An object is kept in front of a concave mirror of focal length 20 cm. The image formed is three times the size of the object. Calculate the two possible distances of the object from the mirror.

We are given a concave mirror.
So, focal length, f = - 20 cm
Also given the image formed is three times the size of the object. 

(a) When the image formed is real:

Given, 
Magnification,  $\mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = -\frac{\mathrm{v}}{\mathrm{u}} = -3$ 

$\Rightarrow$ Image distance, $\mathrm{v} = + 34$ 

Now, using the mirror formula,
                  $\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}$ 
we have, 

$\therefore$            $\frac{1}{\mathrm{u}}+\frac{1}{3\mathrm{u}} = -\frac{1}{20}$ 


$\Rightarrow$ $\mathrm{u} = \frac{-20 \times 4}{3} = \frac{-80}{3} = -26.67 \mathrm{cm}$ is the object distance measured from the left side of the mirror.

(b) When the image formed is virtual:

Magnification, $\mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = -\frac{\mathrm{v}}{\mathrm{u}} = +3$ 

$\Rightarrow$ Image distance, $\mathrm{v} = -3\mathrm{u}$ 

Now, using the mirror formula, 

                      $\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}$ 
we have,
                  $\frac{1}{\mathrm{u}}-\frac{1}{3\mathrm{u}} = -\frac{1}{20}$

$\Rightarrow$      $\mathrm{u} = \frac{-20 \times 2}{3} = \frac{-40}{3} = -13.33 \mathrm{cm}$

Therefore, the two possible distance of objects from the mirror are 26.67 cm and 13.33 cm from the left side.

Refraction of light: The phenomenon of bending of light from its straight line path as it passes obliquely from one transparent medium to another is called refraction of light.
The path of the ray of light in the first medium is called incident ray.
The path of the ray of light in the second medium is called refracted ray.
The angle between the incident ray and the normal at the surface of separation is called angle of incidence (i).
The angle between the refracted ray and the normal at the surface of separation is called angle of refraction (r).
Whenever a ray of light passes from one medium another, the following three situations are possible:

Fig.Refraction of light

(i) When a ray of light passes from an optically rarer medium to a denser medium, it bends towards the normal and ∠r < ∠i, as shown in Fig.(a).
(ii) When a ray of light passes from an optically denser to a rarer medium, it bends away form the normal and Lr > ∠i as shown in Fig.(b).
(iii) A ray of light travelling along the normal passes undeflected. Here ∠ i = ∠ r = 0°.
Some examples of refraction of light:
(i) When a pencil is partly immersed in water in a glass tumbler, it appears to be displaced at the interface of air and water. As shown in Fig(a) rays starting from point A in the optically denser medium bend away from normal after passing through water-air interface. On entering the eye, the ray appears to come from A'. Point A appears to be at A' and the portion AO appears to be at OA'. Thus the light reaching our eye from the portion of the pencil inside water appears to come from different direction, compared to the part above water. That is why the pencil appears bent at the interface.

Fig.(a). A straight pencil partially put in water appears bent
If instead of water, we use liquids like kerosene or tarpentine, the pencil will appear to be displaced or bent by a different extent.

(ii) If we see a water tank, its botton appears to be raised. It also appears to be concave shaped although it is flat. Rays from A (Fig.(b)) after striking water-air interface bend away from normal. Thus point A appears to be at A' i.e., the bottom appears to be raised and concave.

Fig.(b) The bottom of tank of water appears raised and concave shaped
For the same reason, when a thick glass slab is placed over some printed matter, its letters appear raised when seen through the slab.

Refraction through a rectangular glass slab: Consider a rectangular glass slab PQRS, as shown in Fig. A ray AB is incident on the face PQ at an angle of incidence i₁. On entering the glass slab, it bends towards the normal and travels along BC at an angle of refraction r₁. The refracted ray BC is incident on the face SR at an angle of incidence i₂. The emergent ray CD bends away from the normal at an angle of refraction r₂.
Using Snell’s law for refraction from air to glass at face PQ,
                                                   ....(1)

Fig. Refraction through a glass slab Using Snell’s law for refraction from glass to air at face SR,
                         
But                            therefore
                                                             ...(2)
Multiplying equations (1) and (2), we get
                     
or                          
Thus, the emergent ray CD is parallel to the incident ray AB, but it has been laterally displaced by a perpendicular distance CN with respect to the incident ray. This lateral shift in the path of light on emerging from a medium with parallel faces is called lateral displacement.
It is found that the lateral displacement is directly proportional to the thickness of the glass slab.

The ability to refract light is represented by the optical density of a medium. A medium having larger refractive index is called optically denser medium. The other medium having lower refractive index is called optically rarer medium. 

The speed of light is higher in a rarer medium as compared to than in a denser medium. That is the reason, a ray of light travelling from a rarer medium to a denser medium slows down and bends towards the normal. When the ray travels from a denser medium to a rarer medium, it speeds up and bends away from the normal. 

Table: Refractive indices of some material media (with respect to vacuum)

Given, a convex mirror.
 
We have, 

Radius of curvature, R = +3.00 m    [R is +ve for a convex mirror]
Object distance, u = - 4.5 m
Image distance, v = ?
Image size,  h' = ? 

Using the relationship between focal length and radius of curvature, we have 

Focal length, $\mathrm{f} = \frac{\mathrm{R}}{2} = +\frac{3.00 \mathrm{m}}{2} = + 1.50 \mathrm{m}$          [ f is +ve for a convex mirror] 

Using the mirror formula, 

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

$\therefore$                     $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ = \frac{1}{1.50} - \frac{1}{-4.5} \\ = \frac{1}{1.50}+\frac{1}{4.5} \\ = \frac{3+1}{4.5} \\ = \frac{4}{4.5} \\ = \frac{8}{9} \end{gathered}$$

Image distance,  $\mathrm{v} = +\frac{9}{8}\mathrm{m}.$ 

The image is formed at a distance of 9/8 m behind the mirror. 

Magnification,
                      $$\begin{gathered} \mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = -\frac{\mathrm{v}}{\mathrm{u}} \\ = \frac{9/8}{4.5} \\ = + 0.25 \end{gathered}$$ 

The image formed is virtual, erect and smaller in size by a factor of 0.25 (one-fourth) than the object.