Class 10 REFRACTION OF LIGHT

### Topic Covered

color{red} ♦ REFRACTION OF LIGHT

### REFRACTION OF LIGHT

Light seems to travel along straight-line paths in a transparent medium. What happens when light enters from one transparent medium to another?

Does it still move along a straight-line path or change its direction? We shall recall some of our day-to-day experiences.
You might have observed that the bottom of a tank or a pond containing water appears to be raised.

Similarly, when a thick glass slab is placed over some printed matter, the letters appear raised when viewed through the glass slab. Why does it happen? Have you seen a pencil partly immersed in water in a glass tumbler?

It appears to be displaced at the interface of air and water. You might have observed that a lemon kept in water in a glass tumbler appears to be bigger than its actual size, when viewed from the sides. How can you account for such experiences?

Let us consider the case of the apparent displacement of a pencil, partly immersed in water. The light reaching you from the portion of the pencil inside water seems to come from a different direction, compared to the part above water.

This makes the pencil appear to be displaced at the interface. For similar reasons, the letters appear to be raised, when seen through a
glass slab placed over it.

Does a pencil appear to be displaced to the same extent, if instead of water, we use liquids like kerosene or turpentine? Will the letters appear to rise to the same height if we replace a glass slab with a transparent plastic slab?

You will find that the extent of the effect is different for different pair of media. These observations indicate that light does not travel in the same direction in all media.

It appears that when travelling obliquely from one medium to another, the direction of propagation of light in the second medium changes. This phenomenon is known as refraction of light. Let us understand this phenomenon further by doing a few activities.

ul"Activity 10.7"

♦ Place a coin at the bottom of a bucket filled with water.
♦ With your eye to a side above water, try to pick up the coin in one go. Did you succeed in picking up the coin?
♦ Repeat the Activity. Why did you not succeed in doing it in one go?

ul"Activity 10.8"

♦ Place a large shallow bowl on a Table and put a coin in it.
♦ Move away slowly from the bowl. Stop when the coin just disappears from your sight.
♦ Ask a friend to pour water gently into the bowl without disturbing the coin.
♦ Keep looking for the coin from your position. Does the coin becomes visible again from your position? How could this happen?

The coin becomes visible again on pouring water into the bowl. The coin appears slightly raised above its actual position due to refraction of light.

ul"Activity 10.9"

♦ Draw a thick straight line in ink, over a sheet of white paper placed on a Table.
♦ Place a glass slab over the line in such a way that one of its edges makes an angle with the line.
♦ Look at the portion of the line under the slab from the sides. What do you observe? Does the line under the glass slab appear to be bent at the edges?
♦ Next, place the glass slab such that it is normal to the line. What do you observe now? Does the part of the line under the glass slab appear bent?
♦ Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?

ul"Refraction through a Rectangular Glass Slab"

To understand the phenomenon of refraction of light through a glass slab, let us do an Activity.

ul"Activity 10.10"

♦ Fix a sheet of white paper on a drawing board using drawing pins.
♦ Place a rectangular glass slab over the sheet in the middle.
♦ Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
♦ Take four identical pins.
♦ Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
♦ Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
♦ Remove the pins and the slab.
♦ Join the positions of tip of the pins E and F and produce the line up to AB. Let E F meet  AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O′.
♦ Join O and O′. Also produce E F up to P, as shown by a dotted line in Fig. 10.10.

In this Activity, you will note, the light ray has changed its direction at points O and O′.

"Note :" that both the points O and O′ lie on surfaces separating two transparent media. Draw a perpendicular NN’ to AB at O and another perpendicular MM’ to CD at O′.

The light ray at point O has entered from a rarer medium to a denser medium, that is, from air to glass. Note that the light ray has bent towards the normal. At O′, the light ray has entered from glass to air, that is, from a denser medium to a rarer medium.

The light here has bent away from the normal. Compare the angle of incidence with the angle of refraction at both refracting surfaces AB and CD.

In Fig. 10.10, EO is the incident ray, OO′ is the refracted ray and O′ H is the emergent ray. You may observe that the emergent ray is parallel to the direction of the incident ray. Why does it happen so?

The extent of bending of the ray of light at the opposite parallel faces AB (air-glass interface) and CD (glass-air interface) of the rectangular glass slab is equal and opposite. This is why the ray emerges parallel to the incident ray.

However, the light ray is shifted sideward slightly. What happens when a light ray is incident normally to the interface of two media? Try and find out.

Now you are familiar with the refraction of light. Refraction is due to change in the speed of light as it enters from one transparent medium to another. Experiments show that refraction of light occurs according to certain laws.

The following are the laws of refraction of light.

(i) 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.

(ii) 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.

If i is the angle of incidence and r is the angle of refraction, then,

 (sin i)/(sin r) = constant ...........(10.4)

This constant value is called the refractive index of the second medium with respect to the first. Let us study about refractive index in some detail.

ul" The Refractive Index"

You have already studied that a ray of light that travels obliquely from one transparent medium into another will change its direction in the second medium.

The extent of the change in direction that takes place in a given pair of media is expressed in terms of the refractive index, the “constant” appearing on the right-hand side of Eq.(10.4).

The refractive index can be linked to an important physical quantity, the relative speed of propagation of light in different media.

It turns out that light propagates with different speeds in different media. Light travels the fastest in vacuum with the highest speed of 3 xx 10^8 m s^(–1).

In air, the speed of light is only marginally less, compared to that in vacuum. It reduces considerably in glass or water.
The value of the refractive index for a given pair of media depends upon the speed of light in the two media, as given below.

Consider a ray of light travelling from medium 1 into medium 2, as shown in Fig.10.11. Let v_1 be the speed of light in medium 1 and v_2 be the speed of light in medium 2.

The refractive index of medium  2 with respect to medium 1 is given by the ratio of the speed of light in medium 1 and the speed of light in medium 2. This is usually represented by the symbol n_(21). This can be expressed in an equation form as

n_(21) = text( Speed of light in medium 1)/text( Speed of light in medium 2) = v_1/v_2 ......(10.5)

By the same argument, the refractive index of medium 1 with respect to medium 2 is represented as n_(12). It is given by

n_(12) = text( Speed of light in medium 2)/text(Speed of light in medium 1) = v_2/v_1  ..........(10.6)

If medium 1 is vacuum or air, then the refractive index of medium 2 is considered with respect to vacuum. This is called the absolute refractive index of the medium. It is simply represented as n_2.

If c is the speed of light in air and v is the speed of light in the medium, then, the refractive index of the medium n_m is given by

 n_m = text( Speed of light in air)/text( Speed of light in the medium) = c/v ........(10.7)

The absolute refractive index of a medium is simply called its refractive index. The refractive index of several media is given in Table 10.3. From the Table you can know that the refractive index of water, n_w = 1.33.

This means that the ratio of the speed of light in air and the speed of light in water is equal to 1.33. Similarly, the refractive index of crown glass, n_g =1.52. Such data are helpful in many places.

However, you need not memorise the data.

Note from Table 10.3 that an optically denser medium may not possess greater mass density. For example, kerosene having higher refractive index, is optically denser than water, although its mass density is less than water.

ul"Refraction by Spherical Lenses"

You might have seen people using spectacles for reading. The watchmakers use a small magnifying glass to see tiny parts. Have you ever touched the surface of a magnifying glass with your hand?

Is it plane surface or curved? Is it thicker in the middle or at the edges? The glasses used in spectacles and that by a watchmaker are examples of lenses. What is a lens? How does it bend light rays? We shall discuss these in this section.

A transparent material bound by two surfaces, of which one or both surfaces are spherical, forms a lens. This means that a lens is bound by at least one spherical surface. In such lenses, the other surface would be plane.

A lens may have two spherical surfaces, bulging outwards. Such a lens is called a double convex lens. It is simply called a convex lens. It is thicker at the middle as compared to the edges.

Convex lens converges light rays as shown in Fig. 10.12 (a). Hence convex lenses are called converging lenses. Similarly, a double concave lens is bounded by two spherical surfaces, curved inwards.

It is thicker at the edges than at the middle. Such lenses diverge light rays as shown in Fig. 10.12 (b). Such lenses are called diverging lenses. A double concave lens is simply called a concave lens.

A lens, either a convex lens or a concave lens, has two spherical surfaces. Each of these surfaces forms a part of a sphere. The centres of these spheres are called centres of curvature of the lens. The centre of curvature of a lens is usually represented by the letter C.

Since there are two centres of curvature, we may represent them as C_1 and C_2. An imaginary straight line passing through the two centres of curvature of a lens is called its principal axis.

The central point of a lens is its optical centre. It is usually represented by the letter O. A ray of light through the optical centre of a lens passes without suffering any deviation.

The effective diameter of the circular outline of a spherical lens is called its aperture.

We shall confine our discussion in this Chapter to such lenses whose aperture is much less than its radius of curvature. Such lenses are called thin lenses with small apertures.
What happens when parallel rays of light are incident on a lens? Let us do an Activity to understand this.

ul"Activity 10.11"

CAUTION: Do not look at the Sun directly or through a lens while doing this Activity or otherwise. You may damage your eyes if you do so.

♦ Hold a convex lens in your hand. Direct it towards the Sun.

♦ Focus the light from the Sun on a sheet of paper. Obtain a sharp bright image of the Sun.

♦ Hold the paper and the lens in the same position for a while. Keep observing the paper. What happened? Why? Recall your experience in Activity 10.2.

The paper begins to burn producing smoke. It may even catch fire after a while. Why does this happen? The light from the Sun constitutes parallel rays of light.

These rays were converged by the lens at the sharp bright spot formed on the paper. In fact, the bright spot you got on the paper is a real image of the Sun. The concentration of the sunlight at a point generated heat. This caused the paper to burn.

Now, we shall consider rays of light parallel to the principal axis of a lens. What happens when you pass such rays of light through a lens? This is illustrated for a convex lens in Fig.10.12 (a) and for a concave lens in Fig.10.12 (b).

Observe Fig.10.12 (a) carefully. Several rays of light parallel to the principal axis are falling on a convex lens. These rays, after refraction from the lens, are converging to a point on the principal axis.

This point on the principal axis is called the principal focus of the lens. Let us see now the action of a concave lens.

Observe Fig.10.12 (b) carefully. Several rays of light parallel to the principal axis are falling on a concave lens. These rays, after refraction from the lens, are appearing to diverge from a point on the principal axis.

This point on the principal axis is called the principal focus of the concave lens.

If you pass parallel rays from the opposite surface of the lens, you get another principal focus on the opposite side. Letter F is usually used to represent principal focus.

However, a lens has two principal foci. They are represented by F_1 and F_2. The distance of the principal focus from the optical centre of a lens is called its focal length.

The letter f is used to represent the focal length. How can you find the focal length of a convex lens? Recall the Activity 10.11. In this Activity, the distance between the position of the lens and the position of the image of the Sun gives the approximate focal length of the lens.

ul "Image Formation by Lenses"

Lenses form images by refracting light. How do lenses form images? What is their nature? Let us study this for a convex lens first.

♦ Take a convex lens. Find its approximate focal length in a way described in Activity 10.11.

♦ Draw five parallel straight lines, using chalk, on a long Table such that the distance between the successive lines is equal to the focal length of the lens.

♦ Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.

♦ The two lines on either side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as
2 F_1, F_1, F_2 and 2F_2, respectively.

♦ Place a burning candle, far beyond 2 F_1 to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.

♦ Note down the nature, position and relative size of the image.

♦ Repeat this Activity by placing object just behind 2F_1, between F_1 and 2F_1 at F_1, between F_1 and O. Note down and tabulate your observations.

The nature, position and relative size of the image formed by convex lens for various positions of the object is summarised in Table 10.4.

Let us now do an Activity to study the nature, position and relative size of the image formed by a concave lens.

ul"Activity 10.13"

♦ Take a concave lens. Place it on a lens stand.

♦ Place a burning candle on one side of the lens.

♦ Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible. If not, observe the image directly through the lens.

♦ Note down the nature, relative size and approximate position of the image.

♦ Move the candle away from the lens. Note the change in the size of the image. What happens to the size of the image when the candle is placed too far away from the lens.

The summary of the above Activity is given in Table 10.5 below.

What conclusion can you draw from this Activity? A concave lens will always give a virtual, erect and diminished image, irrespective of the position of the object.

ul"Image Formation in Lenses Using Ray Diagrams"

We can represent image formation by lenses using ray diagrams. Ray diagrams will also help us to study the nature, position and relative size of the image formed by lenses.

For drawing ray diagrams in lenses, alike of spherical mirrors, we consider any two of the following rays –

(i) A ray of light from the object, parallel to the principal axis, after refraction from a convex lens, passes through the principal focus on the other side of the lens, as shown in Fig. 10.13 (a).

In case of a concave lens, the ray appears to diverge from the principal focus located on the same side of the lens, as shown in Fig. 10.13 (b).

(ii) A ray of light passing through a principal focus, after refraction from a convex lens, will emerge parallel to the principal axis. This is shown in Fig. 10.14 (a).

A ray of light appearing to meet at the principal focus of a concave lens, after refraction, will emerge parallel to the principal axis. This is shown in Fig.10.14 (b).

(iii) A ray of light passing through the optical centre of a lens will emerge without any deviation. This is illustrated in Fig.10.15(a) and Fig.10.15 (b).

The ray diagrams for the image formation in a convex lens for a few positions of the object are shown in Fig. 10.16. The ray diagrams representing the image formation in a concave lens for various positions of the object are shown in Fig. 10.17.

"Sign Convention for Spherical Lenses"

For lenses, we follow sign conventions, similar to the one used for spherical mirrors. We apply the rules for signs of distances, except that all measurements are taken from the optical centre of the lens.

According to the convention, the focal length of a convex lens is positive and that of a concave lens is negative. You must take care to apply appropriate signs for the values of u, v, f, object height h and image height h′.

ul" Lens Formula and Magnification"

As we have a formula for spherical mirrors, we also have formula for spherical lenses. This formula gives the relationship between object distance (u), image-distance (v) and the focal length (f ). The lens formula is expressed as

 1/v - 1/u = 1/f ......(10.8)

The lens formula given above is general and is valid in all situations for any spherical lens. Take proper care of the signs of different quantities, while putting numerical values for solving problems relating to lenses.

"Magnification :"

The magnification produced by a lens, similar to that for spherical mirrors, is defined as the ratio of the height of the image and the height of the object. It is represented by the letter m.

If h is the height of the object and h′ is the height of the image given by a lens, then the magnification produced by the lens is given by,

 m = text( Height of the Image)/text( Height of the object) = (h')/h .......(10.9)

Magnification produced by a lens is also related to the object-distance u, and the image-distance v. This relationship is given by

Magnification (m ) = h′//h = v//u ........(10.10)
Q 3254191054

A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also, find the magnification produced by the lens.
Class 10 Chapter 10 Example 3
Solution:

A concave lens always forms a virtual, erect image on the same side of the object.

Image - distance v = –10 cm;

Focal length f = –15 cm;

Object - distance u = ?

Since  1/v - 1/u = 1/f

or,  1/u = 1/v - 1/f

 1/u = 1/(-10) - 1/(-15) = - 1/(10) + 1/(15)

 1/u = (-3 + 2)/(30) = 1/(-30)

or, u = – 30 cm
Thus, the object-distance is 30 cm.
Magnification m = v//u

 m = (- 10 cm)/(- 30 cm) = 1/3 = + 0.33
The positive sign shows that the image is erect and virtual. The image is one-third of the size of the object.
Q 3264191055

A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification.
Class 10 Chapter 10 Example 4
Solution:

Height of the object h = + 2.0 cm;

Focal length f = + 10 cm;

object-distance u = –15 cm;

Image-distance v = ?

Height of the image h′ = ?

Since  1/v - 1/u = 1/f

or  1/v = 1/u + 1/f

 1/c = (-2 + 3) /(30) = 1/(30)

or, v = + 30 cm

The positive sign of v shows that the image is formed at a distance of 30 cm on the other side of the optical centre. The image is real and inverted.

Magnification  m = (h')/h = v/u

or, h′ = h (v//u)

Height of the image, h′ = (2.0) (+30//–15) = – 4.0 cm

Magnification m = v//u

or  m = (+ 30 cm)/(- 15 cm) = -2

The negative signs of m and h′ show that the image is inverted and real. It is formed below the principal axis. Thus, a real, inverted image, 4 cm tall, is formed at a distance of 30 cm on the other side of the lens. The image is two times enlarged.

### Power of a Lens :

You have already learnt that the ability of a lens to converge or diverge light rays depends on its focal length. For example, a convex lens of short focal length bends the light rays through large angles, by focussing them closer to the optical centre.

Similarly, concave lens of very short focal length causes higher divergence than the one with longer focal length. The degree of convergence or divergence of light rays achieved by a lens is expressed in terms of its power.

The power of a lens is defined as the reciprocal of its focal length. It is represented by the letter P. The power P of a lens of focal length f is given by

 P = 1/f ........(10.11)

The SI unit of power of a lens is ‘dioptre’. It is denoted by the letter D.

If f is expressed in metres, then, power is expressed in dioptres. Thus, 1 dioptre is the power of a lens whose focal length is 1 metre. 1 D = 1 m^(–1). You may note that the power of a convex lens is positive and that of a concave lens is negative.

Opticians prescribe corrective lenses indicating their powers. Let us say the lens prescribed has power equal to + 2.0 D. This means the lens prescribed is convex. The focal length of the lens is + 0.50 m.

Similarly, a lens of power – 2.5 D has a focal length of – 0.40 m. The lens is concave.