# 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 (**It can be expressed as:

*v*).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.