Page 156 - Physics_Form_2

Physics for Secondary Schools

trigonometric identity; object sizes are equal. Similarly, from
Figure 4.29, we observe that:
tan(−θ) =−tanθ, h h
0
i
We find that; tanφ = u− R and tan ′ φ = R−v
tanθ =−tan( ′ θ ) Since tan(φ) =−tan( ′ φ ), then,
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h h
tanθ = 0 and tan ′ θ = i h h
u v 0  i
u R Rv


h h
This implies to obtain 0 = i , which can This implies that;
u v
also be rearranged to obtain:  h 0  u  uR
h i = v = m h i v Rv

h 0 u Further algebraic manipulation yields:

This is the image magnification equation uv vR uR uv  
for spherical mirrors. Note that, h ,
0
h , u , v and m are respectively, the 2uv uR vR= + = ( R u v+ )
i
object height, the image height, the 2 uv+ 1 1
object distance, the image distance and R = uv = u + v
magnification.
1 1 1
Since, R = 2 f , it follows that: = +
Therefore: f u v
image height ()h
m = i
object height ()h 0 Alternatively

1 1 1
image distance from the mirror(v) To derive the equation = + , we
m= f u v
object distance from the mirror(u)
analyse the geometry of a concave mirror
Note that, magnification is a ratio and, using the principle of similar triangles.

therefore, has no units. When the value
of m is negative, the image is inverted. Assumptions
The image formed by a curved mirror 1. Distances are measured from the
can be larger, smaller or the same size as mirror’s pole.
the object. When the ratio, m, is greater 2. Sign conventions
than one, the image is enlarged; when it (a) the distance in the direction of
incidence light is positive
is less than one, the image is diminished; (b) distances below the principal
and when it is equal to 1, the image and axis are negative.

150
Student’s Book Form Two

Physics Form 2 Final.indd 150 25/10/2025 10:27