### Refraction of Light at Curved Surfaces

I. Conceptual Understanding
1. A man wants to get the picture of a zebra. He photographed a white donkey after fitting a glass with black stripes on to the lens of his camera. What photo will he get? Explain.     (AS - 1) (2 Marks)
A: * The man gets the picture of a zebra which has dark and white stripes on the body of the white donkey.
* The reflected light rays from the white donkey enter the camera and the black stripes fitted on to the lens do not allow this light to pass through to the screen.
* As such on the screen the image is formed with dark and white stripes on the body of the donkey giving it the appearence of a zebra.

2. Two converging lenses are to be placed in the path of parallel rays so that the rays remain parallel after passing through both lenses. How should the lenses be arranged? Explain with a neat ray diagram. (AS - 1) (4 Marks)
A: * The two lenses L1 and L2 are to be arranged as shown in the figure.
* L1 and L2 are the two convergent lenses and P1 and P2 are the optic centres of the lenses respectively.
* P1F = f1 is the focal length of the lens L1 and P2F = f2 is the focal length of the lens L2.
* When the two lenses are kept at a distance of (f1 + f2) from each other than the parallel rays passing through the two lenses remain parallel after passing through them.

3. The focal length of a converging lens is 20 cm. An object is 60 cm from the lens. Where will the image be formed and what kind of image is it? (AS - 1) (2 Marks)
A: Given: Focal length f = 20 cm, object distance u = 60 cm, image distance v = ? ... Image distance v = 30 cm
* A real, diminished, inverted image is formed at a distance of 30 cm from the lens.

4. A double convex lens has two surfaces of equal radii 'R' and refractive index n = 1.5. Find the focal length 'f'. (AS - 1) (2 Marks)
A: Given: R1 = R2 = R, Refractive index n = 1.5 focal length f = ? * The focal length of the double convex lens is equal to its radius of curvature.

5. Write the lens makers formula and explain the terms in it. (AS - 1) (2 Marks)
A: Lens makers formula: * Here f is the focal length of the lens, n is the refractive index of the material of the lens.
* R1 and R2 are the radii of curvature of the two surfaces of the lens.

6. How do you verify experimentally that the focal length of a convex lens is increased when it is kept in water?   (AS - 1) (4 Marks)
A: * Take a convex lens and find its focal length. Let it be 'f'.
* Take a cylindrical vessel such as a glass tumbler, whose height must be nearly four times of the focal length of the lens.
* Keep a black stone inside the vessel at its bottom.
* Now pour water into the vessel upto a height such that the height of the water level from the top of the stone is greater than focal length of the lens.
* Now dip the lens horizontally using a circular lens holder as shown in the figure.
* Set the distance between stone and lens that is equal to or less than focal length of lens.
* Now look at the stone through the lens.
* We can see the image of the stone if the distance between lens and stone is less than the focal length of the lens.
* Now increase the distance between lens and stone until you cannot see the image of the stone.
* We have dipped the lens to a certain height which is greater than the focal length of lens in air but we can see the image. This shows that the focal length of lens has increased in water.

7. How do you find the focal length of a lens experimentally?   (AS - 1) (4 Marks)
A: * Place a V-stand at the middle on a long table.
* Place a convex lens on the V-stand. Imagine the principal axis of the lens.
* Take a candle and light it. Take the candle far away from the lens along the principal axis.
* Adjust the screen which is on the other side of the lens until we get an image on it.
* Measure the distance of the image from the V-stand of lens and measure the distance of the candle from the stand of the lens.
* Record the values in the table below. Put the object at different distances like 50 cm, 40 cm, 30 cm etc. and measure the image distances in all cases by adjusting the screen. Note all these values in the table.

 S.No. Object distance u image distance v Focal length 'f'

The focal length 'f' of the convex lens is found out using the formula: . The average of all the values gives the focal length of the convex lens.

II. Asking questions and making hypothesis

8. Harsha tells Siddhu that the double convex lens behaves like a convergent lens. But Siddhu knows that Harsha's assertion is wrong and corrected Harsha by asking some questions. What are the questions asked by Siddhu?    (AS - 2) (4 Marks)
A: * Did you observe the course of rays of an object when it is placed between the pole and focus of the lens?
* Do you know how does an air bubble behave in water?
* What is the impact on the behaviour of the lens when put in a medium whose refractive index is more than the refractive index of the material of the lens?
* What happens to the focal length of the lens when it is put in water?
* Is it possible to convert a convergent lens into a divergent lens?

9. Assertion (A): A person standing on the land appears taller than his actual height to a fish inside a pond.
Reason: (R): Light bends away from the normal as it enters 'air' from water.
Which of the following is correct explain?     (AS - 2) (2 Marks)
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true and R is not the correct explanation of A.
c) A is true but R is false.
d) Both A and R are false.
e) A is false but R is true.
A: Explanation: c is correct.
* Light from the rarer medium when enters the denser medium it bends towards the normal.
* As such the person standing on the land appears taller than his height to a fish inside the pond.

10. A convex lens is made up of three different materials as shown in the figure. How many of the images does it form?     (AS - 2) (2 Marks) A: * As the lens is made up of three different materials they will have three different refractive indicies.
* So this type of lens forms three different images of the given object.

11. Can a virtual image be photographed by a camera?    (AS - 2) (1 Mark)
A: * We can photograph the image formed in a plane mirror, which is a virtual image.
* So it is possible to photograph a virtual image. Just as our eye can see the virtual image. The camera can also record the virtual image.

III. Experimentation and Field Investigation

12. You have a lens. Suggest an experiment to find out the focal length of the lens. (AS - 3) (4 Marks)
A: * Place a V-stand at the middle on a long table.
* Place a convex lens on the V-stand, imagine the principal axis of the lens.
* Take a candle and light it. Take the candle far away from the lens along with principal axis.
* Adjust the screen which is on the other side of the lens until we get an image on it.
* Measure the distance of the image from the V-stand of lens. And measure the distance of the candle from the stand of the lens.
* Record the values in the table below. Put the object at different distances like 50 cm, 40 cm, 30 cm etc and measure the image distance in all cases by adjusting the screen. Note all these values in the table.

 S.No. Object distance u image distance v Focal length 'f'

The focal length 'f' of the convex lens is found out using the formula: . The average of all the values gives the focal length of the convex lens.

13. Let us assume a system that consists of two lenses with focal lengths f1 and f2 respectively. How do you find the focal length of the system experimentally, when
i) two lenses are touching each other
ii) they are separated by a distance 'd' with common principal axis.  (AS - 3) (4 Marks)
A:
i) Let f1 and f2 be the focal lengths of the two thin convex lenses L1 and L2 respectively. * These two lenses are put in contact.
* Let the object O be placed beyond the focus of the first lens L1 at a distance u from it. The first lens L1 produces an image I1 of this object at a distance v1.
* This serves as a virtual object for the second lens producing the final image I at a distance v.
* For the image formed by the first lens L1 * For the image formed by the second lens L2. * Adding equations (1) and (2). * If the two lenses are regarded as equivalent to a single lens of focal length f then  Q) Combined focal length of two thin lenses separated by a distance 'd'
A: * L1 and L2 are two convergent lenses separated by a distance 'd'. * Let f1 and f2 be the respective focal lengths of the lenses L1 and L2.
* The lens L1 forms the image I1 of the object O as shown in the figure.
* The image I1, serves as an object for the lens L2 and it forms the final image at I.
* The combined focal length F of these two thin lenses separated by a distance 'd' is given by the equation: IV. Information Skills and Projects

14. Collect the information about the lenses available in an optical shop. Find out how the focal length of a locus may be determined by the given 'Power' of the lens.      (AS - 4) (4 Marks)
A: * In an optical shop the following lenses are available.

 a) Plano convex lens b) Plano concave lens c) Double convex lens d) Double concave lens e) Convexo Concave lens f) Concavo convexo lens g) Achromatic Lenses h) Aspheric Lenses i) Cylindrical Lenses j) Atoric Lenses k) UV Lenses l) IR lenses

Determination of the focal length:
1) We know the power of a lens P = where f is in centimeter.
2) So the focal length of the lens f = cm.
3) If P is positive then the lens is biconvex lens and if P is negative then the lens is biconcave lens.

15. Collect the information about lenses used by Galileo in his telescope.    (AS - 4) (2 Marks)
A: * Two lenses of different focal lengths were used by Galileo in his telescope.
* One of these lenses is a biconvex lens used as an objective lens.
* The second lens, a biconcave lens served as an eyepiece lens. This is nearer to the eye of the observer.
* The objective is focused towards the object. It has greater focal length compared to the focal length of the eye piece.
* The image of the distant object is formed at the focus of the objective. This image behaves as an object for the eye piece.
* The final image is formed as an erect and magnified image which is seen by the observer.
* The figure drawn shows the arrangement of the Lenses in the Galileo telescope.

V. Communication through drawing, Model making

16. Use the data obtained by activity - 2 in table 1 of this lesson and draw the graphs of u vs v and vs .   (AS - 5) (4 Marks)
A: a) Graph of u - v:
* The following data is obtained in activity - 2. The focal length of the given lens is found to be 20 cms. S.No. Object Distance (u) Image Distance (v) Focal Length 1) 25 cm 100 cm 20 cm 2) 30 cm 60 cm 20 cm 3) 40 cm 40 cm 20 cm 4) 60 cm 30 cm 20 cm

* The shape of u - v graph is shown in the figure. The shape of the graph is rectangular hyperbola.

b) Graph of - : S.No. u v  1) 25 cm 100 cm .04 .01 2) 30 cm 60 cm .033 .016 3) 40 cm 40 cm .025 .025 4) 60 cm 30 cm 0.016 .016

* If we draw a graph between and the graph is a straight line which touch the axis as shown in the figure.

17. Figure shows ray AB that has passed through a divergent Lens. Construct the path of the ray up to the lens if the position of the foci is known.  (AS - 5) (2 Marks)
A: * A ray parallel to the principal axis of a divergent lens after refraction through the lens appears to be coming from the focus of the lens. This is shown in fig (2).

18. Figure shows a point light source and its image produced by a lens with an optical axis N1N2. Find the position of the lens and its foci using a ray diagram.   (AS - 5) (2 Marks)
A: * The object 'O' is placed in between the focus and optic centre of the lens.
* The image formed by the lens is shown in the figure.
* The image formed is erect, virtual and magnified and on the same side of the object.

19. Find the focus by drawing a ray diagram using the position of source S and the image S' given in the figure below.   (AS - 5) (2 Marks) A: * As shown in the figure (2) when the object OS is placed in between F1 and C1 its image is formed beyond C2.
* The image formed is real, inverted and magnified.

20. A parallel beam of rays is incident on a convergent lens with a focal length of 40 cm where should a divergent lens with a focal length of 15 cm be placed for the beam of rays to remain parallel after passing through the two lenses? Draw a ray diagram.             (AS - 5) (4 Marks)
A: * A parallel beam of rays incident on a convergent lens after refraction pass through the focus of the lens.
* A beam of rays incident on a divergent lens after refraction pass parallel to the poincipal axis, if the rays seem to come from the focus of the divergent lens.
* Rays incident parallel to the principal axis of the convergent lens after refraction meet at the focus of the lens which is 40 cm away from the lens. See figure 1. * The divergent lens of focal length 15 cm is put at a distance of 15 cm from the focus of the convergent Lens towards the convergent lens, so that these rays meeting at the focus of the convergent lens incident on the divergent lens and diverge parallel to the principal axis as shown in figure. 2. * PF = 40 cm (Focal length of the convergent Lens)
P′F = 15 cm : (Focal length of the divergent lens)
PP′ = (40 _ 15) = 25 cm (Distance between the two lenses)
* To get the parallel beam of rays to emerge out as parallel beam of rays, the convergent and the divergent lenses should be placed at a distance of 25 cm. from each other as shown in figure - 2.

21. Draw ray diagrams for the following positions and explain the nature and position of image.
a) Object is placed at C2 of a Convergent lens.
b) Object is placed between F2 and optic centre P of a Convergent lens.  (AS - 5) (4 Marks) A: a) * When the object is placed at C2 of a convergent lens then the image is formed at C1 of the same lens on the other side.
* The image formed is real, inverted and is of the same size of the object.
Object OJ is placed at C2 and the image is formed at C1 as shown in figure.

b) Object is placed between F2 and optic centre P. * The object OJ is placed in between F2 and optic centre P.
* The image IG is formed behind the object. This image is magnified, virtual and erect.

VI: Appreciation and Aesthetic sense, values

22. How do you appreciate the coincidence of experimental facts with the results obtained by a ray diagram in terms of behaviour of images formed by Lenses.    (AS - 6) (4 Marks)
A: * Theoritically some rules are framed to obtain the image formed by a lens for an object.
* These rules are framed keeping in mind two principles.
a) Light travels in straight lines
b) Light selects the shortest path for its travel. (Fermat's Principle)
* Ray diagrams are drawn for the different positions of the object from the lens to find out the positions of the respective images.
* Surprisingly, the ray diagrams results are totally coincident with the experimental results.
* This proved that theoritical results exactly coincide with the experimental results which is to be appreciated.
* This indirectly give us the impression that theory and experiments go hand in hand which is the basis for the development of Science.

VII. Application to Daily Life, Concern to Biodiversity

23. Find the refractive index of glass which is a symmetrical convergent lens, if its focal length is equal to the radius of curvature of its surface.   (AS - 7) (4 Marks)
A: Given: Refractive index of glass (convex lens) n = ?
* Focal length f = R
* As the lens is a symmetrical one: R1 = R2 = R 24. Find the radii of curvature of a convexo - concave convergent lens made of glass with refractive index 1.5, having focal length of 24 cm. One of the radii of curvature is double the other.   (AS - 7) (4 Marks)
A: Given: Refractive index n = 1.5
Local Length f = 24 cm
Radius of curvature R1 = R, and R2 = 2R
Value of R = ? 25. The distance between two point sources of light is 24 cm. Where should a convergent lens with a focal length of f = 9 cm be placed between them to obtain the images of both sources at the same point?   (AS - 7) (4 Marks)
A: * The distance between the two sources is = 24 cm.
* The convergent len's focal length f = 9 cm. * Let the lens be placed at a distance of x cm from the source S, to obtain the images of both sources at the same point.
Now for the source: S1
a) Object distance u1 = -x
* Focal length f = 9 cm
* Image distance v1 = ? * For the Source S2:
a) * Object distance u2 = - (24 - x)
* Focal length f = 9 cm
* Image distance v2 = ?
b) As the Image I1 of the source S1 and image I2 of the source S2 coincide v2 = -v1 (or) 24x - x2 = 108
(or) x2 - 24x + 108 = 0 (i.e.) (x - 18)(x - 6) = 0
∴  x = 6 or 18
It means the convex lens may be placed at a distance of 6 cm from one source so that it will be at a distance of 18 cm from the other source.

26. Suppose you are inside the water in a swimming pool near an edge. A friend is standing on the edge. Do you find your friend taller or shorter than his usual height?       (AS - 7) (2 Marks)
A: * The friend seems to be taller than his actual height.
* This is because the light ray travels from a rarer medium (air) to denser medium gets refraction and bends towards the normal.
* This seems to be coming from a distance point for the swimmer. So the person appears taller than his usual height.

Questions and answers given in the Lesson

1. What happens to a ray that is incident on a curved interface separating the two media? Are the laws of refraction still valid?  (AS - 2) (1 Mark)
A: The incident ray gets deviated from its path.
* The laws of refraction are still valid.

2. How do rays bend when they are incident on a curved surface?       (AS - 1) (1 Mark)
A: A ray bends towards the normal when it travels from a rarer medium to a denser medium.
* A ray bends away from the normal when it travels from a denser medium to a rarer medium.

3. What happens to a ray that moves along the principal axis? Similarly a ray that travels through the centre of curvature.  (AS - 1) (1 Mark)
A: * According to Snell's law, the ray that moves along the normal drawn to the surface does not deviate.
* Hence the ray that moves along the principal axis and the ray that travels through the centre of curvature go undeviated.

4. You might have observed that a lemon in a glass of water appears bigger than its actual size, when viewed from the sides of tumbler.
a) How can you explain this change in the size of lemon?
b) Is the lemon that appears bigger in size an image of lemon or is it the real lemon?
c) Can you draw a ray diagram to explain this phenomenon? (AS - 3) (4 Marks)
A: a) Due to refraction of light when it travels from one medium to another medium. We notice a change in the size of the lemon.
b) The lemon that appears bigger in size is the image of the lemon. A: Lemon
B: Image of lemon
RS: Lemon visible
PQ: Surface of water

5. What happens to the light ray when a transparent material with two curved surfaces is placed in its path?  (AS - 1) (1 Mark)
A: The light ray undergoes refraction.

6. Have you heard about lenses? How a lens is formed?    (AS - 1) (2 Marks)
A: * I have heard about lenses.
* A lens is formed when a transparent material is bounded by two surfaces of which one (or) both surfaces are spherical.

7. How does a light ray behave when it is passed through a lens?    (AS - 1) (1 Mark)
A: * A light ray deviates when it passes through a lens.
* In some directions the light ray does not deviate.

8. How does the lens form an image?           (AS - 1) (1 mark)
A: * A lens forms an image of an object when light rays converge through it.
* A lens forms an image of an object when light rays diverge through it.

9. If we allow a light ray to pass through the focus of a lens which path does it take? (AS - 1) (1 Mark)
A: The light ray that passes through the focus incidents of the lens and travels parallel to the principal axis of the lens after refraction.

10. What happens when parallel rays of light fall on a lens making some angle with the principal axis?        (AS - 1) (2 Marks)
A: * When parallel rays, making an angle with principal axis, fall on a lens as shown in figures
(1) & (2), the rays converge at a point or appear to diverge from a point laying on the focal plane.
* Focal plane is the plane perpendicular to the principal axis at the focus.

11. What do you mean by an object at infinity? What type of rays fall on the lens?  (AS - 1) (2 Marks)
A: * Compared to the size of the object if the distance of the object from the lens is very much greater, then the object is said to be at infinity.
* When the object is at infinity, rays from the object travel parallel to the principal axis and incident on the lens.

12. Could you get an image on the screen for every object distance?       (AS - 1) (1 Mark)
A: * We get an image of an object on the screen when it is placed at the focus and beyond it from a convex lens.
* We do not get the image of an object on the screen when the object is placed in between the optic centre and focus of the convex lens.

13. Why do not you get an image for certain object distances?                  (AS - 1) (1 Mark)
A: When the object is at a certain distance or beyond it, light rays from it diverge without reaching the lens. So we do not get an image for such objects by the lens.

14. Can you find the minimum limiting object distance for obtaining a real image? What do you call this minimum limiting object distance for real image?    (AS - 1) (2 Marks)
A: * To get the real image (image on the screen) the object should be at least placed at the focus or beyond the focus of the convex lens.
* This minimum limiting object distance for real image is called the focal length of the convex lens.

15. When you do not get an image on the screen, try to see the image with your eye directly from the place of the screen. Could you see the image? What type of image do you see?  (AS - 1) (2 Marks)
A: * We can see the image with our eye.
* The image formed is a virtual image and it cannot be caught on a screen.

16. Can you find the image distance of this virtual image? How? (AS - 1) (1 Mark)
A: * We have lens formula: * As we know the values of f and u we can find the value of v which is the distance of the virtual image from the lens.

17. What difference do you notice in the refracted rays in figures 1 and 2 below? What could be the reason for that difference? (AS - 1) (2 Marks)
A: * In both the figures the incident ray travels parallel to the principal axis.
* In figure 1, the incident ray from the rarer medium enters the denser medium and after refraction the convex surface the refracted ray bends towards the normal.
* In figure 2, the incident ray from the denser medium enters the rarer medium and after refraction at the convex surface, the refracted ray bends away from the normal.

18. What difference to you notice in the refracted rays in the below figures 1 and 2? What could be the reason for that difference? (AS - 1) (2 Marks)
A: * In figure - 1, an incident ray is incident on the concave surface from a denser medium and this after refraction bends away from the normal.
* In figure - 2, an incident ray is incident on the concave surface from a rarer medium and this after refraction bends towards the normal.

ACTIVITIES

ACTIVITY - 1
1. Describe an activity to explain the refraction of light at curved surface.       (AS - 3) (4 Marks)
Case 1:
A: * Draw an arrow of length 4 cm using a black sketch pen on a thick sheet of paper.
* Take an empty cylindrical-shaped transparent vessel such as glass tumbler. Keep it on the table.
* Ask your friend to bring the sheet of paper on which arrow was drawn behind the vessel while you look at it from the other side.
* The arrow mark should be in horizontal position.
* We will see a diminished (small-sized) image of the arrow.
* The reason for this is that rays of light from the arrow mark refract through the curved surface and travel through glass.
* These rays again refract when pass in to air from glass and as such a diminished image is formed.

Case 2:
* Now the vessel is filled with water.
* The arrow mark is observed from the same position as before.
* We observe an inverted image.
* This is because light enters the curved surface, moves through water, emerges out of the glass. So it forms an inverted image.

2. Describe with an activity the characteristics of the images formed for an object placed at different distances from the convex lens.
A: * Take a V-stand and place it on a long (nearly 2 m) table at the middle.
* Place a convex lens on the V-stand. Imagine the principal axis of the lens.
* Light a candle and ask your friend to take the candle far away from the lens along the principal axis.
* Adjust a screen (a sheet of white paper placed perpendicular to the axis) which is on other side of the lens until you get an image on it.
* Measure the distance of the image from the V-stand of lens and also measure the distance between the candle and stand of lens.
* Record the values in a table 1.

 S.No. Object Distance (u) Image Distance (v) Focal Length (f) Characteristics of Image 1. 25 cm 100 cm 20 cm Real, Inverted, enlarged 2. 30 cm 60 cm 20 cm Real, Inverted, enlarged 3. 40 cm 40 cm 20 cm Real, Inverted, same size 4. 60 cm 30 cm 20 cm eal, Inverted, diminished 5. 50 cm 60 cm 20 cm Virtual, erect, enlarged

* Now place the candle at a distance of 60 cm from the lens, such that the flame of the candle lies on the principal axis of the lens.
* Try to get an image of the candle flame on the other side on a screen.
* Adjust the screen till you get a clear image.
* Measure the image distance (v) from lens and record the values of 'u' and 'v' in table 1.
* Repeat this for various object distances like 50 cm, 40 cm, 30 cm etc.
* Measure image distances in all the cases and note them in table 1.

ACTIVITY - 3
3. On what factors does the focal length of the lens depend? Narrate an activity to find the factors?   (AS - 3) (4 Marks)
A: * Take a convex lens and find the average focal length of the lens.
* Take a cylindrical vessel such as glass tumbler. Its height must be much greater than the focal length of the lens.
* Keep a black stone inside the vessel at its bottom. Now pour water into the vessel up to a height such that the height of the water level from the top of the stone is greater than focal length of lens.
* Now dip the lens horizontally using a circular lens holder as shown in the figure above the stone.
* Set the distance between stone and lens that is equal to or less than focal length of lens.
* Now look at the stone through the lens.
* We can see the image of the stone if the distance between lens and stone is less than the focal length of the lens in air.
* Now increase the distance between the lens and the stone until the image of the stone is not seen.
* We have dipped the lens to a certain height which is greater than the focal length of lens in air. But we can see the image.
* This shows that the focal length of the lens has increased in water.
* Thus we conclude that the focal length of lens depends upon the surrounding medium in which it is kept.

I. Conceptual Understanding

1. What is a lens?   (AS - 1) (1 Mark)
A: A lens is formed when a transparent material is bounded by two surfaces of which one (or) both surfaces are spherical.

2. What is a double convex lens?   (1 Mark)
A: * A lens may have two spherical surfaces bulging outwards such a lens is called double convex lens or biconvex lens.
* It is thick at the middle as compared to edges.

3. What is a double concave lens?    (1 Mark)
A: * A double concave lens is bounded by two spherical surfaces curved inwards (Biconcave lens).
* It is thin at the middle and thicker at the edges.

4. Explain with the help of a ray diagram the term 'focal length of a lens'.    (2 Marks)
A: * A parallel beam of light incident on a lens converges to a point as shown in figure (a) or seems to emanate from a point on the principal axis as shown in figure (b). * The point of convergence (or) the point from which rays seem to emanate is called focal point or focus (F).
* Every lens has two focal points.
* The distance between the focal point and optic centre is called the focal length of lens denoted by 'F'.

5. Mention the lens formula and explain the terms in it.   (2 Marks)
A: * Lens formula is: * f is the focal length of the lens, v is image distance from the lens and u is object distance from the lens.

6. Define the following with the help of a ray diagram.  (4 Marks)
a) Centre of curvature b) Radius of curvature
c) Principal axis d) Optic centre A: a) Centre of curvature:
Each curved surface of a lens is a part of a sphere.
The centre of the sphere which contains the part of the curved surface is called centre of curvature. It is denoted by a letter 'C'.
The distance between the centre of curvature and curved surface is called radius of curvature (R).
c) Principal axis:
If a lens contains two curved surfaces then their centres of curvature are denoted as C1 and C2. The line joining the points C1 and C2 is called principal axis.
d) Optic centre (P):
The mid point of a thin lens is called optic centre (P) of a lens.

7. Write the sign convention for refraction through a lenses.   (4 Marks)
A: For all purposes of applications of refraction at curved surfaces and through lenses following conventions are used.
* All distances are measured from the pole (or optic centre).
* Distances measured along the direction of the incident light ray are taken as positive.
* Distances measured opposite to the direction of the incident light ray are taken as negative.
* The heights measured vertically above from the points on axis are taken as positive.
* The heights measured vertically down from points on axis are taken as negative.

8. Obtain a relation between refractive indices of two media (n1, n2), object distance (u), image distance (v) and radius of curvature (R) for a curved surface. (4 Marks)
A: Relation between u, v, R, n1 and n2:
* Consider a curved surface separating two media of refractive indices n1 and n2 as shown in the figure.
* A point object is placed on the principal axis at point O.
* The ray, which travels along the principal axis passes through the pole undeviated. The second ray, which forms an angle α with principal axes, meets the interface (surface) at A. The angle of incidence is θ1.
* The ray bends and passes through the second medium along the line AI. The angle of refraction is θ2.
* The two refracted rays meet at I and the image is formed there.
* Let the angle made by the second refracted ray with principal axis be γ and the angle between the normal and principal axis be β (see figure).
* In figure PO is the object distance 'u'.
* PI is image distance 'v'.
* PC is radius of curvature 'R'.
* n1, n2 are refractive indices of two media. * In the triangle ACO, θ1 = α + β
and in the triangle ACI, β = θ2 + γ ⇒ β - γ = θ2
* According to Snell's law, we know
n1 sin θ1 = n2 sin θ2
substituting the values of θ1 and θ1, we get
n1 sin (α + β) = n2 sin (β _ γ) .......... (1)
* If the rays move very close the principal axis, the rays can be treated as parallel and are called paraxial rays. Then the angles α, β and γ become very small. This approximation is called paraxial approximation.
* sin (α + β) = α + β and sin (β _ γ) = β _ γ
substituting in equation (1)
n1(α + β) = n2(β _ γ)  n1α + n1β = n2β _ n2γ .......... (2)
* Since all angles are small, we can write
tan α = AN/ NO = α, tan β = AN/ NC = β, tan γ = AN/ NI = γ
* Substitute these in equation (2), we get
n1 AN/NO + n1AN/NC = n2 AN/NC _ n2 AN/NI .......... (3)
* As the rays moved very close to the principal axis, the point N coincides with pole of the interface (P). Therefore NI, NO, NC can be replaced by PI, PO and PC respectively.
* After substituing these values in equation (3), we get
n1/PO + n1/PC = n2/PC _ n2/PI
n1/PO + n2/PI = (n2 _ n1)/PC ........ (4)
Equation (4) shows the relation between refractive indices of two media, object distance, image distance and radius of curvature.
* Here PO is object distance (u)
PI is image distance (v)
PC is radius of curvature (R)
* According to sign convention
PO = _ u, PI = v, PC = R
* Substituting these values in as (4) 9. What are the rules to be followed to draw ray diagrams for the formation of an image of an object by a lens?  (4 Marks)
A: Following are the basic rules to draw ray diagrams to locate the position of images formed by a lens for any position of the object on the princiapl axis.
* Select a point on the object placed at a point on the principal axis.
* Draw two rays that were chosen by you from rays mentioned in different situations.
* Extend both rays to intersect at a point. This point gives position of the image.
* Draw a normal from point of intersection to the principal axis.
* Length of this normal represents the size of the image.

10. Derive the relation (or) Deduce the Lens formula.    (4 Marks) A: To show: (Lens formula)
* OO' is the object placed on the principal axis of a lens at a distance 'u'.
* As shown in the ray diagram the image II' is formed at a distance v on the other side of the lens.
* Let f be the focal length of the lens.
* Triangles PP'F1 and F1II' are similar
(See the diagram) * But F1I = PI - PF1 * This equation is derived in the case of convex lens. To convert this in to a general equation, we need to use sign convention.
* According to sign convention
PO = _ u, PI = v and PF1 = f
* Substituting these values in equation (4), we get This equation is called lens formula.

11. Derive lens makers formula (or) A: * Imagine a point object 'O' placed on the principal axis of the thin lens as shown in figure.
* Let this lens be placed in a medium of refractive index na and let refractive index of lens medium be nb.
* Consider a ray, from 'O' which is incident on the convex surface of the lens with radius of curvature R1 at A as shown in figure. * The incident ray refracts at A. Let us assume that it forms image at Q, if there were no concave surface.
* From the figure object distance PO = _ u
Image distance v = PQ = x
Radius of curvature R = R1
n1 = na and n2 = nb * But the ray that has refracted at A suffers another refraction at B on the concave surface with radius of curvature (R2), At B the ray is refracted and reaches I on the principal axis.
* The image Q of the object due to the convex surface is taken as object for the concave surface. So, we can say that I is the image of Q for concave surface.
See figure
Object distance u = PQ = +x
Image distance PI = v
Radius of curvature R = _ R2
* For refraction, the concave surface of the lens is considered as medium 1 and surrounding medium is considered as medium 2. Hence the suffixes of refractive indices interchange. Then we get,
n1 = nb and n2 = na  * This is derived for specific case for the convex lens so we need to generalize this relation. For this we use sign convention. Applying sign convention to this specific case we get, * We know  * If surrounding medium is air, then refractive index could be absolute refractive index of the lens. This can be used only when the lens is kept in air. This is the lens maker's formula.

V. Communication Through Drawing, Model Making

12. What are the different types of lenses available. Draw their diagrams.    (2 Marks)
A: Following are the 5 different types of lenses available. 13. Show with the help of a ray diagram the centre of curvature, normal, pole, principal axis of a curved surface separating two different media. (2 Marks)
A: * Consider a curved surface separating two media as shown in the figure.
* The centre of the sphere of which the curved surface is a part, is called the centre of curvature 'C'.
* Any line drawn from the centre of curvature to a point on the curved surface becomes normal to the curved surface at that point.
* The centre of the curved surface is called the pole (P) of the curved surface.
* The line that joins the centre of curvature and the pole is called 'principal axis'.

14. Explain the behaviour of certain light rays when they are incident on a lens, with the help of ray diagrams.   (4 Marks)

A: Behaviour of a light ray when it passes through a lens: * Any ray passing along the principal axis is undeviated.
b) Any ray passing through the optic centre is also undeviated. c) Rays passing parallel to the principal axis converge at the focus or appear to diverge from the focus. d) Rays passing through focus will take a path parallel to principal axis after refraction. e) When parallel rays, making an angle with the principal axis fall on a lens, the rays converge at a point or appear to diverge from a point lying on the focal plane.
* Focal plane is the plane perpendicular to the principal axis at the focus. 15. For drawing ray diagrams related to lenses, how do you represent a convex lens and a concave lens?   (1 Mark)
A: For drawing ray diagrams related to lenses we represented convex lens with a symbol and concave lens with .

16. Describe how do you represent image formation by a convex lens for various positions of the object with the help of a ray diagram.  (4 Marks)
A: a) Object at infinity:
* Rays falling on the lens from an object at infinity are parallel to the principal axis.
* They converge to the focal point. So point sized image is formed at the focal point. b) Object placed beyond the centre of curvature on the principal axis. * Two rays are chosen.
* One ray passing parallel to the principal axis and another ray passing through the optic centre.
* The image is formed in between the points F1 and C1 on the principal axis.
* The image formed is real, inverted and diminished.
c) Object placed at the centre of curvature (C2)
* The image formed is real, inverted and of the same size of the object.
* The image is formed at C1 on the principal axis. d) Object placed between the centre of curvature and focal point.
* The image formed is real, inverted and magnified.
* The image is formed beyond C1. e) Object located at the focal point
* The image will be formed at infinity.
* We cannot discuss the size and nature of the image. f) Object placed between focal point and optic centre.
* When the object is placed between focus and optic centre of the convex lens, we get virtual, erect and magnified image on the same side of the lens where the object is placed. * The ray diagram is shown.
g) Object placed between F1 and C1 of a concave lens.
* Irrespective of the position of the object, on the principal axis we get an erect, virtual image, diminished in size between the focal point and optic centre of a concave lens. 17. What do you understand when you study the image formation by a convex lens when the object is placed between focal point and optic centre.    (AS - 1) (2 Marks)
A: I understand two things.
* As the image formed is virtual, we can see it with our eyes. In all other cases the image is real which we can't see with our eyes but can be viewed if the image is captured on screen.
* A magnified virtual image is formed on the same side of the lens where the object is placed. Thus the image you are seeing through a lens is not real, it is a virtual image of the object.

Worked out examples

1. A bird is flying down vertically towards the surface of water in a pond with constant speed. There is a fish inside the water. If that fish is exactly vertically below the bird, then the bird will appear to the fish to be:
a) farther away than its actual distance
b) closer than its actual distance
c) moving faster than its actual speed
d) moving slower than its actual speed
Which of the four options are true? How can you prove it? A: For refraction at a plane surface, * Let x be the height of the bird above the water surface at an instant and 'n' be the refractive index of water.
n1 = refractive index of air,
Then n1 = 1, n2 = n, u = -x and
let v = -y (see figure)
* Substituting these values in equation (1) * In the above equation, we know that n is greater than 1. Hence y is greater than x. Thus the bird appears to the fish to be farther away than its actual distance.
* We have assumed that bird is flying vertically down with constant speed. For the observer on the ground, bird appears that it has covered 'x' distance for certain time.
* But for fish, it appears that bird has covered a distance 'y' in the same time.
* As y is greater than x, we can conclude that the speed of the bird, observed by the fish, is greater than its equal speed.
* So, options (a) and (c) are correct.

2. A transparent sphere of radius R and refractive index n is kept in air. At what distance from the surface of the sphere should a point object be placed on the principal axis so as to form a real image at the same distance from the second surface of the sphere? (2 Marks)
A: * From the symmetry of figure, the rays must pass through the sphere parallel to the principal axis.
* from the figure,
u = _ x, v = ∞ (refracted ray is parallel to the optical axis after refraction at first surface)
* n1 = 1 and n2 = n, (where n1 is refractive index of air) 3. A transparent (glass) sphere has a small, opaque dot at its centre. Does the apparent position of the dot appear to be the same as its actual position when observed from outside.   (4 Marks)
A: * Let refractive index of glass n1 = n
Refractive index of air n2 = 1
* Then u = _ R (radius of sphere); Radius of curavature R = _ R * After solving this equation, we get
Image distance v = _ R
* Thus we can say that the image distance and object distance are equal and that the apparent position of dot is the same as its actual position.
* It is independent of the refractive index of the material of the sphere.

4. Draw a ray diagram to locate the position of image when a point source (S) is placed on optical axis MN of a convex lens, in such a way that it is beyond focal point (F). See figure (4 Marks)
A: * Draw a perpendicular line to principal axis passing through the focus (F').
* Draw a ray from point source (S) in any direction to meet lens at point (P'). * Now draw another line parallel to the ray drawn from the point source (S) such that it passes through the optic centre (P). This line intersects the normal at point F0.
* Now draw a line passing from point P' to pass through the point F0 such that it meets principal axis at a point say (I).
* 'I' is the image point for the point source (S).

5. Complete the ray diagram to show the paths of the rays after refraction through the lenses shown in the figures (a) and (b)?  (2 Marks) A: * Follow the steps mentioned in example (4) above to complete the ray diagrams.
* We will notice that the paths of the rays are shown in figures (c) and (d). 6. An electric lamp and a screen are placed on the table, in a line at a distance of 1 m. In what positions of convex lens of focal length of f = 21 cm will the image of lamp be sharp?   (4 Marks)
A: * Let 'd' be distance between the lamp and screen and 'x' be the distance between lamp and lens. From figure, we have u = _ x and v = d _ x
* By substituting these in lens formula, we get * After solving the equation, we get x2 _ dx + fd = 0 * It is a quadratic equation. Hence we get two solutions. The solutions of the above equation are Given that f = 21 cm and d = 1 m = 100 cm
Substituting these values in equation (1), we get x1 = 70 cm and x2 = 30 cm.

7. What is the focal length of double concave lens kept in air with two spherical surfaces of radii R1 = 30 cm and R2 = 60 cm. Take refractive index of lens as n = 1.5   (4 Marks)
A: * From the figure using sign convention we get R1 = -30 cm, R2 = 60 cm and also given that n = 1.5 * Solving this, we get f = _120 cm
* Here minus indicates that the lens is divergent.

Writer: C.V. Sarveswara Sarma

Posted Date : 29-12-2020

గమనిక : ప్రతిభ.ఈనాడు.నెట్‌లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.