real.
Height
of the image, h′ = – 6.0 cm
The image is inverted and enlarged.
