Light waves
Refraction of Light through Lenses
Subject: Physics
Theme: Waves-Motion Without Material Transfer
Topic: Lenses
Sub Topic: Refraction of Light through Lenses
Date: dd/mm/yyyy
Class: S.S.S 2
Duration: 40 Minutes
No of Learners: 40
Learning Objectives: By the end of the lesson learners should be able to:
1. Defined a lens
A lens is any transparent material with two faces, of which at least one is curved. It causes a beam of Ire passing through it to either converge or diverge.2. List and explain different types of lens
(i) Converging or convex lens and(ii) Diverging or concave lens.
(i) A converging (convex) lens makes a parallel beam of light to converge to a point. Such lenses thicker at the centre than at the edges.
(ii) A diverging (concave) lens makes a parallel beam of light to appear to diverge from a point. Such lenses are thinner at the centre than at the edges.
3. List and Explain Important Parameters (Technical terms) of a Lens
The following are technical terms of a lens: optical centre, principal axis, principal foci (F & F'), focal length and the power of the lens.The principal focus of a converging lens is real.
The principal focus of a diverging lens is virtual.
Because of the principle of reversibility of light, rays originating from F and passing through the convex lens will emerge parallel to the principal axis.
(i) The optical centre (C) of a lens is the point through which rays of light pass without being deviated by the lens. The line passing through the optical centre of the lens and joining the centres of curvature of its surfaces is termed the principal axis.
(ii) The principal focus (F) of a converging or convex lens is the point to which all rays parallel and close to the principal axis converge after refraction through the lens.
(iii) The principal focus of a diverging or concave lens is the point from which all rays parallel and close to the principal axis appear to diverge after refraction through the lens.
A lens has two principal foci since light may pass through a lens in either direction.
(iv) The focal leagth (l) is the distance between the optical centre and the principal focus of the lens.
(v) The power (P) of a lens is equal to the reciprocal of the focal length and is measured in diopres when f is in metres.
P = 1/f
4. Explain construction of ray diagrams
Three particular classes of rays are used in geometrical constructions of locate the image formed by converging lens:1. A ray from the object, parallel to the principal axis which refracts through the lens and passes through the principal focus.
2. A ray from the object passing through the optical centre of the lens which passes through undeviated. The intersection of the two rays corresponds to the position of the top of the image of the object.
3• Sometimes, a third ray may be used. A ray from the object through the principal focus which emerges parallel to the principal x after refraction. The three emergent rays s intersect at a point, the image position.
Example
An object 10mm tall stands vertically on the principal axis of a converginglens of focal length of 10mm, and at a distance of 17mm from the lens. By graphical construction find the position, size and nature of the image.
SOLUTION
Let;
Principal axis = The horizontal straight line OCI
Lens = LCL'
Object = OA (10 units long, 17 units from the lens)
Focal lenght = CF (10 units)
5. Explain formation of images by a converging or convex lens
(i) Behind the object
(ii) Virtual
(iii) Erect
(iv) Longer than object
(i) Beyond 2F
(ii) Real
(iii) Inverted
(iv) Longer than object
(i) At 2F
(ii) Real
(iii) Inverted
(iv) The same size as object
(i) Between F and 2F
(ii) Real
(iii) Inverted
(iv) Smaller than object
(i) At F
(ii) Real
(iii) Inverted
(iv) Smaller than object
6. Explain formation of images by a diverging or concave lens
Images formed by Diverging or Concave Lens: For all positions of the object, the image is virtual, erect and diminished.7. Explain The Lens Formula and Sign Convention
Experiments show that there is a relation between, the object distance, u, the image distance, v. and the focal length, of a lens. The relation is given by two Sign Convention of the lens formula.(1.) 1/v - 1/u = 1/f (New Cartesian Sign Convention)
All distances are measured from the optical centre, being positive if in the same direction as the incident light and negative if against it.
The formula may be applied to both types of lenses, diverging or concave lens and converging or convex lens.
The focal length of a converging lens is +ve;
a diverging lens -ve
This authomatically brings all signs into agrements with the ordinary Cartesian graphical convention used in mathematics, hence its name.
(2.) 1/v + 1/u = 1/f (Real-is-Positive Sign Convension)
(i) All distances are measured from th optical centre.
(ii) Distances of real objects and real images are positive.
(iii) Distance of virtual objects and images are negative
(iv) The focal length of a converging lens is +ve (real focus).
(v) The focal length of a diverging lens is -ve (virtual focus).
MAGNIFICATION (M) is defined as,
M = Image height / Object height = distance of image from lens / distance of object from lens
M = v/u
8. Worked calculations on using The Lens Formula and Sign Convention
EXAMPLE 1.An object is placed 20cm from a converging lens of focal length 15cm. Find the nature, position and magnification of the image.
SOLUTION
NEW CARTESIAN
u = -20cm (object being left of lens)
f = +15cm (converging lens)
1/v - 1/u = 1/f
1/v - (-1/20) = 1/15
1/v = 1/15 - 1/20
1/v = (4-3)/60
1/v = 1/60
v = 60cm (on right, real)
Also m = image distance / object distance
m = 60/20
m = 3
:. A real image is formed 60cm from lens on side opposite to object, of magnification 3
Alternatively;
REAL-IS-POSITIVE
u = +20cm (real object)
f = +15cm (converging lens)
1/v + 1/u = 1/f
1/v + 1/20 = 1/15
1/v = 1/15 - 1/20
1/v = (4-3)/60
1/v = 1/60
v = 60cm (real image)
Also m = image distance / object distance
m = 60/20
m = 3
:. A real image is formed 60cm from lens on side opposite to object, of magnification 3
EXAMPLE 2.
An object is placed 5cm from a converging lens of focal length 15cm. Find the nature, position and magnification of the image.
SOLUTION
NEW CARTESIAN
u = -5cm (object being left of lens)
f = +15cm (converging lens)
1/v - 1/u = 1/f
1/v - (-1/5) = 1/15
1/v = 1/15 - 1/5
1/v = (1-3)/15
1/v = -2/15
v = -7.5cm (on left, virtual)
Also m = image distance / object distance
m = 7.5/5
m = 1.5
:. A virtual image is formed 7.5cm from lens on the same side as object, of magnification 1.5
Alternatively;
REAL-IS-POSITIVE
u = 5cm (real object)
f = +15cm (converging lens)
1/v + 1/u = 1/f
1/v + 1/5 = 1/15
1/v = 1/15 - 1/5
1/v = (1-3)/15
1/v = -2/15
v = -7.5cm (virtual image)
Also m = image distance / object distance
m = 7.5/5
m = 1.5
:. A virtual image is formed 7.5cm from lens on the same side as object, of magnification 1.5
EXAMPLE 3.
A four times magnified virtual image is formed of an object placed 12cm from a converging lens. Calculate the position of the image and the focal length of the lens.
SOLUTION
Magnification (m) = image distace / object distace
4 = image distance / 12
image distance = 48cm
NEW CARTESIAN
u = -12cm (object being left of lens)
u = -48cm (virtual, on left)
1/f = 1/v - 1/u
1/f = (-1/48) - (-1/12)
1/f = (-1/48) + 1/12
1/f = (-1+4)/48
1/f = 3/48
1/f = 1/16cm
f = 16cm
:. Image distance = 48cm from lens on the same side as object.
Focal length of lens = 16cm
Alternatively;
REAL-IS-POSITIVE
u = +12cm (real object)
u = -48cm (virtual image)
1/f = 1/v + 1/u
1/f = (-1/48) + 1/12
1/f = (-1+4)/48
1/f = 3/48
1/f = 1/16cm
f = 16cm
:. Image distance = 48cm from lens on the same side as object.
Focal length of lens = 16cm
EXAMPLE 4.
Find the nature and position of the image of an object placed 10cm from a diverging lens of focal length 15cm.
SOLUTION
NEW CARTESIAN
u = -10cm (object being left of lens)
u = -15cm (diverging)
1/v - 1/u = 1/f
1/v - (-1/10) = (-1/15)
1/v = (-1/15) - 1/10
1/v = (-2-3)/30
1/v = -5/30
1/v = -1/6cm
v = -6cm
:. A virtual image is formed 6cm from lens on the same side as object.
Alternatively;
REAL-IS-POSITIVE
u = +10cm (real object)
u = -15cm (diverging)
1/v + 1/u = 1/f
1/v + 1/10 = (-1/15)
1/v = (-1/15) - 1/10
1/v = (-2-3)/30
1/v = -5/30
1/v = -1/6cm
v = -6cm
:. A virtual image is formed 6cm from lens on the same side as object.
Rationale:
Formation of images by lenses: Construction of ray diagrams light rays incident on a converging lens converge to form a real image after passing through a converging or convex lens. For a diverging lens, the rays from an object diverge to form a virtual image of the object. If we know the distance of the object from the lens and the focal length of the lens we can obtain the position and nature of the image by the construction of ray diagrams as outlined below.Prerequisite/ Previous knowledge:
Dispersion of white lightLearning Materials:
lenses, graph, mirror, etcReference Materials:
1. New School Physic for Senior Secondary Schools, by M. W. Anyakowa.2. Lamlas's SSCE and UTME, by O. Ajaja and H. B. Olaniyi.
3. Okeke P.N and etal (2011), Macminian, Senior Secondary School Physics, New Edition.
4. Farinde E.O and etal (2015), Essential Physics. 6th Edition,
5. New System Physics by Dr. Charles Chow.
Lesson Development:
| STAGE | TEACHER'S ACTIVITY | LEARNER'S ACTIVITY | LEARNING POINTS |
|---|---|---|---|
| STEP I: PREVIOUS KNOWLEDGE full class session |
The teacher asks learners questions based on previous knowledge. 1. What is Dispersion of light? 2. What are Primary and Secondary colours? |
Learners' response to teacher's question. 1. Dispersion of light is the separation of white light into its component colours of red, orange, yellow, green, blue, indigo and violet. It is due to the fact that the different colours of white light travel at different speeds through the glass. 2. Primary colours are red ,green and blue. Secondary colours are yellow, cyan and magenta. All the three primary colours combined to produce white. |
Confirming previous knowledge. |
| STEP II: INTRODUCTION full class session |
The teacher introduces the topic for discussion. Images formed by a concave or diverging lens are always virtual, erect and diminished for all positions of the object. Images formed by a convex or converging lens may either be real or virtual depending on the position of the object. |
Learners listen to teacher, ask questions and make contribution. | Developing the idea of the topic, reflection of light through lenses. |
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| STEP III: EXPLORATION Grouping |
The teacher presents the instructional material to the students and divides them into four groups. The representative of a group is asked to discuss the various types of lenses. | Learners' expected answer (i) Converging or convex lens and (ii) Diverging or concave lens. |
Types of lenses. |
| STEP IV: DISCUSSION Laws of Refraction |
The teacher guides the students as follow: Three particular classes of rays are used in geometrical constructions of locate the image formed by converging lens: (i) Rays parallel to the principal axis which pass through the principal focus after refraction through the lens. (ii) Rays through the principal focus which emerge parallel to the principal axis after refraction through the lens. (iii) Rays through the optical centre which are undeviated. two of these rays only are sufficient to lacate an image, and which particular pair is chosen is merely a matter of convenience. |
The students listen, contribute and ask question | geometrical constructions of locate the image formed by converging lens. |
| STEP V: APPLICATION | The teacher guides learners to calculate the position, size and nature of image of an object placed in front of lens. | The students calculate the position, size and nature of image of an object placed in front of lens. 1. An object 5cm high is placed at a distance 12cm from a convex lens of focal length 8cm. Calculate the position, size and nature of the image. Solution Using the real-is-positive convention f = +8cm u = +12cm 1/v + 1/u = 1/f 1/v = 1/f - 1/u 1/v = 1/8 -1/12 1/v = 1/24 v = 24cm (eal image) m = v/u m = 24/12 m = 2 Alternatively Using the New Cartesian convention f = +8cm u = -12cm (object left of lens) 1/v - 1/u = 1/f 1/v = 1/f + 1/u 1/v = 1/8 + (-1/12) 1/v = 1/8 - 1/12 1/v = 1/24 v = 24cm (on the right, :. real image) m = image height / object height 2 = h/5 h = 2 x 5 h = 10cm :. The image is real, inverted, magnified and at a distance of 24cm from the lens on the side opposite to object. The height of the image is 10cm. |
Calculate the position, size and nature of the image |
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| STEP VI: EVALUATION | The teacher asks the learners questions. 1. An object placed 5 cm in front of a converging lens of focal length 10 cm produces a magnified image 8 cm high. Find the size and the position of the image. 2.Find the nature and position of the image of an object placed 6 cm from a diverging lens of focal length 15 cm. 3. A converging lens produces a four times magnified and upright image of an object placed in front of it. If the focal length of the lens is 11 cm, calculate the object distance. |
The learners expected answers. (1.) Using the real-is-positive convention 1/v + 1/u = 1/f 1/v + 1/5 = 1/10 1/v = 1/10 - 1/5 1/v = (1 - 2)/10 1/v = -1/10 v = -10cm (The image is virtual) m = v/u = height of image / height of object 10/5 = height of image / height ofobject 2 = 8/height of object h = 8/2 h = 4cm (2.) u = 6cm f = -15cm 1/v + 1/u = 1/f 1/v = (-1/6) - 1/15 v = - 30/7 v = -4.28cm :. The image is virtual and 4.28cm from the lens on the same side as the object 3. m = v/u = 4 v = 4u An upright image is virtual. Hence image distance, v is -ve. 1/v + 1/u = 1/f (-1/v) + 1/u = 1/10 (-1/4u) + 1/u = 1/10 (-1 + 4)/4u = 1/10 3/4u = 1/10 4u = 3 x 10 u = 30/4 u = 7.5cm |
Ask the learners questions to assess the achievement of the set objectives. |
| CONCLUSION | The teacher concluded the lesson with the students. To find the position and nature of the image graphically, do the following: (1) Choose a suitable scale for the construction. (2) Draw a straight vertical line MN to represent the position of the lens and mark the position of the focus F according to the scale. C represents the focal length. (3) Draw an arrowcd line OA perpendicular to OC to represent the object so that OC represents the distance of the object from the lens and OA its height according to the scale. (4) From the top A of the object, draw a ray AP parallel to the principal axis. AP is refracted through F'. (5) Draw another ray AC through the centre of the lens C. This ray passes straight through without changing direction. The point of intersection B of the two emerging rays PB and CB represents the top of the image. (6) Complete the image by drawing B1 normal to the principal axis. |
The students listen and make contribute | To find the position and nature of the image graphically. |
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| ASSIGNMENT | The teacher gives learners a take home. | 1. Defined principal focus, focal leight of a converging lens, and explain the meaning of real image and virtual image. 2. A small object is placed 6 cm away from a converging lens of focal lenght 10 cm. By means of a carefully drawn scale diagram find the nature, position, and magnification of the image. 3. A small object is placed on the axis of a coverging lens of focal length 10 cm so that it is (a) 15 cm and (b) 5 cm from the lens. Show by ray diagrams the nature and positions of the images formed and either from your drawing or by calculation find their exact positions. 4. Where must an object be placed so that the image formed by a converging lens will be, (i) at infinity (ii) the same size as the object (iii) erect (iv) inverted abd enlarged, and (v) as near the object as possible? |
Improving their level of understanding. |