Integrated Science SME

Key Unit competence: Interpret and solve problems solving in thin
                                               lenses and glass prism

Introductory Activity 5
Observe the images below and answer the following questions

a). Name image A,B,C and D and identify where they can be used in our
daily life activities
b). What is the effect of light on A, B, C and D?
c). Analyze the phenomenon in figure C and write your observations in
your notebook
d). How the images are formed on each of the above figures? Explain.

5.1. Types of lenses and their characteristics
Activity 5.1
1. Observe closely the lenses below and answer the asked questions:

a). Predict the materials used to make the above lenses. Explain your
answer
b). How are the lenses different?
c). Classify these lenses according to your observation and explain the
criteria used to classify them.

A lensis an optical device with perfect or approximate axial symmetry which transmits and refractslight, by converging or diverging the beam. Most lenses are spherical lenses: their two surfaces are parts of the surfaces of spheres. Lenses are classified by the curvature of the two optical surfaces.

The lens which is thicker at the centre than at the edges is called a convex lens while the one which is thinner at its centre is known as a concave lens. The curved surface of the lens is called a meniscus. The lens in the human eye is thicker in the centre, and therefore it is a convex lens.

The light rays from the ray box change the direction after passing through the lens. They are therefore refracted by the lens. Hence, lenses form images of objects by refracting light.

You can see that the rays from the convex lens are getting closer and closer to a point. The rays are thus converging, and hence a convex lens is called a converging lens. You can also see that the refracted rays from the concave lens are spreading out. This kind of lens is called diverging lens. When light passes through a lens, refraction occurs at the two lens surfaces.

Using Snell’s law and geometry, you can predict the paths of rays passing through lenses. To simplify such problems, assume that all refraction occurs on a plane, called the principal plane that passes through the center of the lens. This approximation, called the thin lens model, applies to all the lenses.

5.2. Refraction of light through lenses

Activity 5.2
i. Hold a hand lens about 2 m from the window. Look through the lens.
(CAUTION: Do not look at the sun).
What do you see?
ii. Move the lens farther away from your eye.
What changes do you notice?
iii. Now, hold the lens between the window and a white sheet of paper,
but closer to paper.
iv. Slowly move the lens away from the paper towards the window. Keep
watching the paper.

What do you see? What happens as you move the paper?

Do you see that an inverted image of trees outside is formed on the
paper? How do you think the image is formed?

Lenses can be thought of as a series of tiny refracting prisms, each of
which refracts light to produce an image. These prisms are near each other (truncated) and when they act together, they produce a bright image focused at a point.

5.2.1. Ray diagrams and properties of images formed by lenses

Notice that an image cannot be seen on the screen irrespective of the
position of the object. The nature of the image formed by a convex lens depends on the position of the object along the principal axis of the lens.

The principal focus of a lens plays an important part in the formation of
an image by a lens since parallel rays from the object converge to it, and
thus, we consider points F and 2F when describing the nature of the images formed by the lens.

These images can be larger or smaller than the object or same size as the
object. When an image is larger than the object, we say that it is magnified and when it is smaller, we say that it is diminished.

Images which can be formed on the screen are Real images. Because light rays pass through these images, real images can be formed on the screen. All real images formed by the convex lens are inverted.

To determine an image point, we need to consider only the three rays
indicated if Fig 4, which uses an arrow (on the left) as the object, and a
converging lens forming an image to the right. These rays, emanating from a single point on the object, are drawn as if the lens were infinitely thin, and we show only a single sharp bend at the centre line of the lens instead of refraction at each surface. These three rays are drawn as follows:

The point where these three rays cross is the image point for that object
point. Actually, any two of these rays will suffice to locate the image point, but drawing the third ray can serve as a check.

5.2.2. Graphical determination of focal length of lenses
If the lens is biconvex or plano-convex, a collimated parallel or diverging
beam of light travelling through the lens will be converged to a spot behind the lens. In this case, the lens is called a converging lens.

When an object is placed at a distance greater than the focal length from the lens, the image is real and inverted on the other side of the lens. When an object is placed at a distance equal to twice the focal length from the lens, the image is the same size as the object and inverted.

A convex lens forms a virtual image that is upright and larger compared to the object when the object is located between the lens and the focal point.

Concave lenses produce only virtual images that are upright and smaller
compared to their objects.

When an object is between F and the lens, there is no image formed on
the screen. The image formed is not real and is only seen by removing the screen and placing an eye in its position. We say that it is a virtual image. For a virtual image, rays appear to come from its position.

Unlike for a convex lens where the nature of the image depends on the
position of the object, a concave lens gives only an upright, small, virtual
image, and is situated between the principal focus and the lens for all
positions of the object.

Application activity 5.2
1. Design an experiment to study images formed by convex lenses of
various focal lengths. How does the focal length affect the position
and size of the image produced?

2. An object of length 5 cm is placed at a distance 25 cm in front of a
lens of focal length 10 cm.

Use a ray diagram to construct the image of this object and state its
properties if the lens is:
a). Converging or convex;
b). Diverging or concave.

5.3. The thin lens equations
Activity 5.3
Task 1:
i. Place a converging lens on a table while facing a window.
ii. Place a white screen behind the lens. Move the screen to and fro
(forwards and backwards) until a sharp image of a distant object
is seen on the screen.
iii. Discuss and write down the observation in your notebook.
iv. Measure the distance from the lens to the screen. What is this
distance called?

Task 2:
v. Draw a ray diagram to determine the nature and position of the
image of an object placed 10cm from a diverging lens of focal
length 15cm.
vi. Using the above information, find the nature and position of the
image using a lens formula. (Assign f a negative sign during your
substitution).
What is the location of the image?

5.3.1. Convex lens
We now derive an equation that relates the image distance to the object
distance and focal length of a thin lens. This equation will make the
determination of image position quicker and more accurate than doing ray tracing. Let  be the object distance, the distance of the object from the center of the lens, and  the distance of the image from the center of the lens. And let  and  refer to the heights of object and image. Consider two rays shown in Figure below for a converging lens, assumed to be very thin.

Will be valid for both converging and diverging lenses, and for all situations, if we use the following sign conventions:
1. The focal length is positive for converging lenses and negative for
diverging lenses.
2. The object distance is positive if the object is on the side of the lens
from which light is coming; otherwise, it is negative.
3. The image distance is positive if the image is on the opposite side
from where light is coming; if it is on the same side, it is negative.
Equivalently, the image distance is positive for real image and
negative for a virtual image.
4. The height of the image, is positive if the image is upright, and
negative if the image is inverted relative to the object. (ho is always
taken as positive).

5.3.3. Magnification
The magnification, m, of a lens is defined as the ratio of the image height
to object height. From the above figures and the sign conventions, we have

for an upright image the magnification is positive, and for an inverted image the magnification is negative.

5.3.4. The power of lenses
Whenever a ray of light passes through a lens it bends except when it passes through the optical centre. The degree of convergence or divergence of a lens is expressed as power. A lens of short focal length deviates the rays more while a lens of large focal length deviates the rays less. Thus the power of a lens is defined as the reciprocal of its focal length.

From sign convention 1, it follows that the power of converging lens, in
diopters, is positive, whereas the power of a diverging lens is negative. A
converging lens is sometimes referred to as positive lens, and a diverging
lens as a negative lens.
In case the lenses are combined

Example 1
An object is placed 32.0 cm from a convex lens that has a focal length of 8.0 cm.
a). Where is the image?
b). If the object is 3.0 cm high, how tall is the image?
c). What is the orientation of the image?

Solution
1. Analyse and sketch the problem
• Sketch the situation, locating the object and the lens.
• Draw the two principal rays

5.3.5. Lens maker formula
The following assumptions are made for the derivation:
• The lens is thin, so that distances measured from the poles of its
surfaces can be taken as equal to the distances from the optical centre
of the lens.
• The aperture of the lens is small.
• Object point is considered.
• Incident and refracted rays make small angles.

This equation gives the lens maker’s formula
Note:
• The lens maker’s formula indicates that a convex lens can behave like
a diverging one if i t n > n i.e., if the lens is placed in a medium whose
index is greater than the index of lens. Similarly a concave lens can be
made convergent.
• The lens maker’s formula can be derived for a concave lens in the
same way.
Sign convention states that real is positive while virtual is negative. This
should be put under consideration when one is using the lens formula to
solve problems.

Application activity 5.3
1. Compute the composition and focal length of the converging lens
which will project the image lamp, magnified 4 times, upon a screen
10.0 m from the lamp.
2. A lens has a convex surface of radius 20 cm and a concave surface of
radius 40 cm and is made of glass of refractive index 1.54. Compute
the focal length and the power of the lens, and state whether it is a
converging or a diverging lens.

5.4. Defects of lenses, their corrections and applications of lenses
Activity 5.4
i. Place a white sheet of paper on a horizontal ground.
ii. Hold a glass ruler above the paper so as to focus rays from the
sun on to the paper.
iii. Observe carefully the image formed on the sheet of paper.
iv. Repeat the above with the convex lens. What have you observed?

Aberration
A lens made of a uniform glass with spherical surface cannot form a perfect image. The spherical aberration is prominent image defect for a point source on the optical axis of such a lens. It arises because all rays through the lens are not focused to a common point.

The dependence of index of refraction of wavelength also causes the focal length, and thereby the image position, to depend on the color of the light. All simple lenses suffer from such chromatic aberration.

There are several different types of aberration which can cause the image to be an imperfect replica of the object (spherical aberration, chromatic aberration, coma, field curvature, barrel, pincushion distortion, astigmatism, etc).

Note that the image has coloured patches. This defect where by an image formed has coloured patches is called chromatic aberration. There are two kinds of defects; spherical aberration and chromatic aberration.

5.4.1. Spherical aberration
This arises in lenses of larger aperture when a wide beam of light incident on the lens, not all rays is brought to one focus.

As a result, the image of the object becomes distorted. The defect is due to the fact that the focal length of the lens for rays far from the principal axis are less than for rays closer to a property of a spherical surface and as a result, they converge to a point closer to the lens.

This defect can be minimized (reduced) by surrounding the lens with an
aperture disc having a hole in the middle so that rays fall on the lens at a
point closer to its principal axis. However, this reduces the brightness of the image since it reduces the amount of light energy passing through the lens.

5.4.2. Chromatic aberration
Chromatic aberration is caused by the dispersion of the lens material. Since, from the lens formulae, f is dependent upon n, it follows that different colours of light will be focused to different positions.

Chromatic aberration can be minimised by using an achromatic doublet
(or achromat) in which two materials with differing dispersion are bonded together to form a single lens.

Application activity 5.4
i. How spherical and chromatic aberration can be reduced?
ii. If a plano-convex lens is used as objective lens in a telescope, how
is its convex surface faced to minimize spherical aberration? Explain

5.5. Refraction through prisms

Activity 5.5
1. Have you ever heard of a prism? How does it look like?
2. Observe clearly the shapes of the glasses provided below and identify
those with the shape of a prism.
Explain your choices

3. With the help of a teacher, touch, observe and identify the real shape
of the prism.
4. Examine the features of the one selected as a prism.

5.5.1. Terms associated with refraction through prism
In optics, a prism is transparent material like glass or plastic that refracts light. At least two of the flat surfaces must have an angle less than
between them. The exact angle between the surfaces depends on the application.

the edges and so bring the parallel rays to a focus.

The truncated prisms of the diverging lens point the opposite way to those of the converging lens, and so a divergent beam is obtained when parallel rays are refracted by this lens because the deviation of the light is in the opposite direction.

The middle part of the lens acts like a rectangular piece of glass and a ray
incident to it strikes it normally, and thus passes un-deviated.

Dispersion of light by a prism
Dispersion of light is the separation of a beam of light into its constituent
colours. This takes place when a light beam passes through a dispersive
medium. A beam of white light incident on a prism splits into its constituent colours to form “a visible spectrum”consisted of colours violet, indigo, blue, green, yellow, orange and red.

When a polychromatic light is incident on the first surface of the prism, each constituent colour gets refracted through a different angle. When these colours are incident on the second surface of the prism they are again refracted further.

Application activity 5.5.1
1. (i) Place a prism in the centre of a piece of paper so that its refracting
surface is directly facing the windows in order to receive light from
the sun.
(ii) Place a white screen on the far side of the prism so that the
refracted rays hit it.
(iii) Observe what is formed on the screen.
(iv) In brief, write in your notebook the observation.
2. Do a research to find other terms associated to the refraction light
through prism.

5.5.2. Deviation by a prism and determination of refractive index

Activity 5.5.2
1. Have you ever heard of the word deviation? List down in your notebook at least two ways in which light can be deviated.
2. You are provided with a glass prism of refracting angle , four
optical pins, a white sheet of paper, a soft board and fixing pins.

i. Place a prism on a white sheet of paper and mark its outline ABC.

1. Deviation of light by a glass prism
Light can be deviated by reflection and refraction. Since a prism refracts
light, it therefore changes its direction.

A prism deviates light on both faces. These deviations do not cancel out as in a parallel sided block where the emergent ray, although displaced, is parallel to the incident ray surface. The total deviation of a ray due to refraction at both faces of the prism is the sum of the deviation of the ray due to refraction at the first surface and its deviation at the second face.

N.B: The deviation D for a small angle prism is D = A (n-1)
The expression D = A (n – 1) shows that for a given angle A, all rays entering a small angle prism at small angles of incidence suffer the same deviation.

2. Angle of minimum deviation
From the variation figure below, there is one angle of incidence which gives a minimum deviation. The experiment shows that this minimum deviation occurs when the angle of emergence is exactly equal to the angle of incidence and the two internal angles of refraction are equal. At this value, a ray passes symmetrically through the prism and the ray inside the prism is perpendicular to the directing plane.

And (The refracting angle); so,     

4. Applications of total internal reflection of light by a prism
Many optical instruments use right -angled prisms to reflect a beam of light through 90° or 180° (By total internal reflection) such as cameras, binoculars, periscope and telescope. One of the angles of right angled prism is 90°. When a ray of light strikes a face of prism perpendicular, it enters the prim without deviation and strikes the hypotenuse at an angle of 45°. Since the angle of incidence 45° is greater than critical angle of the glass which is 42°, the light is totally reflected by the prism through an angle of 90°. Two such prisms are used in periscope. The light is totally reflected by the prism by an angle of 180°. Two such prisms are used in binoculars.

Application activity 5.5.2
1. A ray of light is refracted through a 60 degree prism of ordinary glass
making an incident of 35 degree with the normal i.e incident ray
=35°.What will the angle of emergent ray and the angle of deviation
measure?
2. What are the conditions for minimum deviation when a ray of light
passes through a prism?
3. Show that the deviation produced by a small angled prism is
independent with angle if incidence.

Skills lab 5
Determination of the focal length of a lens
Apparatus required:
1 torch bulb fitted in a bulb holder
1 switch
3 Torch cells
2 Cell holders
Connecting wires about 50cm long.
Wire gauze
Converging lens, of focal length 15cm.
Lens holder,
White screen
Metter rule.
Retort stand with its holding accessories.

Instructions
In this experiment you will determine the focal length of the converging lens provided.
a). Mount the lens on the holder and place it facing a window
b). Place the screen behind the lens and adjust the screen until a clear
image of a distant object is obtained
c). Measure and record the distance, x, between the lens and the screen.

End unit assessment 5
1. How does a convex lens work? Explain your reason
2. What will happen when a light ray travelling in glass is incident on an
air surface? Explain your answer
3. What will happen When a light ray makes an angle of 900
 while entering a glass slab? Explain your reason
4. What are factors responsible for the angle of deviation through prism
depends?
5. A beam of monochromatic light is incident at i = 50° on one face of an
equilateral prism, the angle of emergence is 40°.Calculate the angle
of minimum deviation.
6. A 600 prism has a refractive index of 1.5. Calculate the angle of
incidence for minimum deviation, the angle of emergence of light at
maximum deviation and angle of maximum deviation.