Issues:

1. How to determine the image of an object, placed in front of a spherical mirror?

2. What will be the position of the image?

3. Will the image be larger or smaller than the object?

4. Will the image be read or virtual?

5. Will the image be 'eract' or inverted?

Intersection of any two of the 4 different rays:

1. Ray parallel to the principal axis.

2. Ray incident at the pole, P

3. Ray passing through the principal focus, F

4. Ray passing through the centre of curvature, C

consider a linear object AB- located in front of a concave mirror

$B P$- Object distance, $u$

$B'P$ - Image distance, ${v}$

CP - Radius of curvature, $R$

$F P$ - Focal length, $f$

$A B-$ Height (size) of abject, $h$

$N^{\prime} B^{\prime}$ - Height (sigh) of the image, $h^{\prime}$

Distances measured w.r.t the pole (x=0, y=0)

Coordinates of the points B,C,B and F

Thus, BP = -u, CP = -R, FP = -f

AB = h, A'B' = -h'

For small aperture, M is clear to the axis.

$\Rightarrow$ D is close to P

$\therefore$ FD $\simeq$ FP

Also, B'F = BP - FP

Eq(3) becomes

$\frac{B'P-FP}{FP}$ = $\frac{B'P}{BP}$ $\rightarrow$ (4)

Applying 'sign' convention ;

$\frac{(-v)-(-f)}{(-f)}$ = $\frac{-v}{-u}$

gives $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

The mirror equation

• $\frac{-v}{-f}-1=\frac{v}{u}$

$\frac{v}{f}-\frac{v}{u}$=

$1 \Rightarrow v(\frac{1}{f}-\frac{1}{u})=1$

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\triangle A' B' F$ and $\triangle F M D$ are similar.

$\Rightarrow \frac{A' B'}{M D}=\frac{B' F}{F D}$

or $\frac{A'B'}{A B}=\frac{B'F}{F D}$ (1)

$\triangle$ ABP and $\triangle$ A'B'P are similiar

$\therefore$ $\frac{A'B'}{AB} = \frac{B'P}{BP}$ (2)

From (1) and (2),

$\frac{B'F}{F D} = \frac{B'P}{BP}$

$m = \frac{\text{size of image}}{\text{size of object}}$ = $\frac{h'}{h}$ Also called "Linear Magnification"

Eq(2): $\frac{A'B'}{AB} = \frac{B'P}{BP}$

Applying sign convention,

$\frac{h'}{h} = \frac{-v}{u}$

or $m = \frac{h'}{h} = \frac{-v}{u}$

for the present case, v<u $\Rightarrow$ |m|<1

m-negative $\rightarrow$ inverted image.

$m = \frac{h'}{h} > 1 \Rightarrow \text{Magnified image}$

m - positive $\Rightarrow$ Eract virtual image

How about getting a magnified real image?

BP = -u , B'P = V

FP = f, CP = R

AB = h, A'B' = h'

Substituting in Eq(2).

$\frac{A'B'}{AB} = \frac{B'P}{BP}$

$\frac{h'}{h} = \frac{v}{-u}$

or $m = \frac{h'}{h} = -\frac{v}{u}$ as before

A linear object is placed in front of a concave mirror of radius of curvature $15 {cm}$. What will be the position and magnification of the image if the object distance is -

1. $10 {cm}$,

2. $5 {cm}$

in front of the mirror.

R = -15 cm

f = -7.5 cm

v = ?, m = ?

1. u = -10 cm

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{7.5} - \frac{1}{-10}$

2. u = -5 ${cm}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$=\frac{1}{-7.5}-\frac{1}{(-5)}=\frac{1}{5}-\frac{1}{7.5}$

$=\frac{7.5-5}{37.5}=\frac{2.5}{37.5}=\frac{1}{15}$

$v=\frac{375}{25}= 15$

$\therefore m=\frac{-v}{u}=\frac{-15}{-5}=3$

magnified virtual image

3. u = -15 cm

4. u = -20 cm

(3) $\Rightarrow \frac{1}{v} =\frac{1}{f}-\frac{1}{u}$

$\frac{1}{v} =\frac{1}{-7.5} - \frac{1}{-15} = \frac{1}{15} - \frac{1}{7.5}$

$\frac{1}{v} =\frac{1-2}{15}=\frac{-1}{15}=$

$\Rightarrow v = -15 {cm}$

$\frac{1}{v} =\frac{1}{f}-\frac{1}{u}$

$\Rightarrow \frac{1}{v}=\frac{1}{-7.5}-\frac{1}{-20} =\frac{1}{20}-\frac{1}{7.5}$

$=\frac{7.5-20}{150}$

$\frac{-12.5}{150}$

$v = - \frac{150}{12.5} = -12 cm$

$m = \frac{(-12)}{(-20)} = -0.6$

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

was obtained by substituting the coordinates corresponding to the positions of the object, image and the principal focus.

Therefore, in using the 'mirror equation' to determine the unknown parameter ( $u$ or $v$ or $f$ the coordinates corresponding to the known parameters must be substituted.