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Human Eye and Colourful World

Improve Your Learning
 

I. CONCEPTUAL UNDERSTANDING
1. How do you correct the eye defect 'Myopia'? (AS - 1) (4 Marks)
Ans: * Some people cannot see objects at long distances but can see nearby objects clearly. This type of defect in vision is called 'Myopia'. It is also called 'near sightedness'.
* For these people the maximum focal length is less than 2.5 cm. In such cases the rays coming from distant objects, after refraction through the eye lens, form an image before the retina as shown in figures (a) and (b).

* Let 'M' be the point where from an object appears clearly to a person with the defect of vision 'Myopia'. This is called far point for him.
* 'L' is the least distance of distinct vision.
* If the object is at M or in between M and point of least distance of distinct vision (L), the eye lens can form an image on the retina.
* The defect, in which people cannot see objects beyond far point is called 'Myopia'.
* To correct ones Myopia, we need to select a lens which forms an image at the far point for an object at infinity.
* We need to select a bi-concave lens to achieve this.

* This image acts like an object for the eye lens. Hence the final image is formed on the retina.
* The object distance (u) = ∞ (infinity)
Image distance (v) = -D
(D is distance of far point)
Focal length of the bi-concave lens f = ?

*  f = -D. (-ve sign shows that the lens is a concave lens)

2. Explain the correction of the eye defect Hypermetropia.  (AS - 1) (4 Marks)
Ans: Hypermetropia: It is also called 'far sightedness'. A person with hypermetropia can see distant objects clearly but cannot see objects at near distances.
* This is because the minimum focal length of eye lens for the person of hypermetropia is greater than 2.27 cm.
* In such cases the rays coming from a nearby object, after refraction at the eye lens, forms an image beyond the retina as shown in the figure below.

* If an object is at H or beyond H, the eye can form its image on retina. This point H is called the near point.
* If the object is between H and the point of least distance of distinct vision (L), then it cannot form an image on the retina as shown in fig (a) above.
Correction of the defect Hypermetropia:
* Eye lens can form a clear image on the retina when an object is placed beyond near point.
* To correct this defect, we need to use a lens which forms an image of an object beyond near point, when the object is between near point (H) and the least distance of distinct vision (L).
* This is possible only when a double convex lens is used.
To find the focal length of the double convex lens:

* Let the object be at the point of least distance of distinct vision. (25 cm)
* Object distance (u) = -25 cm
* This defect of vision is corrected by the biconvex lens if the image of the object could be formed at the near point H.
* This image acts like an object to the eye lens and the final image is formed on the retina.
* Object distance u = -25 cm
Image distance v = distance of near point = -d
Focal length of biconvex lens f = ?

f is measured in centimetres.
* If d > 25 cm then f is +ve and we need to use biconvex lens to correct defect of hypermetropia.

3. How do you find experimentally the refractive index of material of a prism?    (AS - 1) (4 Marks)
Ans: Aim: Finding the refractive index of a prism.
Material required: Prism, piece of white chart, pencil, pins, scale and protractor.
Procedure: * Take a prism and place it on the white chart in such a way that the triangular base of the prism is on the chart. Draw a line around the prism (boundary) using a pencil.
* Remove the prism and name the vertices of the boundary triangle as P, Q and R.
* Find the angle between the refracting surfaces PQ and PR. This is the angle of the prism (A).
* Mark M on the side of triangle PQ and also draw a perpendicular to PQ at M. Place the centre of the protractor at M and along the normal.
* Mark an angle of 30° and then draw a line up to M. This line denotes the incident ray. This angle is called angle of incidence (i1). Note it in the table below.
Draw a small arrow on it as shown in figure (a).

 

S.No. Angle of incidence (i1) Angle of emergence (i2) Angle of deviation (d)
       

* Place the prism in its position (Triangle) again. Now fix two pins vertically on the line at point A and B as shown in figure (a).
* Look for the images of pins through the prism from the other side (PR) and fix another two pins at point C and D in such a way that all the four pins appear to lie along the same straight line.
* Now remove the prism and take out pins. Draw a line joining the two pin-holes formed by the pins to meet surface 'PR', this is the emergent ray which 'emerges from' the surface PR at a point 'N'.
* The angle between the normal at N and the emergent ray is the angle of emergence (i2). Measure this angle and note its value in the table.
* Join the points M and N by a straight line. The line passing through the points A, B, M, N, C and D represents the path of light when it suffers refraction through the prism.
* Extent both incident and emergent rays till they meet at point 'O'. Measure the angle between these two rays. This is the angle of deviation 'd'. Note its value in the table.
* Repeat this experiment and procedure for various angles of incidence 40°, 50°.. etc. Find the corresponding angles of deviation and emergence and note them in the table.
* We notice that angle of deviation first decreases and then increases with increase in the angle of incidence.
* If a graph is drawn taking angle of incidence i1 along X - axis and angle of deviation along Y-axis we get a curve as shown in figure (b).


* A tangent line is drawn to the cure parallel to X-axis at its lowest point.
* The point where it cuts the Y-axis gives the angle of minimum deviation 'D'.
* The refractive index of the material of the prism is calculated from the formula.


                                                     
4. Explain the formation of rainbow.  (AS - 1) (4 marks)
Ans:

                     
Formation of rainbow:
* The beautiful colours of the rainbow are due to dispersion of the Sun light by millions of tiny water droplets.
* Let us consider the case of an individual water drop.
* Observe the figure. The rays of Sun light enter the drop near its top surface.
* At this first refraction, the white light is dispersed into its spectrum of colours, violet being deviated the most and red the least. Reaching the opposite side of the drop, each colour is reflected back into the drop because of total internal reflection.
* Arriving at the surface of the drop, each colour is again refracted into air.
* At the second refraction the angle between red and violet rays further increases when compared to the angle between those at first refraction.
* We observe a bright rainbow when the angle between the incoming and outgoing rays is near the maximum angle of 42°.

 

5. Explain briefly the reason for the blue of the sky.   (AS - 1) (2 Marks)
Ans: * When we look at the sky in a direction perpendicular to the direction of the Sun rays, the sky appears blue. This is because the intensity is maximum at 90° angle of scattering.
* Our atmosphere contains different types of molecules and atoms. The reason for blue sky is due to the molecules of N2 and O2.
* The sizes of these molecules are comparable to the wavelength of blue light. These molecules act as scattering centres for scattering of blue light.

 

6. Explain two activities for formation of artificial rainbow. (AS - 1) (4 Marks)
(Each activity 2 Marks)
Ans: Activity 1: (This is Activity 4 in the Text Book)
* Take a metal tray and fill it with water.
* Place a mirror in the water such that it makes an angle to the water surface.
* Now focus white light on the mirror through the water as shown in figure.
* Try to obtain colour on a white card board sheet kept above the water surface.
* We can see the seven colours viz VIBGYOR on the card board sheet.


Acvitity 2:
* Select a white coloured wall on which the Sun rays fall. Stand in front of a wall in such a way that the Sun rays fall on your back.
* Hold a tube through which water is flowing. Place your finger in the tube to obstruct the flow of water.
* Water comes out through the small gaps between the tube and your finger like a fountain.
* Observe the changes on the wall while the water shower is maintained.
* We see different colours on the wall, which are similar colours of VIBGYOR.

7. Derive an expression for the refractive index of the material of a prism. (AS - 1) (4 Marks)
Ans: To derive:
                           
* Observe the ray diagram in figure (a).
* From the triangle OMN,
we get d = (i1 - r1 + i2 - r2 ) or
d = (i1 + i2) - (r1 + r2) ................(1)
* From the triangle PMN,
we get A + (90° - r1) + (90° - r2) = 180°
It means r1 + r2 = A ................(2)
* From equations (1) and (2)
d = (i1 + i2) - A (or)
A + d = i1 + i2 ................(3)
* We know from Snell's law n1 . sin i = n2 . sin r
* Let n be the refractive index of the material of the prism.
* Using Snells law at M with refractive index of air.
n1 = 1, i = i1, n2 = n, r = r1 gives
sin i1 = n sin r1 ................(3)
* Similarly at N;
n1 = n, i = r2, n2 = 1, r = i2 gives
n . sin r2 = sin i2 ................(5)
* At the angle of minimum deviation,
angle of incident i1 = Angle of emergence i2
* Further MN is parallel to the base (QR),
So d = D (minimum deviation)
when i1 = i2 = i
∴ Equation (3) becomes A + D = 2i
        or i = 
* It is clear that r1 = r2 = r when d = D, so from equation (2),
A we get 2r = A or r = 
* Substituting the values of i and r in equation (4)
  = n . sin 
∴ Refractive index of material of the prism
   


8. Light of wavelength λ1 enters a medium with refractive index n2 from a medium with refractive index n1. What is the wavelength of light in the second medium?    (AS - 1) (2 Marks)
Ans: * Wavelength of the first medium = λ1
Refractive index of the first medium = n1
* Wavelength of the second medium = λ2 = ?
Refractive index of the second medium = n2
* Formula: As light enters from first medium to second medium.


                       
II. ASKING QUESTIONS AND MAKING HYPOTHESIS
Note: For questions 9 and 10 the following options are given. Choose the correct option by making hypothesis based on given assertion and reason. Give an explanation.
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.

9. Assertion (A): The refractive index of a prism depends only on the kind of glass of which it is made of and the colour of light.
Reason (R): The refractive index of a prism depends on the refracting angle of the prism and the angle of minimum deviation.   (AS - 2)(2 Marks)
Ans: Answer (b)
1) Both A and R are true.
2) R is not the correct explanation of A.

 

10. Assertion (A): Blue colour of sky appears due to scattering of light.
Reason (R): Blue colour has shortest wavelength among all colours of white light. (AS - 2)(2 Marks)
Ans: Answer (c) is correct.
1) A is true but R is false.

 

III. EXPERIMENTATION AND FIELD INVESTIGATION

11. Suggest an experiment to produce a rainbow in your class room and explain the procedure.  (AS - 3) (4 Marks)
Ans: Experiment to produce a rainbow in the classroom: (Activity - 3)
* Take a prism and place it on the table near a vertical white wall.
* Take a thin wooden plank. Make a small hole in it and fix it vertically on the table.
* Place the prism between the wooden plank and wall.
* Place a white light source behind the hole of the wooden plank. Switch on the light.
* The rays coming out of the hole of plank become a narrow beam of light.
* Adjust the height of the prism such that the light falls on one of the lateral surfaces.
* Observe the change in emerged rays of the prism.
* Adjust the prism by slightly rotating it till you get an image on the wall.
* We find a band of different colours on the wall.
* These colours are the similar colours (VIBGYOR). We observe in the rainbow.

 

IV. INFORMATION SKILLS AND PROJECTS

12. Prisms are used in binoculars. Collect information why prisms are used in binoculars. (4 Marks)
Ans: * To observe distant objects to appear closer binoculars are used.
* Generally two lenses which move parallel to each other are arranged in binoculars.
* Prisms are used in binoculars to get more reflections.
* By using prisms the size of the binoculars are reduced.
* The intensity of the field of view and the size of the object can be increased by using prisms in the binoculars.
* The reflection capacity of the binoculars can be increased to 95% by using prisms in the binoculars.

 

V. COMMUNICATION THROUGH DRAWING, MODEL MAKING

13. Incident ray on one of the face (AB) of a prism and emergent ray from the face AC are given in the figure. Complete the ray diagram.  (2 Marks)
* AB and AC are refracting surfaces
* BC is reflecting surface
* i1: angle of incidence
* i2: angle of emergence
* θ, θ : angle of incidence and angle of reflection
* N1, N2 and N3 are the normals at the respective points.

VI. APPRECIATION AND AESTHETIC SENSE, VALUES


14. How do you appreciate the role of molecules in the atmosphere for the blue colour of the sky? (2 Marks)
Ans: * Scattering of light gives blue colour to the sky.
* The reason for blue sky is due to the molecules N2 and O2 in the atmosphere.
* The size of these molecules are comparable to the wavelength of blue light.
* These molecules act as scattering centres for scattering of blue light.
* We appreciate the role of O2 and N2 molecules in the atmosphere is showing us the sky as beautiful in blue colour.

 

15. Eye is the only organ to visualise the colourful world around us. This is possible due to accommodation of eye lens. Prepare a six line stanza expressing your wonderful feelings. (4 Marks)
Ans: Of all the organs in our body
undoubtedly eye is the most important one
one can see all the objects far and near
thanks to the accommodation nature of the eye
The beauty of nature in all colours
one can enjoy due to this wonderful organ
one should safe guard this organ for ever
to keep himself enjoying the beauty of nature

 

16. How do you appreciate the working of ciliary muscles in the eye? (2 Marks)
Ans: * The ciliary muscle to which eye lens is attached helps the eye lens to change its focal length by changing the radii of curvature of the eye lens.
* When the eye is focused on a distant object, the ciliary muscles are relaxed so that the focal length has its maximum value which is equal to its distance from the retina.
* When the eye is focused on a closer object, the ciliary muscles are strained and the focal length of the eye-lens decreases.
* The ciliary muscles adjust the focal length in such away that the image is formed on retina and we see the objects clearly.
* This is a wonderful adjustment of the eye lens made by the Ciliary muscles and I appreciate the working of these muscles.
* This is a gift of nature to the human beings which cannot be substituted by any artifificial means.

VII. APPLICATION TO DAILY LIFE, CONCERN TO BIODIVERSITY

17. Why does the sky some times appear white? (4 Marks)
Ans: * Our atmosphere contains atoms and molecules of different sizes.
* According to their sizes, they are able to scatter different wavelengths of light.
* For example, the size of the water molecule is greater than the size of the N2 or O2.
* It acts as a scattering centre for other frequencies which are lower than the frequency of blue light.
* On a hot day, due to rise in the temperature water vapour enters into atmosphere which leads to abundant presence of water molecules in the atmosphere.
* These water molecules scatter the colours of other frequencies (other than blue)
* All such colours of other frequencies reach your eye and the sky appears white.

 

18. Glass is known to be a transparent material. But ground glass is opaque and white in colour. Why? (2 Marks)
Ans: * Glass is a transparent material and so it allows the rays of light to pass through it.
* One surface of a ground glass is made rough and as such this surface contains ups and downs making it a unplane surface.
* This unplane surface gives irregular reflections of the light rays incidenting on it.
* This surface becomes opaque and do not allow rays of light to pass through it.
* As such the ground glass appears to be white in colour.

 

19. If a white sheet of paper is stained with oil, the paper turns transparent. Why? (2 Marks)
Ans: * A White sheet of paper is opaque. It is a solid material and is an absorbent.
* When this white sheet of paper absorbs water it becomes partly transparent till the water in it dries up.
* White sheet of paper absorbs oil, and oil do not quickly dry up like water. So the paper stained with oil turns transparent until the oil in it dries up.

 

PROBLEMS


1. A ray of light falls on one of the faces of a Prism of angle 40°, so that it suffers angle of minimum deviation of 30°. Find the angle of prism and angle of refraction at the given surface. (4 Marks)
Ans: * Given: Angle of incidence on the face of Prism: i = 40°, angle of minimum
deviation D = 30°
Angle of prism A = ?
Angle of refraction r = ?
Formula: i =  
or A = 2i - D
∴ Angle of Prism: A = 2 × 40 - 30 = 50°
Formula: Angle of refraction r = 
∴ r =  = 25°

2. The focal length of a lens suggested to a person with hypermetropia is 100 cm. Find the distance of near point and power of the lens. (4 Marks)
Ans: Given: Object distance u = -25 cm.
Image distance v = distance of near point = -d
Focal length f = 100 cm.

3. A person is viewing an extended object. If a converging lens is placed in front of his eyes will he feel that the size of the object has increased? Why? (2 Marks)
Ans: * If a convergent lens is placed in front of the eyes of a person viewing an extended object he feels that the size of the object is increased.
* The image of the object is formed when the object is in between the focus and optic centre of the lens.
* The virtual image formed is erect and bigger in size.

QUESTIONS AND ANSWERS GIVEN IN THE LESSON

1. Why do the values of least distance of distinct vision and angle of vision change with person and age?  (AS - 1) (2 Marks)
Ans: * The ciliary muscle to which eye lens is attached helps the eye lens to change its focal length by changing the radii of curvature of the eye lens.
* When the eye is focused on a distant object, the Ciliary muscles are relaxed so that the focal length of eye lens has its maximum value which is equal to its distance from retina.
* The working capacity of this ciliary muscle changes from person to person and with age also.
* So the values of least distance of distinct vision and angle of vision changes from person to person.

 

2. How can we get same image distance for various positions of objects? (AS - 1) (1 Mark)
Ans: We get same image distance for various positions of objects only when focal length of eye lens changes. This is possible only when the eye lens is able to change its shape.

 

3. Can you answer the above question using concepts of refraction through lenses?  (AS - 1) (1 Mark)
Ans: * The focal length of a lens depends on the material of the lens and for a given lens its focal length remains constant.
* To get same image distance for various positions of objects we have to change the lens of suitable focal length.

 

4. How does eye lens change its focal length?   (AS - 1) (1 Mark)
Ans: The ciliary muscle of which eye lens is attached helps the eye lens to change its focal length by changing the radil of curvature of the eye lens.

 

5. How does this change take place in the eye ball?  (AS - 1) (2 Marks)
Ans: * When the eye is focused on a distant object, the ciliary muscles are relaxed so that the focal length of eye lens has its maximum value which is equal to its distance from the retina.
* The parallel rays coming in to the eye are then focused on to the retina and we see the object clearly.

 

6. Does eye lens form a real image or virtual image?  (1 Mark)
Ans: The eye-lens forms a real and inverted image of an object on the retina.

 

7. How does the image formed on retina helps us to perceive the object without change in its shape, size and colour?  (AS - 1) (4 Marks)
Ans: * The retina is a delicate membrane, which contains about 125 million receptors called 'rods' and 'cones' which receive the light signal.
* Rods identify the colour, cones identify the intensity of light.
* These signals are transmitted to the brain through about 1 million optic-nerve fibres.
* The brain interprets these signals and finally process the information so that we perceive the object in terms of its shape, size and colour.

 

8. Is there any limit to the change of focal length of the eye-lens?  (AS - 1) (2 Marks)
Ans: * When the object is at infinity, the parallel rays from the object falling on the eye-lens are refracted and they form a point-sized image on retina.
* In this situation the eye-lens has a maximum focal length.

 

9. What are the maximum and minimum focal length of the eye lens? How can we find them?   (AS - 1) (4 Marks)
Ans: * When the object is at infinity the eye lens has maximum focal length.
u = ∞, v = 2.5 cm (image distance)
This image distance is equal to distance between eye-lens and retina.


* When the object is placed at a distance of 25 cm from our eye, the eye lens has minimum focal length.
* Here u = -25 cm, v = 2.5 cm, fmin = ?

10. What happens if the eye lens is not able to adjust its focal length? (AS - 1) (1 Mark)
Ans: If the eye lens is not able to adjust its focal length then one cannot see the objects clearly and comfortably.

 

11. What happens if the focal length of eye lens is beyond the range of 2.5 cm to 2.27 cm?  (AS - 1) (1 Mark)
Ans: The vision becomes blurred due to defects of the eye lens.

 

12. What can we do the correct Myopia?  (AS - 1) (1 Mark)
Ans: To correct Myopia, we use a concave lens.

 

13. How can you decide the focal length of the lens to be used to correct Myopia?  (AS - 1) (4 Marks)
Ans: * Object distance u = -∞
Image distance v = distance of far point = -D
Focal length of Bi - concave lens f = ?

(or) f = -D
        f is -ve means it is a concave lens.

 

14. What happens when the eye has a minimum focal length greater than 2.27 cm?  (AS - 1) (1 Mark)
Ans: In this case the rays coming from the nearby object after refraction at the eye lens from the image beyond the retina.

 

15. How can you correct the defect of Hypermetropia?             (AS - 1) (1 Mark)
Ans: * To correct the defect of hypermetropia, we need to use a lens which forms an image of an object beyond near point, when the object is between near point and least distance of distinct vision.
* This is possible only when a double convex lens is used.

 

16. How can you decide the focal length of the convex lens to be used for correcting the defect of hypermetropia?   (AS - 1) (4 Marks)
Ans: * Object distance (u) = -25 cm
Image distance (v) = distance of near point = -d
focal length of bi-convex lens = f = ?


We know d > 25 cm, so f is +ve.
* It means we have to use bi convex lens to correct defect of hypermetropia.

 

17. Have you ever observed details in the prescription of an eye doctor? (AS - 1) (1 Mark)
Ans: * The prescription contains the defect of vision, the corrective lenses indicating their power.

 

18. You might have heard people saying 'my sight is increased or decreased' what does it mean?   (AS - 1) (1 Mark)
Ans: It means the increase or decrease in the vision of the eye.

 

19. What do you mean by power of a lens?  (AS - 1) (2 Marks)
Ans: * The degree of convergence or divergence of light rays that can be achieved by a lens is expressed in terms of its power.
* The reciprocal of focal length is called power of lens.
* Power of lens 
* The unit of power is dioptre. (D).

 

20. How could the white light of the Sun gives us various colours of the rainbow? (AS - 1) (2 Marks)
Ans: * The speed of light is constant in vacuum for all colours, it depends on the wave length of light when it passes through a medium.
* When white light passes through a medium, each colour selects its least time path and we have refraction of different colours to different extents.
* This results in separation of colours, producing a spectrum of different colours.
* This is how a rainbow colours are formed when white light of the Sun passes through rain drops.

 

21. What happens to a light ray when it passes through a transparent medium bounded by plane surfaces which are inclined to each other?  (AS - 1) (1 Mark)
Ans: Light ray is incident on one of the plain surfaces emerges from the other plan surface.

 

22. What is a Prism?    (AS - 1) (1 Mark)
Ans: A prism is a transparent medium separated from the surrounding medium by at least two plane surfaces which are inclined at a certain angle in such a way, light incident on one of the plane surfaces emerges from the other plane surface.

 

23. What is the shape of the outline drawn for a prism?   (AS - 1) (1 Mark)
Ans: It is a triangle.

 

24. How do you find the angle of deviation?  (AS - 1) (1 Mark)
Ans: The angle between the extended incident ray and emergent ray is called the angle of deviation.

 

25. What do you notice from the angles of deviation in the case of a prism? (AS - 1) (1 Mark)
Ans: The angle of deviation decreases first and then increases with the increase of the angle of incidence.

 

26. Can you draw a graph between angle of incidence and angle of deviation? (AS - 1) (1 Mark)
Ans: We can draw a graph between angle of incidence and angle of deviation.

 

27. From the graph, can you find the minimum of the angles of deviation? (AS - 1) (2 Marks)
Ans:


* The graph drawn for angle of incidence and angle of deviation is shown in the figure.
* Draw a tangent line to the curve, parallel to X - axis at its lowest point of the graph.
* The point where this line cuts the Y- axis gives the angle of minimum deviation.

 

28. Is there any relation between angle of incidence and angle of emergence and angle of deviation?  (AS - 1) (1 Mark)
Ans: A + D = (i1 + i2)
Where A is angle of prism, D : angle of deviation
i1 = angle of incidence and i2 : angle of emergence.

 

29. Is this splitting of white light in to colours explained by using ray theory? (AS - 1) (1 Mark)
Ans: It is not possible to explain the splitting of white light into different colours using ray theory.

 

30. In activity-3, we noticed that light has chosen different paths. Does this mean that the refractive index of the prism varies from colour to colour?  (AS - 1) (1 Mark)
Ans: Refractive index of the prism varies from colour to colour.

 

31. Is the speed of light of each colour different?  (AS - 1) (1 Mark)
Ans: * Speed of light for all colours is the same in vacuum.
* Speed of light for each colour is different in a medium.

 

32. Can you guess, why white light splits in to different colours when it passes through a prism?  (AS - 1) (1 Mark)
Ans: White light splits into different colours due to
a) dispersion of light and
b) speed of light for different colours is different in a medium.

 

33. Does white light split into more colours? Why?  (AS - 1) (2 Marks)
Ans: We know that the frequency of light is the property of the source and it is equal to number of waves leaving the source per second.
* This cannot be changed by any medium.
* Hence frequency does not change due to refraction.
* Thus coloured light passing through any transparent medium retains its colour.

 

34. Can you give an example in nature, where you observe colours as seen in activity 3?  (AS - 1) (1 Mark)
Ans: Rainbow is an example in nature.

 

35. When do you seen a rainbow in the sky? (AS - 1) (1 Mark)
Ans: * On a rainy day we can see a rainbow in the sky.
* The beautiful colours of the rainbow are due to dispersion of Sun light by million of tiny water droplets.

 

36. Can we create a rainbow artificially?  (AS - 1) (1 mark)
Ans: We can create a rainbow artificially.

 

37. Why does the light dispersed by the rain drops appear as a bow? (AS - 1) (4 Marks)
Ans: * A rainbow is not the flat two dimensional arc as it appears to us in the sky.
* The rainbow we see is actually a three dimensional cone with the tip at our eye as shown in figure (a).
* All the drops that disperse the light towards us lie in the shape of the cone - acone of different layers.
* The drops that dispersered Red colour to our eye are on the outer most layer of the cone.
* Similarly the drops that disperse orange colour to our eye are on the layer of the cone beneath the red colour cone.
* In this way the cone responsible for yellow lies beneath orange and so on till the violet colour cone becomes the inner most cone. Refer figure (b).

38. Why is the sky blue?  (AS - 1) (2 Marks)
Ans: * The sky is blue because the molecules of N2 and O2 act as scattering centres in the sky for scattering of blue light.
* The sizes of these molecules are comparable to the wavelength of blue light. So these setter blue light.

 

39. What is scattering?  (AS - 1) (1 Mark)
Ans: The process of re-emission of absorbed light in all directions with different intensities by atoms or molecules, is called scattering of light.

 

40. Do you know what happens to the free atom or molecule when it is exposed to certain frequency of light?  (AS - 1) (2 Marks)
Ans: * Atoms or molecules which are exposed to light energy and emit some part of the light energy in different directions.
* This is the basic process happens in scattering of light.
* The effect of light on a molecule or an atom depends on the size of atom or molecule
* If the size of the atom or molecule is small it will be affected by higher frequency of light and vice versa.

 

41. Why is that the sky appears white some times when you view it in a certain direction on hot days?  (AS - 1) (4 Marks)
Ans: * Our atmosphere contains atoms and molecules of different sizes.
* According to their sizes, they are able to scatter different wavelengths of light.
* For example the size of the water molecule is greater than the size of N2 and O2.
* Water molecules act as scattering centres for other frequencies which are lower than the frequency of blue light.
* On a hot day, due to rise in temperature water vapour enters into atmosphere which leads to abundant presence of water molecules inthe atmosphere.
* These water molecules scatter the colours of other frequencies, other than blue.
* All such colours of other frequencies reach our eye and the sky appears white.

 

42. Do you know the reasons for appearance the red colour of Sun during Sunrise and at Sunset? (AS - 1) (4 Marks)
Ans: * The atmosphere contains free molecules and atoms with different sizes.
* These molecules and atoms scatter light of different wavelengths which are comparable to their size.
* Molecules having a size that is comparable to the wavelength of red light are less in the atmosphere.
* Hence scattering of red light is less when compared to the other colours of light.
* The light from the Sun needs to travel more distance in atmosphere during Sunrise and Sunset to reach our eye.
* In morning and evening times, during sunrise and sunset, except red light all colours scatter more and vanish before they reach us.
* Since scattering of red light is very small, it reaches us.
* As a result Sun appears red in colour during Sunrise and Sunset.

 

43. Can you guess the reason why Sun does not appear red during noon hours? (AS - 1) (2 Marks)
Ans: * During noon hours, the distance travelled by the Sun rays in the atmosphere is less than that compared to morning and evening hours.
* Therefore all colours reach our eyes without much scattering.
* Hence the Sun appears white during noon hours.

 

ACTIVITIES

Activity 1:
1. Describe an activity to find the least distance of distinct vision.  (AS - 3) (2 Marks)
Ans: * Take a text book and hold it with your hands in front of you at a certain distance.
* Try to read the contents on the page slowly move the book towards your eye till it is very close to your eyes.
* You may see that printed letters on the page of the text book appear blurred or you feel strain in the eye.
* Now slowly move the book backwards to a position where you can see clear printed letters without straining your eye.
* Ask your friend to measure the distance between your eye and text book at this position.
* Note down its value. Repeat the activity with other friends and note down the distances for distinct vision in each case.
* Find the average of all these distances of clear vision.
* From this activity we will come to know that to see an object comfortably and distinctly, one must hold it at a distance about 25 cm from our eyes.
* This distance is called least distance of distinct vision.

 

Activity 2:
2. Are you able to see the top and bottom of an object placed at a distance of about 25 cm from your eyes irrespective of its shape? Find out with the help of an activity. (AS - 3) (4 Marks)
Ans: * Collect a few wooden sticks used in cloth roller in clothes store (or) collect waste PVC pipes that are used for electric wiring.
* Prepare sticks or pipes of 20 cm, 30 cm, 35 cm, 40 cm, 50 cm from them.
* Place a retort stand on a table and stand near the table such that your head is beside the vertical stand. (see fig).
* Adjust the clamp on the horizontal rod and fix it at a distance of 25 cm from your eyes.
* Ask one of your friends to fix a wooden stick of 30 cm height to the clamp in a vertical position as shown in fig. (a).
* Now keeping your vision parallel to horizontal rod of the stand, try to see the top and bottom of wooden stick kept in vertical position.
* As the least distance of distinct vision varies from person to person, if you are not able to see both ends of the stick at this distance (25 cm), adjust the vertical stick on the horizontal rod till you are able to see both ends of the stick at the smallest possible distance from your eye.
* Fix the vertical stick at this position with the help of the clamp.
* Without changing the position of the clamp on the horizontal rod, replace this stick of 30 cm length with other sticks of various lengths one by one and try to see the top and bottom of the stick simultaneously without any change in the position of eye either upwards downwards or side ways.
* Observe the figure (b) you can see the whole object AB which is at a distance of 25 cm (least distance of distinct vision) because the rays coming from the ends A and B of the object AB will enter the eye.
* Similarly you can also see whole object CD with eye as explained above. Let us assume that AB moves closer to the eyes to a position A'B' as shown in figure (b).
* From the figure (b) you notice that you will be able to see only the part (EF) of the object A' B' because the rays coming from E and F enter your eye. The rays coming from A' and B' cannot enter your eye.
* The rays coming from the extreme ends of an object from an angle at the eye. If this angle is below 60°, we can see the whole object. If this angle is above 60°, then we can see only the part of the object.
* This maximum angle, at which we are able to see the whole object is called angle of vision. The angle of vision for a healthy human being is about 60°. It varies from person to person and with age.
* You have learnt that the value of least distance of distinct vision is about 25 cm and the value of angle of vision of human beings is about 60°. You also learnt that these values change from person to person and with age of person.

 

Activity 3:
3. How can you explain the phenomenon of dispersion of light by an activity? (AS - 3) (4 Marks)
Ans: * Do this experiment in the dark room. Take a prism and place it on the table near a vertical white wall.
* Take a thin wooden plant. Make a small hole in it and fix it vertically on the table.
* Place the prism between the wooden plank and wall. Place a white light source behind the hole of the wooden plank. Switch on the light.
* The rays coming out of the hold of the plank become a narrow beam of light.
* Adjust the height of the prism such that the light falls on one of the lateral surfaces.
* Observe the changes in emerged rays of the prism. Adjust the prism by slightly rotating it till you get an image on the wall.
* You observe a spectrum of seven colours on the wall.
* White light in split up into seven colours (VIBGYOR) by the prism.
* This phenomenon of splitting of white light into different colours is called dispersion of light.
* You can note that the angle of deviation is different for each colour.
* The angle of deviation is maximum for violet colour and minimum for red colour.

 

Activity 4:
4. Describe an activity for the formation of different colours of light with the help of white light. (AS - 3) (2 Marks)
Ans: * Take a metal tray and fill it with water.
* Place a mirror in the water such that it makes an angle to the water surface.
* Now focus white light on the mirror through the water as shown in figure.
* Try to obtain colour on a white card board sheet kept above the water surface.
* Note the names of the colours could see in your book.
* We observe different colours on the wall. These colours are the same as we observe in a rainbow.
* VIBGYOR Colours are observed.


 

Activity 5:
5. How would you form the seven colours of light in Sun light on a white wall? Describe an activity.  (AS - 3) (2 Marks)
Ans: * Select a white coloured wall on which the Sun rays fall. Stand in front of a wall in such a way that the Sun rays fall in your back.
* Hold a tube through which water is flowing. Place your finger in the tube to obstruct the flow of water.
* Water comes out through the small gaps between the tube and your finger like a fountain.
* Observe the changes on the wall while the water shower is maintained.
* We observe the seven colours on the wall. These colours are VIBGYOR.

 

Activity 6:
6. Describe an activity to explain the phenomenon of scattering. (AS - 3) (4 Marks)
Ans: * Take a solution of sodium-thio-sulphate (hypo) and sulphuric acid in a glass beaker. Place the beaker in an open place where abundant Sun light is available. Watch the formation of grains of sulphur and observe changes in the beaker.
* We will notice that sulphur precipitates as the reaction is in progress. At the beginning, the grains of sulphur are smaller in size and as the reaction progresses, their size increases due to precipitation.
* Sulphur grains appear blue in colour at the beginning and slowly their colour becomes white as their size increases.
* The reason for this is scattering of light. At the beginning, the size of grains is small and almost comparable to the wave length of blue light. Hence they appear blue in the beginning.
* As the size of grains increases, their size becomes comparable to wave lengths of other colours. As a result of this, they act as scattering centres for other colours.
* The combination of all these colours appears as white.

 

Additional Questions

I. Conceptual Understanding

1. What is least distance of distinct vision? Mention its value?  (2 Marks)
Ans: * The least distance at which an object is to be held to see it comfortably and distinctly is called the least distance of distinct vision.
* Generally the value of least distance of distinct vision is 25 cm. However it varies from person to person and with age.

 

2. What is angle of vision? What is its value? (2 Marks)
Ans: *  The maximum angle, at which we are able to see the whole object is called angle of vision.
*  The angle of vision for a healthy human being is about 60°. It varies from person to person and with age.

 

3. Describe the basic components of human eye. (4 Marks)
Ans: *  The eye ball is nearly spherical in shape. The front portion is more sharply curved and is covered by a transparent protective membrane called 'cornea'. This portion is visible outside.
*  Behind the cornea, there is a place filled with a liquid called aqueous humour and behind this a crystalline lens which is responsible for image formation.
*  Between the aqueous humour and the lens, we have a muscular diaphragm called 'iris' which has a small hole in it called pupil.
*  Iris is the coloured part that we see in an eye.

 

4. Explain the role of 'iris' in eye?  (2 Marks)
Ans: *  Iris helps in controlling the amount of light entering the eye through pupil.
*  In low light condition, Iris makes the pupil to expand so that more light is allowed to go in.
*  In excess light condition, it makes the pupil contract and thereby prevent the excess light not to go in to eye.
* Thus 'iris' enables pupil to act as 'variable aperture' for entry of light into the eye.

 

5. Why should the eye lens change its shape? (2 Marks)
Ans: * For different positions of object, the image distance remains constant only when there is a change in focal length of lens.
* The focal length of a lens depends on the material by which it has been made and radii of curvature of lens.
* We need to change focal length of eye lens to get same image distance for various positions of object in front of the eye.
* This is only possible when the eye lens is able to change its shape.

 

6. Describe the role of retina to perceive the object?  (2 Marks)
Ans: * The retina is a delicate membrane, which contains about 125 million receptors called 'rods' and 'cones'.
* These rods identify the colour and cones identify the intensity of light when light signal is received.
* These signals are transmitted to the brain through about 1 million optic-nerve fibres.
* The brain interprets these signals and finally process the information so that we perceive the object in terms of its shape, size, and colour.

 

7. Explain the term 'accommodation' relating to eye.   (2 Marks)
Ans: * The ciliary muscles adjust the focal length of eye lens in such a way that image is formed on retina to see the object clearly irrespective of its position.
* This process of adjusting focal length of eye lens is called 'accommodation'.

 

8. What is 'Myopia'? How is it corrected? (2 Marks)
Ans: * Some people cannot see objects at long distances but can see nearby objects clearly.
* This type of defect of vision is called 'Myopia' or 'near sightedness'.
* This defect of vision can be corrected by using a concave lens of suitable focal length.

 

9. What is 'far point' and 'near point'?   (2 Marks)
Ans: * Far point: The point of maximum distance at which the eye lens can form an image on the retina is called 'far point'.
* Near point: The point of minimum distance at which the eye lens can form an image on the retina is called near point.

 

10. What is hypermetropia? How is it corrected?  (2 Marks)
Ans: Hypermetropia: It is also called 'far sightedness'. A person with hypermetropia can see distant objects clearly, but cannot see objects at near distances.
Hyper metropia can be corrected by using a convex lens of suitable focal length.

 

11. Define intensity of light. (1 Mark)
Ans: The intensity of light is the energy of light passing through unit area of plane, taken normal to the direction of propogation of light in one second.

 

12. Describe the structure of the eye with a neat diagram.  (4 Marks)


Ans: * The eye ball is nearly spherical in shape.
* The front portion is more sharply curved and is covered by a transparent protective membrane called the cornea.
* It is this portion which is visible from outside. Behind the cornea, there is place filled with a liquid called aqueous humour and behind this a crystalline lens which is responsible for the image formation.
* Between the aqueous humour and the lens, we have a muscular diaphragm called 'iris' which has a small hole in it called pupil. Iris is the coloured part that we see in an eye.
* Iris helps in controlling the amount of light entering the eye through 'pupil'. It enables pupil to act as a 'variable aperture' for entry of light in to the eye.
* The light that enters the eye forms an image on the retina.
* The distance between the lens and retina is about 2.5 cm.

 

13. What is Presbyopia? How is it corrected?   (4 Marks)
Ans: * Presbyopia is vision defect when the ability of accommodation of the eye usually decreases with ageing.
* For most people the near point gradually recedes away. They find it difficult to see nearby objects clearly and distinctly.
* This happens due to gradual weakening of ciliary muscles and diminishing flexibility of the eye lens.
* This effect can be seen in aged people. Sometimes a person may suffer from both Myopia and hypermetropia with ageing.
* To correct this type of defect of vision we need bi-focal lenses which are formed using both concave and convex lenses. Its upper portion consists of the concave lens and lower portion consists of the convex lens.

 

14. Why does the sky appear dark if it is seen from above atmosphere? (2 Marks)
Ans: * Due to the atmospheric molecules, the Sun light is scattered and the sky appears blue.
* In the absence of atmosphere, no scattering of Sun light takes place and so the sky appears black.

 

15. What is scattering? Explain the process of scattering?   (4 Marks)
Ans: Scattering: The process of reemission of absorbed light in all directions with different intensities by atoms or molecules, is called scattering of light.


* Let us consider that the free atom or free molecule is somewhere in space as shown in figure.
* Light of certain frequency falls on that atom or molecule. This atom or molecule responds to the light whenever the size of the atom or molecule is comparable to the wavelength of light.
* If this condition is satisfied, the atom absorbs light and vibrates. Due to these vibrations, the atom re-emits a certain fraction of absorbed energy in all directions with different intensities.
* The re-emitted light is called scattered light and the process of re-emission of light in all directions with different intensity is called scattering of light.
* The atoms or molecules are called scattering centre.

 

16. A glass prism causes dispersion of white light. Why the same thing cannot be done by a glass slab?  (4 Marks)
Ans: * In a prism refraction of light occurs at two plane surfaces inclined to each other at an angle.
* Dispersion of white light takes place at one the first surface of the prism and the constituent colours are deviated through different angles.
* At the second surface, the different split colours undergo refraction but get further separation.
* In the case of a rectangular glass slab, the refraction process takes place at two parallel surfaces.
* Though there is split of white light is to constituent colours at the first surface on refraction, these split colours when undergo refraction at the second surface emerge out as a parallel beam, in the form of white light only.

II. Asking Questions and Making Hypothesis

17. Why a rainbow is formed in the sky after rain? (2 Marks)
Ans: * A rainbow is a natural spectrum of sun light formed in the sky after rain.
* The rainbow is formed due to reflection, refraction and dispersion of Sun light incident on tiny rain droplets present in the atmosphere.

 

18. Explain why a rainbow always forms opposite to the direction of Sun. (1 Mark)
Ans: A rainbow always forms opposite to the direction of sun because the Sun light is to be incident on the rain droplets.

 

19. When a monochromatic ray of light ray is sent through a Prism, does it split in to different colours? Give reasons for your answer.
Ans: * Monochromatic ray of light means a single colour. The frequency of this colour cannot be changed by any medium.
* Due to refraction in the Prism, the frequency of the monochromatic ray of light does not change.
* So the monochromatic ray of light passing through the prism can not change its colour and emerges out of the prism with no change of colour of the ray of light.

 

20. How does a rainbow appear for an observer in an aeroplane flying high? (2 Marks)
Ans: * An observer in an aeroplane observes the rainbow as a complete circle.
* A rainbow is a three dimensional cone of dispersed white light of the Sun and it appears as a circle which is the base of the cone.

 

21. Do the rainbow appear the same for two observers at different places viewing it?  (2 Marks)
Ans: * All the rain droplets that disperse the Sun light to form the rainbow lie with in a cone of semi vertical angle of 40° to 42°.
* As the observers are at different places they observe the different parts of the rainbow on the surface of the cone.
* As the observers view the rainbow from the vertex of the cone, the rainbow appear to each observer will be different.

 

22. Why does the Sun light dispersed by the rain droplets appear as a bow?  (4 Marks)
                        
Ans: * A rainbow is not the flat two dimensional arc as it appears to us.
* The rainbow we see is actually a three dimensional cone with the tip at our eye as shown in the figure.
* All the drops that disperse the light towards us lie in the shape of the cone - a cone of different layers.
* The drops that disperse red colour are on the outer most layer of the cone, similarly the drops that disperse orange colour are on the layer of the cone beneath the red colour cone.
* In this way the cone responsible for yellow lies beneath orange and so on till the violet colour cone becomes the innermost cone.

V. COMMUNICATION THROUGH DRAWING MODEL MAKING

20. With the help of a diagram define the following terms with respect to a prism.  (4 Marks)
a) Incident ray, refracted ray, emergent ray
b) Angle of incidence, Angle of emergence
c) Angle of deviation
d) Normal


Ans: a) PQR is the trace of the Prism.
* Incident ray: The ray of light AB is incident on the plane surface PQ. This ray is called the incident ray.
* Refracted ray: The incident ray AB is refracted along MN. This ray MN is called the refracted ray and it moves through Prism.
* Emergent ray: The refracted ray MN comes out of the surface PR of Prism as an emergent ray along CD. So CD is emergent ray.
b) * Angle of incidence: A perpendicular is drawn at M to the surface PQ of prism.
This is the normal to the surface PQ. The angle between the incident ray and normal at M is called angle of incidence (i1).
* Angle of emergence: A perpendicular is drawn at N to the surface PR. This is the normal to surface PR. The angle between the emergent ray and normal is called angle of emergence (i2).
c) Angle of deviation: The angle between the incident ray and emergent ray is called angle of deviation (d).
d) Normal: A perpendicular drawn to the surface of the prism at any point is normal to that surface at that point.

 

21. Describe a triangular prism with the help of a diagram.               (4 Marks)
Ans: * Put the prism on a white chart in such a way that the triangular base is on the chart.
* Draw a line around the prism using a pencil. Remove the prism. It is a triangle PQR as shown is the diagram.
* A prism is a transparent medium separated from the surrounding medium by atleast two plane surfaces which are inclined at a certain angle in such a way that, light incident on one of the plane surfaces emerges from the other plane surface.
* A triangular glass prism contains two triangular bases and three rectangular plane lateral surfaces. These lateral surfaces are inclined to each other.

 

VI. APPRECIATION AND AESTHETIC SENSE, VALUES


22. How do you appreciate the working of ciliary muscles in the eye? (4 Marks)


Ans: * The ciliary muscles to which eye lens is attached helps the eye lens to change its focal length by changing the radii of curvature of the eye lens.
* When the eye is focused on a distant object, the ciliary muscles are relaxed. So that the focal length of eye lens has its maximum value.
* The parallel rays coming into the eye are then focussed on to the retina and we see the object clearly.
* When the eye is focussed on a closer object, the ciliary muscles are strained and focal length of eye lens decreases.
* The ciliary muscles adjust the focal length in such a way that the image is formed on retina and we see the object clearly. This process of adjusting focal length is called 'accommodation'.
* Really this 'accommodation' is a wonderful phenomenon through which we are able to see the distant and near objects.
* If this mechanism of ciliary muscles is not present, the eye lens cannot adjust its focal length and we cannot see the objects beyond a certain distance.
* If we imagine this, we cannot guess our normal life.
* Hence the role of ciliary muscles is highly appreciable.

 

23. How do you appreciate the twinkling of stars in the sky? Explain the reason for twinkling.  (2 Marks)
Ans: * The twinkling of stars is a good watch for the observers and I appreciate the beauty of nature.
* Infact the twinkling of stars is due to the refraction of the light from the stars by the continuously changing atmosphere.
* Due to varying atmospheric temperature and density the light from the stars is refracted by varying amounts in different directions.
* As a result of this the star light reaches the eye of the observer in one moment and slips off the next moment like that. This gives the impression that the stars are twinkling for the observer.

 

24. The Sunrise and Sunset appear to be a gift of nature. How do you appreciate this. Explain.  (4 Marks)
Ans: * Infact the Sun appears in the sky two minutes before the actual sunrise and also remains visible two minute after the real sunset.
* The Sunrise actually takes place when the Sun is just above the horizon.
* Similarly the Sunset actually takes place when the Sun set just below the horizon.
* Due to refraction of Sun light by the atmosphere the Sun appears two minutes before the Sunrise and the Sun appears two minutes after the actual Sunset.
* This is a wonderful gift of nature and I appreciate the phenomenon of refraction responsible for this.

 

VII. APPLICATION TO DAILY LIFE, CONCERN TO BIO DIVERSITY

25. 'A Short sighted person can easily read a book without wearing spectacles. Justify this.  (2 Marks)
Ans: * A short sighted person can easily see the near objects.
* The person has difficulty in observing far off objects.
* So a short sighted person can easily read a book without wearing spectacles as it in a near object.

 

26. Why is red light used as danger signal? (2 Marks)
Ans: * In the visible light, among all colours, the red colour has more wavelength and hence it is least scattered by the atmosphere or mist.
* So red light can easily penetrate and pass through smoke, mist or fog without getting scattered.
* So red light can be seen from long distances irrespective of atmospheric conditions.
* So red colour is used universally as a danger light signal.

 

27. “From the chimney of a coal burning industry the smoke released appears blue on a mist day”. Explain the reason. (2 Marks)
Ans: * On a misty day, the atmosphere contains more number of tiny water droplets, dust, and carbon particles.
* These particles scatter the blue colour of white light passing through it.
* This scattered blue colour light reaches the eye of the observer, giving the impression that the smoke is blue coloured.

 

28. Mention two examples, which appear blue on a misty day.          (2 Marks)
Ans: * The hills at a distance with a crowd of trees appear blue for an observer on a misty day.
* The smoke of agarbatti on a misty day appear to be blue in colour.

 

29. “Why Orange lights are arranged for vehicles which move in areas where frequent fog is present.   (4 Marks)
Ans: * In areas where frequent fog is present more number of water droplets are in the atmosphere.
* In such areas if vehicles are provided with white light, it is scattered by the water droplets in the atmosphere and for the vehicle driver the path is not clearly visible.
* For vehicle drivers, driving becomes very difficult.
* As the wavelength of orange light is comparatively more, it is not scattered much and as such the path becomes more visible.
* So for such vehicles orange lights are arranged.

 

30. What is tyndall effect? Give examples of tyndall effect you experience in your daily life.  (4 Marks)
Ans: a) Tyndall effect: The phenomenon of scattering of white light by colloidal particles is known as 'Tyndall effect'.
* Sun light passing through the trees and falling on the earth. The path of the rays of light is visible due to smoke or tiny droplets of water scattering the Sun light.
* In cinema theatres the path of light from the projector to the screen is visible due to the smoke present in the touring cinima theatres.

 

31. 'Why are lenses used in spectacles to correct the defects of vision'? (2 Marks)
Ans: * Our eye may have defects of vision. Like
a) Myopia b) Hyper metropia c) Presbyopia
* Some times the eye may gradually lose its ability for accommodation. Due to this the above mentioned defects of vision arise.
* A concave lens is used for correction of Myopia a convex lens is used for correction of hypermetropia to correct presbyopia bi-focal lenses which are formed using both concave and convex lenses are used.
* These lenses are fixed in the spectacles to overcome these defects of vision.

Solved Examples

32. If the doctor advised to use 1D lens, what is its focal length? (2 Marks)
Ans: * Power of the lens P = 1D
Formula: Power =  = 1D
∴ f =   = 100 cm
* Focal length of the lens = 100 cm

33. If a glass prism has an angle of 60° and 30° as angle of minimum deviation, calculate its refractive index.
Ans: Given: A = 60º, and D = 30º, n = ?


 

Writer: C.V.Sarveswara Sarma

Posted Date : 05-01-2021

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.

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