Draw a ray diagram to show the formation of image of an object placed between the pole and
focus of a concave mirror. Obtain the relation between u v, and f for a given concave mirror.
State clearly the assumption involved and sign convention used.

Dear Student,

Please find below the solution to the asked query:

The situation will be as shown in the figure.

Image will be behind the mirror. Image is Virtual, Erect and Large in size. Magnification is greater than one (m > + 1).


To understand the tracing of image. Following facts are useful in ray tracing.

• If the incident ray is parallel to the principal axis,  the reflected ray passes through the focus. 
• If the incident ray passes through the focus, then the reflected  ray is parallel to the principal     axis.     
• Incident ray passing through centre of curvature will be reflected back through the centre of curvature (because it is a normally incident ray). 

• It is easy to make the ray tracing of a ray incident at the pole as shown in below.

Now, Let us consider the situation as in the below figure.

Let O is the point object placed at a distance "u" from the pole P of mirror. If we follow the rules mentioned above, we can trace an image I as shown in the figure, which is at a distance "v" from the pole of the mirror. If C is the centre of curvature and F be the focus of the lens,

Here, CB is normal to the mirror at B. By laws of reflection,
$\angle OBC = \angle CBI = \theta$
From the geometry of the diagram,
$\alpha +\theta = \beta , \beta +\theta = \gamma , \alpha +\gamma = 2\beta$
For small aperture of the mirror, α, β and γ can be written as,
$\alpha = \tan \alpha ; \beta = \tan \beta and \gamma = \tan \gamma$
Therefore,
$\tan \alpha + \tan \gamma = 2 \tan \beta \Rightarrow \frac{BP\text{'}}{OP\text{'}} + \frac{BP\text{'}}{IP\text{'}} = 2\frac{BP\text{'}}{CP\text{'}}$
If we apply sign convension,
u = - OP, v = - IP and R = - CP
​So,
$\Rightarrow -\frac{1}{u}+\left(-\frac{1}{v}\right) = -\frac{2}{R}$
If object is at infinite, the image should be at focus, (v=f)
 $u = \infty \Rightarrow \frac{1}{v} = \frac{2}{R}\Rightarrow f = \frac{R}{2}$
Therefore,
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
Above equation is called "Mirror equation".

 

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