Light Reflection And Refraction - Class 10 Physics - Extra Questions

Law of refraction states that:

• The incident ray, reflected ray and the normal, to the interface of any two given mediums; all lie in the same plane.
• The ratio of the sine of the angle of incidence and sine of the angle of refraction is constant.

1. The incident ray,the refracted ray and the normal to the refracting surface at the point of incidence lie in the same plane.
2. For a given pair of media and for a given colour of light the ration between the sine of angle of incidence to the sine of refraction is a constant.This constant is known as refractive index of the second medium with respect to the first medium.

Answer REFLECTED

A light ray striking a surface is called incident ray and the ray that reflects from the surface is called reflected ray.

1. Incident ray, reflected ray, refracted ray and the normal of the system lie in the same plane.
2. Snells Law: $$\frac{sin\ i}{sin\ r}=\frac{n_2}{n_1}=\frac{v_1}{v_2}$$

Snell's law states that the relationship between angle of incidences and refraction for a wave impinging on an interface between two media with different incidences of angle is always constant, The law follows from the boundary condition that a wave be continuous across a boundary, which requires that the phase of the wave be constant on any given plane.

angle of incidence = $$i$$  

angle of refraction = $$r$$  

$$\dfrac{sin i}{sin r}=constant$$

The value of this constant depends upon temperature and wavelength of light. This is also known as Snell's law. 

(2) The incident ray, the refracted ray and the normal at the point of incidence, all the lie in the same plane. 

Refractive index (of a medium)  = speed of light in air (or vaccum) / speed of light in medium 

$$\implies \mu_m=\dfrac{c}{v_m}$$

(b) $$\mu_{glass}=\dfrac{c}{v_{glass}}$$

$$\implies \dfrac{3}{2}=\dfrac{c}{2\times 10^8m/s}$$

(i) $$\implies c=3\times 10^8m/s$$

(ii) $$\mu_{water}=\dfrac{c}{v_{water}}$$

$$\implies v_{water}=\dfrac{3\times 10^8m/s}{4/3}$$

$$=\dfrac{9}{4}\times 10^8m/s$$

For understanding the refraction of light through a glass slab let us consider the above figure consider the figure,

Here AO is the light ray travelling in air and incident on glass slab at point O.

When ray enters the glass medium it bends towards the normal NN’ that is light ray AO gets refracted on entering the glass medium.

When ray gets refracted it now travels through the glass slab and at point B it comes out of the glass slab.

As ray OB goes from glass medium to air it again gets refracted and bends away from normal N1N'1 and goes in direction BC.

Here AO is the incident ray and BC is the emergent ray and both are parallel to each other and OB is the refracted ray.

Emergent ray BC is parallel to incident ray AO because the extent of bending of the ray of light at the opposite parallel faces which are PQ and SR of the rectangular glass slab is equal and opposite.

In the figure i is the angle of incidence, r is the angle of refraction and e is the angle of emergence.

Angle of incidence and angle of emergence are equal as emergent ray and incident ray are parallel to each other.

When a light ray is incident normally to the interface of two media then there is no bending of light ray and it goes straight through the medium.

The pins must not be too close to each other, otherwise, it would be hard to align the light passing through both of them when seen from the other side of the slab. Hence $$I$$ and  $$IV$$  are rejected.

Also, the angle of incidence and the distance of the point of incidence from the mid slab must be optimum for observing the shifted path of light after emergence from the other side of the slab. So, $$II$$ is rejected.

Hence, the correct answer is option  $$III$$ .

A) $$\text{Definition}$$ : Refraction is the bending of light when it travels from one medium to another. 

$$\text{Snell's Law}$$ : Snell's law states that the ratio of the sine of the angle of incidence ( $$sin\ i$$ ) to the sine of the angle of refraction ( $$sin\ r$$ ) is constant. This constant is known as the refractive index of the second medium with respect to the first medium ( $$n_{21}$$ ).

Mathematically, $${ n }_{ 21 }=\dfrac{n_2}{n_1}=\dfrac { sini }{ sinr }$$ , where $$n_1$$ and $$n_2$$ are the absolute refractive index (refractive index with respect to the air medium) of medium- 1 and medium- 2 respectively.

B) Given:

The speed of light in air is $$c=3\times 10^8\ m/s$$

The refractive index of water with respect to air is $$n_{wa}=\dfrac{4}{3}$$

The speed of light in water $$v=?$$

From the definition of refractive index, $$n_{wa}=\dfrac{c}{v}$$

$$\therefore \dfrac{4}{3}=\dfrac{3\times 10^8}{v}$$

$$\Rightarrow v=\dfrac{9\times 10^8}{4}=2.25\times 10^8\ m/s$$

Thus, the speed of light in water is $$2.25\times 10^8\ m/s$$

The principal focus of the convex lens is a point on the principal axis where the light rays appear to be meeting after refraction from the lens as shown in the figure.F in the figure is the principal focus.

Refraction is the bending of a wave of light when it enters a medium where its speed is different. The refraction of light when it passes from a fast medium to a slow medium bends the light ray toward the normal to the boundary between the two media. The amount of bending depends on the indices of refraction of the two media and is described quantitatively by Snell's Law.

Snell's law states that the ratio of the sines of the angles of incidence and refraction is equivalent to the ratio of phase velocities in the two mediums, or equivalent to the reciprocal of the ratio of the index of refraction

$$\dfrac{\sin {{\theta }_{i}}}{\sin {{\theta }_{r}}}=\dfrac{{{v}_{2}}}{{{v}_{1}}}=\dfrac{{{n}_{1}}}{{{n}_{2}}}$$

Given that,

Angle of incident $$i={{72}^{0}}$$

Angle of refraction $$r={{40}^{0}}$$

We know that, according to Snell's law,

Refractive index is

$$n=\dfrac{\sin i}{\sin r}$$

  $$n=\dfrac{0.951}{0.642}$$

  $$n=1.48$$

Hence, the refractive index of substance is $$1.48$$

Given that,

Focal length $$f=15\,cm$$

Distance of the image $$v=-10\,cm$$

We know that,

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

  $$\dfrac{1}{-15}=\dfrac{1}{-10}-\dfrac{1}{u}$$

  $$-\dfrac{1}{u}=\dfrac{1}{-15}+\dfrac{1}{10}$$

  $$-\dfrac{1}{u}=\dfrac{1}{30}$$

  $$u=-30\,cm$$

Hence, the object is placed $$30\ cm$$ away from the concave lans

 We can observe that the lens slightly diverges the path of refracted light. Hence, the lens is concave in nature.Because only concave lens is diverging in nature.

$$\large\textbf{Part (a): Calculate the focal length}$$

Given,

Magnification, m = -3 (Image is real)
Object distance $$,u$$ $$=$$ $$-20cm$$ ; By sign convention as object is placed at the left side of the mirror.
we know,

Magnification,  $$m=\frac{-v}{u}$$
or            $$-3=\frac{-v}{20}$$

Or           $$v=-60cm$$

Thus the image is located at a distance of 60cm in front of the mirror.

Now using the mirror formula,

$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$$
$$\frac{1}{f}=\frac{1}{-60}+\frac{1}{-20}$$
f = -15 cm

$$\large\textbf{Part (b): Apply magnification formula}$$

Given,

Magnification, m = +3 (Image is virtual)

Magnification,  $$m=\frac{-v}{u}=\frac{{{h}_{i}}}{{{h}_{o}}}$$
or            $$+3=\frac{-v}{u}$$

Or $$v=-3u$$

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

Or           $$\frac{1}{-15}=\frac{1}{-3u}+\frac{1}{u}$$
on solving $$u = -10cm$$

Hence, The Object located at a distance of 10 cm in front of the mirror.

$$\large\textbf{Step 1: Calculate the focal length}$$

Given,

Magnification, m = -3 (Image is real)
Object distance $$,u$$ $$=$$ $$-20cm$$ ; By sign convention as object is placed at the left side of the mirror.
we know,

Magnification, $$m=\frac{-v}{u}$$
or            $$-3=\frac{-v}{-20}$$

Or           $$v=-60cm$$

Thus the image is located at a distance of 60cm in front of the mirror.

Now using the mirror formula,

$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$$
$$\frac{1}{f}=\frac{1}{-60}+\frac{1}{-20}$$

$$f = -15 cm$$

$$\large\textbf{Step 2: Apply magnification formula}$$

Given,

Magnification, m = +3 (Image is virtual)

Magnification, $$m=\frac{-v}{u}=\frac{{{h}_{i}}}{{{h}_{o}}}$$
or            $$+3=\frac{-v}{u}$$

Or $$v=-3u$$

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

Or           $$\frac{1}{-15}=\frac{1}{-3u}+\frac{1}{u}$$
on solving $$u = -10cm$$

Hence, The Object located at a distance of 10 cm in front of the mirror.

.

Given,

Radius , $$R = 1.5 m$$

Object distance $$= 10 m$$

Apply mirror formula:

Mirror formula gives the relationship between object distance $$\left( u \right)$$ , image distance $$\left( v \right)$$ and focal length $$\left( f \right)$$ and is written as follows:

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

Here,  $$f\; = \frac{{ - R}}{2} = \; - 0.75\;m$$                 …..(mirror is concave)

All the distances towards the left of the mirror are negative , here object is placed in front of the mirror,(i.e in left)

         $$u\; = \; - 10\;m$$

Therefore,  $$\frac{1}{v} = \frac{1}{{ - 0.75\;m}} - \frac{1}{{\left( { - 10\;m} \right)}}$$

or $$v\; = \; - \;0.81\;m$$

Since $$v\;is\; - ve$$

                                         $$\Rightarrow \;The\;image\;is\;real\;and\;inverted.$$

Hence, a real and inverted image is formed at a distance 0.81 m in front of the mirror.

• When light goes from denser to rarer medium, it bends away from the normal.
• When the angle of incidence is equal to the critical angle, the angle of refraction is $$90^o.$$  
• Grazing emergence is the condition in which the emergent ray is perpendicular to the normal of the emergent surface of medium.

$$\large\textbf{Part i: Calculate the focal length of the Lens}$$

Given,

Position of the image on the screen $$v =+120cm$$ , By sign convention, Image is formed at right side of the lens. so it's istaken as positive

We know,

By Lens formula

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

Or $$\dfrac{1}{f}$$ $$=$$ $$\dfrac{1}{120}$$ $$-$$ $$\dfrac{1}{-60}$$

Or $$f = +40 cm$$

The focal length of the lens convex lens is  $$f = +40 cm$$  

$$\large\textbf{Part ii: Calculate height of the Image}$$

Given,

 Height of the object $$h_o= 5cm$$

we know,

magnification $$,m=\dfrac{v}{u}$$

$$m=\dfrac{120}{-60} =-2$$

Height of the image $$h_i = m \times h_o = -2 \times 5 =-10 cm$$

Height of the image formed is $$10cm$$ . 

As magnification is Negative hence, Image formed is real and inverted.

$$\large \textbf{Part(i) Position of the image }$$

Given,

As the object is placed in front of the mirror,(left of the concave mirror), distance is taken as negative.

Object is placed a distance $$, u=-12cm$$

Concave mirror of radius of curvature $$,R=30cm$$

Or focal length $$, f= -R/2=-30/2 = -15cm$$

We know,

By mirror formula $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

Or $$\frac{1}{{ - 15}} = \frac{1}{v} + \frac{1}{{ - 12}}$$

Or $$\frac{1}{v} =  + \frac{1}{{60}}$$

Or $$v = +60cm$$

Image is formed behind the mirror at a distance of $$60cm$$ from the mirror.

$$\large \textbf{Part(ii) Characteristics  of image }$$

We know,

Magnification, $$m = \frac{{ - v}}{u}$$ ,  

$$m = \frac{{ - v}}{u} =  - \frac{{ + 60}}{{ - 12}} =  + 5$$

Hence , Characteristics of image is:-

$${\mathop{Im}\nolimits} age\,\,\,is\,formed\,behind\,the\,mirror$$

$$Image\,is\,Virtual$$

$$Image\,is\,Erect$$

$${\mathop{ Im}\nolimits} age\,is\,magnified(\,larg er\,than\,object)\,$$

$$\large\underline{\textbf{Part(i): Calculate the focal length}}$$  

Given,

All the distances are measured from the pole and all the distances towards the left of the mirror(i.e in front of the mirror) are taken as negative.

Object is placed at a distance $$, u =-30cm$$

Image formed on screen at $$, v =-60cm$$

We know,

By mirror formula  $$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$$

Or  $$\dfrac{1}{f} = \dfrac{1}{{ - 60}} + \dfrac{1}{{ - 30}}$$

Or  $$\dfrac{1}{f} =  - \dfrac{3}{{60}}$$

Or  $$f =  - 20cm$$

Here, the focal length of the mirror is negative, representing the converging nature of the mirror. Hence the given mirror is a $$\textbf{concave mirror}$$ of focal length $$= 20cm.$$

$$\large\underline{\textbf{Part(ii): Height of the image}}$$  

Given height of object, $${h_o} = 2.4cm$$

Magnification, $$m = \dfrac{{ - v}}{u}$$

Or  $$m = \dfrac{{ - v}}{u} =  - \dfrac{{ - 60}}{{ - 30}} =  - 2$$ (Negative)

Height of the image,  $${h_{im}} = m{h_o} =  - 2 \times 2.4 =  - 4.8cm$$

Hence, Image is Real and inverted. It is magnified too.

$$\large\underline{\textbf{Part(iii): Characteristics of the image}}$$  

Here magnification is negative, so  Image Formed is Real and inverted. It is magnified

Hint: Power and focal length are reciprocal of each other.

Solution:

Step 1: Find focal length of the lens.

Power of lens = $$P=-2.0 \mathrm{D}$$

Focal length of lens = $$f=\frac{1}{P}$$

$$\Rightarrow f=\frac{1}{(-2.0 D)}$$

$$\Rightarrow f=-0.5 m=-50 \mathrm{~cm}$$

Step 2: Find the type of lens.

Now, $$f=-50 \mathrm{~cm}$$

Since, focal length of the lens is negative.

So, this lens is a Concave lens.

Hence, lens with power $$-2.0 D$$ is Concave in nature.

Lens formula: Relation between object distance u, image distance v and focal length f is:

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

Here $$m = +1/3$$ (Concave lens always formed virtual and erect images)

$$u = -30cm$$  

$$\large\textbf{Part (a) Position of the image}$$

$$magnification\,,m{\rm{ }} = \;\frac{v}{u}\,$$

Let image is formed at distance $$= v\  cm$$

Therefore, $$\frac{1}{3}{\rm{ }} = \;\frac{v}{{ - 30}}$$

Or   $$v = -10$$ cm

$$\large\textbf{Part (b) Focal length }$$

By lens formula

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

Or

$$\frac{1}{f} = \frac{1}{{ - 10}} - \frac{1}{{ - 30}}$$

$$f = -15 cm$$

if on decreasing the distance between object and lens , the size of image increases then it is a convex type lens (converging lens)

If we keep on decreasing the distance between the object and the lens, we will not be able to obtain the image on the screen because when object is placed very closed to the convex lens ,the image formed is virtual.

Virtual images cant be obtained on screen.

• Pole of a mirror is the geometrical center of the spherical surface of the mirror.
• Principal axis is the straight line that joins the pole of the mirror to its center of curvature. 
• Center of curvature of mirror is the center of the sphere of which the mirror is part of. 

1. Concave mirror forms virtual image if object is placed between the focus and pole of the mirror. Therefore, for the position of object at $$10\ cm$$ mirror forms the required image.
2. A real and diminished image is formed when object lies beyond C i.e. beyond $$2F$$ . So, for the position of object at $$40\ cm$$ , mirror forms the required image.
3.  An enlarged real image is formed when object lies between $$F$$ and $$2F$$ . So, for the position of object at $$20\ cm$$ , mirror forms the required image.
4. An image of same size of the object is formed when object lies at $$C$$ or $$2F$$ . So, for the position of object at $$30\ cm$$ , mirror forms the required image.

Hint: Use the Snell’s law.

Step 1: Calculate the angle of incidence by applying Snell’s law.

From Snell’s law, we know

$$sini=\mu sinr$$ (where, $$i$$ is the angle of incidence, $$r$$ is the angle of refraction and $$\mu$$ is the refractive index of the glass slab)

When $$\angle i=0$$ , then,

$$sin0=\mu sinr$$

$$\Rightarrow r=0$$

$$\therefore\angle i=\angle r$$

Therefore, when the value of $$i$$ is $$0^{\circ}$$ , the angle of incidence is equal to the angle of refraction in a transparent slab.

Hint: Apply magnification formula and mirror formula.

Derivation:                                                                                

Step 1: List the given data.

·         $$h = 6cm$$                                                              

·         $$f =  15cm$$

·         $$u =  -10cm$$

Step 2: Find the image distance by mirror formula.

By mirror formula, $$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$$ , where $$h$$ is the object height, $$v$$ is the image distance and $$u$$ is the object distance.

$$\Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$$

$$\Rightarrow \dfrac{1}{v} = \dfrac{1}{{ 15}} + \dfrac{1}{{ - 10}}$$

$$\Rightarrow \dfrac{1}{v} = \dfrac{1}{{ - 30}}$$

$$\Rightarrow v =  - 30cm$$

Step 3: Find the image size by magnification formula.

By magnification formula, $$m = \dfrac{{ v}}{u} = \dfrac{{ -30}}{{-10}}$$

$$m =  3$$

Also, $$m = \dfrac{{h'}}{h}$$

$$\Rightarrow h' =  3 \times 6 =  18 cm$$

So, image distance is -30 cm and image height is 18 cm.

Hint: Apply the relation obtained from Snell’s law.

Step 1: 

From Snell’s law of refraction, we can write,

$$sin i=\mu sin r$$ (where, $$\mu$$ is the refractive index or optical density of the medium)

$$\mu =\dfrac{sin 60^o}{sin r}$$

Step 2:

The higher the angle of refraction, the higher will be the value of sin, and the lesser will be the optical density of the medium.

Accordingly optical density of medium A is greater than that of medium B.

When a light ray travels from medium A to medium B, the light will bend away from the normal as when a ray travels from a medium of greater optical density to a medium of lesser optical density, it bends away from the normal.

Hint: Apply the formula from Snell’s law of refraction.

Step 1:

From Snell’s law of refraction, we can write,

$$sin i=\mu sin r$$ (where, $$\mu$$ is the refractive index or optical density of the medium)

$$\mu =\dfrac{sin 60^o}{sin r}$$

Step 2:

According to the above relation, higher the angle of refraction, the higher will be the value of sin, and the lesser will be the optical density of the medium.

Hence, the optical density of medium A is highest.

On entering into the glass medium light ray bends towards the normal that is light ray gets refracted on entering the glass medium. After getting refracted this ray now travels through the glass slab and comes out of the glass slab by refraction from the other interface boundary. Since ray goes from glass medium to air it again gets refracted and bends away from normal. The incident ray and the emergent ray are parallel to each other.

$$i$$  is the angle of incidence, $$r$$  is the angle of refraction and $$e$$  is the angle of emergence. Angle of incidence and angle of emergence are equal as emergent ray and incident ray are parallel to each other. When a light ray is incident normally to the interface of two media then there is no bending of light ray and it goes straight through the medium.

The magnification is given as,

$$m =  - \dfrac{v}{u}$$

$$\dfrac{1}{2} =  - \dfrac{v}{u}$$

$$\dfrac{1}{v} =  - \dfrac{2}{u}$$

The mirror formula is given as,

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

$$- \dfrac{2}{u} + \dfrac{1}{u} = \dfrac{1}{{20}}$$

$$u =  - 20\;{\rm{cm}}$$

Thus, the position of the object is $$- 20\;{\rm{cm}}$$ .

If the angle of incidence and angle of emergence of a light ray falling on a glass slab are given then by applying Snell’s law we can get,

$$\dfrac{{\sin i}}{{\sin r}} = \dfrac{{{n_2}}}{{{n_1}}}$$

$$\dfrac{{\sin r}}{{\sin e}} = \dfrac{{{n_1}}}{{{n_2}}}$$

On multiplying the above two equation we get

$$\dfrac{{\sin i}}{{\sin e}} = 1$$

$$\sin i = \sin e$$

Hence the angle of incidence is equal to the angle of emergence

Magnification $$m = \dfrac{{{h_I}}}{{{h_O}}}$$

                        $$m = \dfrac{{ - 4}}{1} =  - 4$$

Here,  $$u =  - 20\;{\rm{cm}}$$

Magnification can also be written as:

              $$m = \dfrac{{ - v}}{u}$$

            $$- 4 = \dfrac{{ - v}}{{ - 20}}$$

              $$v =  - 80\;{\rm{cm}}$$

Thus image distance is $$80\;{\rm{cm}}$$ .

Now, Using mirror formula,

      $$\dfrac 1f=\dfrac 1v+\dfrac 1u$$

      $$\dfrac 1f=\dfrac 1{-80}+\dfrac 1{-20}=\dfrac{-5}{80}$$

hence,    $$f=-16\ cm$$

• $$u=-30cm$$ and magnification $$m=(-v/u) =-3$$    so    $$v=(3u)$$
• image is real so $$m$$ should be $$negative$$   $$v$$ should be negative so $$v=-90cm$$

By using the formula below and substituting the value of $$v = 2u$$

$$m =  - \dfrac{v}{u}$$

By substituting all the values in mirror formula we get

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

$$\dfrac{1}{{ - 2u}} + \dfrac{1}{{ - u}} = \dfrac{1}{{10}}$$

Then the height is $$15cm$$

The image formed will be erected diminished and virtual.

 Given that,

Height $$h=3\,cm$$

Distance $$u=-15\,cm$$

Radius = $$10\,cm$$

Now, the focus

  $$f=\dfrac{r}{2}$$

  $$f=\dfrac{10}{2}=5\,cm$$

Now, by using mirror equation

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

  $$\dfrac{1}{5}=\dfrac{1}{v}-\dfrac{1}{15}$$

  $$\dfrac{1}{v}=\dfrac{1}{5}+\dfrac{1}{15}$$

  $$v=\dfrac{15}{4}\,cm$$

  $$v=4\,cm$$

So, the image will be formed between focus and pole

Now, the magnification is

  $$m=\dfrac{-v}{u}$$

  $$m=\dfrac{4}{15}$$

 Now,

  $$m=\dfrac{h'}{h}$$

  $$\dfrac{4}{15}=\dfrac{h'}{3}$$

  $$h'=\dfrac{12}{15}$$

  $$h'=0.8$$

Hence, the height of image is $$0.8\ cm$$ and distance of image is $$4\ cm$$

If the angle of incidence and angle of emergence of a light ray falling on a glass slab are given then by applying snell’s law we can get

$$\frac{{\sin i}}{{\sin r}} = \frac{{{n_2}}}{{{n_1}}}$$

When the light travels from the air to glass 

$$\frac{{\sin i}}{{\sin r}} = \mu _g^a$$

Now the light travels withing the glass the angle of incidence is equal to the angle of reflection

When light goes out of the glass into air then according to snells law

$$\frac{{\sin r}}{{\sin e}} = \frac{{{n_1}}}{{{n_2}}}$$

$$\frac{{\sin r}}{{\sin e}} = \mu _a^g$$

On multiplying the above two equation we get

$$\frac{{\sin i}}{{\sin e}} = 1$$

$$\sin i = \sin e$$

$$i=r$$

Hence the angle of incidence is equal to the angle of emergence and the rays are parallel to each other.

Two laws of refraction are as follows.

1. The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
2. The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media. This law is also known as Snell’s law of refraction. (This is true for angle $$0^o < i < 90^o$$ )

When the light ray enters from water to air, it is refracted so that it deviates away from normal because light ray is entering from denser medium to rarer medium.

Thus, an air bubble inside water behaves like a diverging lens.

Refer the diagram for Refracted Ray and Emergent Ray.

The ray of light incident on the glass slab undergoes refraction twice and comes out from the other side of the slab as an emergent ray. 

It is to be noted that the emergent ray shows a lateral shift of the ray from the original path and is parallel to incident ray because it went down refraction two times that too at parallel surfaces PQ and SR.

• When the light bends towards Normal,the medium B must be optically dense than A.
  
• Therefore refractive index of A is less than refractive index of B.
  

$$\large\textbf{Step -1: Calculate of focal length}$$

In the given situation,

Here,

Object Distance ,  $$u\; = \; - \left( {50 - 8} \right) =  - 42\;cm$$   

Image distance ,   $$v\; = \; + \left( {92 - 50} \right) =  + 42\;cm$$  

we know,

According to the lens formula:

                                                    $$\dfrac{1}{v} - \;\dfrac{1}{u}\; = \;\dfrac{1}{f}$$

Therefore,  $$\dfrac{1}{f} = \dfrac{1}{{42\;cm}} - \dfrac{1}{{\left( { - 42\;cm} \right)}}$$

   $$\dfrac{1}{f} = \dfrac{2}{{42\;}}$$

or $$f\; = \;21\;cm$$       Positive sign of focal length indicates, it is a convex Lens.

Hence, the focal length of the Lens is $$21 cm.$$

Hint: Lens formula, $$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \quad$$ where $$v, u, f$$ are image distance, object distace and focal length of the lens

Solution: 

Step-1: Find image distance

Given height of the object $$\left(h_{0}\right)=2 \mathrm{~cm}$$

According to sign convention, object distance from lens $$(u)=-15 \mathrm{~cm}$$

Focal length of the lens $$(f)=+10 \mathrm{~cm}$$

Let image distance $$=v \mathrm{~cm}$$

From lens formula $$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$ we get,

$$\Rightarrow \frac{1}{v}-\frac{1}{(-15)}=\frac{1}{10}$$

$$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{15}$$

$$\Rightarrow \frac{1}{v}=\frac{3-2}{30}=\frac{1}{30}$$

$$\Rightarrow v=30 \mathrm{~cm}$$

As the image distance is positive, hence image is real, formed at 30 cm distance from the lens on  opposite side of lens.

Since image size is $$1 / 3$$ of object size, so image distance is $$1 / 3$$ of object distance because

$$\mathbf{h}^{\prime} / \mathrm{h}=\mathbf{v} / \mathbf{u}$$

Using the mirror formula, we can calculate object distance and image distance

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

or, $$\dfrac{3}{u}-\dfrac{1}{u}=-\dfrac{1}{20}$$

or, $$\dfrac{3-1}{u}=-\dfrac{1}{20}$$

or, $$u=-40 \mathrm{cm}$$

Using this, we can find image distance:

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

or $$, \dfrac{1}{v}-\dfrac{1}{40}=-\dfrac{1}{20}$$

or $$, \dfrac{1}{v}=-\dfrac{1}{20}+\dfrac{1}{40}$$

or $$, \dfrac{-2+1}{40}=-\dfrac{1}{40}$$

But above value of image distance does not match with our initial assumption. This means that the mirror is not a concave mirror but a convex mirror. Let us calculate with the assumptiom that it is a convex mirror.

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

or, $$\dfrac{1}{v}-\dfrac{1}{40}=-\dfrac{1}{20}$$

or $$, \dfrac{1}{v}=-\dfrac{1}{20}+\dfrac{1}{40}=\dfrac{3}{40}$$

or, $$v=\dfrac{40}{3} \mathrm{cm}$$

Nature of mirror: convex mirror.

Nature of image: Smaller than object, erect and virtual.

$$u=-(50-12)=-38 \mathrm{cm}$$

Image distance $$v=88-50=38 \mathrm{cm}$$

Focal length can be calculated using the lens formula;

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

or, $$\dfrac{1}{38}+\dfrac{1}{38}=\dfrac{1}{f}$$

or, $$\dfrac{2}{38}=\dfrac{1}{f}$$

or, $$f=19 \mathrm{cm}$$

Place a rectangular glass slab on a white paper board. Mark the boundaries of the slab as PQRS. Now project a laser beam obliquely on the surface, PQ of the glass. Trace the incident ray and mark a point, A on it.  Mark the point where the incident ray touches the surface, PQ as point B. Now observe the obliquely bend ray passing through the rectangular glass slab. The rays emerges out of the slab from the surface, SR. Trace the emergent ray and mark the point on the opposite surface RS from where it emerges out as point C. Mark one more point on the ray coming out from the point C and name it as point, D. Remove the rectangular glass slab and connect the points A, B, C and D.

Thus, we obtain the incident ray AB, refracted ray BC and the emergent ray CD. The path, ABCD in the figure is the path of refraction of light through the glass slab. The incident ray, AB bends towards the normal as it travels from air (rarer medium) to glass (denser medium) and gets refracted. The ray BC travels away from the normal as it travels from glass (denser medium) to air (rarer medium).