91Ó°ÊÓ

Object image relation for spherical mirror is given by:

$\frac{\mathbf{1}}{\mathbf{u}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{v}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{R}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

Where, u= object distance from the vertex of the mirror,

u is (+) when the object is in front of the reflecting surface of mirror,

v is image distance from the vertex of the mirror,

v is (+) when the image is in front of the reflecting surface of mirror,

R is the radius of curvature of the mirror, f= focal length,

R is (+) in concave mirror and (-) in convex mirror,

u and v are negative if the case is opposite to what is mentioned above

Remember that the focal length of a mirror depends only on the curvature of the mirror and not on the material from which the mirror is made because the formation of the image results from rays reflected from the surface of the material.

Apply: The above rule can be applied on either concave and convex mirrors or even plane mirror when we consider the radius to be infinity. The equation gives one the exact location of an image by knowing the radius or the focal length of a mirror and the location of the object. In addition, the sign for the distance of the image can be negative if the image in the opposite direction of the outgoing rays which means the image is not real image but is virtual one and is located inside the mirror.

Lateral magnification for spherical message is given by:

$\mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{v}}{\mathbf{u}}\mathbf{=}\frac{{\mathbf{h}}_{\mathbf{i}}}{{\mathbf{h}}_{\mathbf{0}}}$

where, m is the magnification,

u is object distance from the vertex of the mirror,

u is (+) when the object is in front of the reflecting surface of mirror,

v is image distance from the vertex of the mirror,

v is (+) when the image is in front of the reflecting surface of mirror,

is the height of the image, is the height of the object,

m is (+) when the image is erect and (-) when the image is inverted

a) Radius of curvature R =32 cm

$f=\frac{R}{2}=\frac{32}{2}=16cm$

Now,$\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}$

Solve for v to get

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

Substitute$f=16cm$ and$u=12cm$

$v=\frac{12.16}{12-16}=-48cm$

Thus, magnification

$$\begin{gathered} m=-\frac{v}{u}=\frac{h_i}{h_o} \\ m=-\frac{-48}{12}=4 \end{gathered}$$

Hence, the magnification is four, i.e., the image is four times as large as face.

b) The image is v distance from the mirror. We got value for v in part (a) as

Now,$\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}$

Solve for v to get

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

Substitute$f=16cm$ and$u=12cm$

$v=\frac{12.16}{12-16}=-48cm$

Thus, the image is 48 cm behind the mirror. The negative sign indicated that the image is at the right of the mirror and thus, image is virtual.

c) To draw the diagram, remember the magnification is four times and the image is to the right of the mirror.