Important Questions
QUESTION AND ANSWERS
1. List out the apparatus required to find the refractive index of a prism experimentally. Explain the procedure with the help of a rough diagram. (AS 3) 4 Marks
A: Finding the refractive index of a prism
Material required: Prism, piece of white chart of size 20 × 20 cm, pencil, pins, scale and protractor.
Procedure:
‣ Let us 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.
‣ Let us draw a line around the prism, using a pencil, having vertices P, Q and R and remove the prism.
‣ Measure the angle between PQ and QR which gives the angle of prism (A).
‣ Let us consider a light ray 'AB' incident on at 'M'. Draw a 'normal' at 'M'.
‣ Let us mark 'M' on PQ and draw a perpendicular to PQ at M.
‣ Let us mark an angle of 30° and draw a line 'AB' at 'M' which gives the incident ray. The angle of incidence is 'i'.
‣ Fix two pins vertically on the line AB.
‣ Now let us look for the images of two pins through the prism on other side and fix another two pins say 'C' and 'D'.
‣ Remove the prism and draw a line to PR which pass through C and D points. This line gives emerging ray.
‣ Draw a normal to 'PR' and measure the angle between CD and normal, which gives the angle of emergence (i2).
‣ Now extend, the both incident and emergent ray till they meet at a point 'O'.
‣ Measure the angle between the extended two rays which gives angle of deviation (d).
‣ As the angle of incidence changes angle of deviation also changes. Repeat this procedure for various angles of incidence such as 40°, 50°, 60° etc.
‣ Now tabulate the reading of angle of incidence (i1), angle of emergence (i2) and angle of deviation (d).
‣ As the angle of incidence is increases, angle of deviation decreases and attains a minimum value (angle of minimum deviation) and further it increases with increasing in angle of incidence.
‣ By taking angle of incidence along X - axis and the angle of deviation along Y - axis draw a graph and from this graph find the angle of minimum deviation D.
‣ Angle of prism is A, Angle of minimum deviation is D
‣ Measure the angle in between PQ, QR and represent it as 'A'.
2. 
The given above diagram shows that the light ray refracting the prism then answer the following questions.
i) Identify the incident and emergent rays? (AS 4) 4 Marks
A: AB is the incident ray and CD is the emergent ray.
ii) Where is angle of deviation? How do you find the angle of deviation?
A: Angle of deviation is 'd'. Extent both incident and emergent rays till they meet at a point 'O'. The angle in between of these two rays is angle of deviation.
iii) If it is an equilateral prism what is the angle of prism?
A: 60°
iv) What does the line ABMNCD indicate?
A: The path of light ray.
3. Write the common defects of vision and explain with diagrams? How can you correct these visual defects by using lens? (AS 1) 4 Marks
A: Sometimes the eye may gradually lose its ability for accommodation. In such conditions the person cannot see an object clearly and comfortably. The vision becomes blurred due to defects of the eye lens. There are mainly three common defects of vision. They are (i) Myopia (ii) Hypermetropia (iii) Presbyopia
Myopia:
‣ Some people cannot see objects at long distances but can see nearly objects clearly. This type of defect in vision is called "Myopia".
‣ For these people the maximum focal length is < 2.5 cm.
‣ Image of distance objects forms before the retina.
‣ The least distance of vision (L) is generally 25 cm.
‣ M is a point of ''far vision" to myopia. He/ She can see the object in between L and M.
Correction Myopia:
‣ The eye lens can form clear on the retina when an object is placed between far point and least point (M and L) of distinct vision.
‣ If we are able to bring the image of the object kept beyond far point, between the far point and the point of least distance of distinct vision using a lens, this image acts as an object for the eye lens.
‣ This can be made possible only when concave lens is used
Hypermetropia:
‣ A person with hypermetropia can see distant objects clearly but cannot see objects at near distances, because the minimum focal length of eye lens for the person of hypermetropia is greater than 2.27 cm.
‣ The image of object nearly forms beyond the retina.
‣ If H is the near point and L is the point of least vision, then he can see the objects beyond the point H.
Correction of Hypermetropia:
‣ Eye lens can form a clear image on the retina when any object is placed beyond near point.
‣ 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 (H) and least distance of distinct vision (L).
‣ This is possible only when a double convex lens is used.
Presbyopia:
‣ 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 near by objects clearly and distinctly.
‣ This happens due to gradual weaking of ciliary muscles and diminishing flexibility of the eye lens. This effect can be seen in aged people.
‣ Some times a person may suffer from both myopia and hypermetropia with ageing.
Correction of Presbyopia:
‣ To correct this type of defect of vision we need bifocal 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.
4. Kishore uses spectacles when you look at him through his spectacles, the size of eyes appears bigger than the original size of eyes.
i) What is the nature of lens used by him?
ii) Explain the eye defect with a ray diagram. (AS 1) 4 Marks
A: Magnification > 1 is possible with convex lens only.
i) So Kishore's spectacles are convex lens in which the image what we have seen is virtual, erect, magnified. The object kept in between O and F of convex lens.
ii) The defect is hypermetropia also known as far sightedness. The person can see distant object clearly but can not see objects at near distances. The image formed beyond the retina. So, using convex lens the rays can be converged on retina.
5. Draw the diagram showing how to correct the Myopia by using concave lens? (AS 5) 4 Marks
A:
6. Suggest an experiment to produce a rainbow in your classroom and explain the procedure. (AS 3) 4 Marks
A: Aim: To produce rainbow.
Apparatus: Metal tray, water, mirror.
Procedure:
* 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 the white light on the mirror, through the water as shown in the figure.
* Try to obtain colour on a white card board sheet kept above the water surface.
* Note the names of the colours you could see in note book.
* We know that white light is splitting into certain different colours as rainbow.
7. Sky appears sometimes white and sometimes blue in colour, and sun appears red in colour during sunrise and sunset. State the reason. (AS 1) 4 Marks
A: * The sky appears sometimes white when you view in certain direction on hot days.
* On the 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 our eye and the sky appear white.
Sky appears in blue colour: Sky appears in blue colour due to scattering of light. Atoms or molecules which are exposed to light absorb light energy and emit some part of the light energy in different directions. This is the scattering of light.
* We know that our atmosphere contains different types of molecules and atoms. The reason for blue sky is due to the molecules 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.
The sun appears the red colour during sunrise and sunset:
* Sun light has to travel more distance through the atmosphere during sunset/ sunrise than during other times of the day. A molecule which scatter a light of given colour when its size is comparable to the wavelength of that colour. The number of molecules that can scatter red light is less. So red colour light travels through the atmosphere unscattered where as other colour scatters away making the sun look red during sunset and sunrise.
* During noon hours sun light travels less distance through the atmosphere than the morning and evening times. Therefore all the colours reach us without much scattering thus making the noon sun to appear white.
8. When we observe the rainbow in the sky most of the times appears to be semi circular. Explain. (AS 1) 4 Marks
A:
* 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 of your eye. All the drops that disperse the light towards we lie in the shape of the cone - a cone of different layers. The drops that disperse orange colour to our eye are on the layer of the cone beneath the cone that disperse red colour. In this way the cone responsible for yellow lies beneath orange and so on it continues till to violet colour which the innermost cone. In rainbow if the angle of vision is below 60° we can see the whole object, if the angle of vision is more than 60° than we can see only the part of the object. Due to this the rainbow looks like semi circular shape.
9. Image is formed always on the retina for the object which is at any distance before the eye. Answer how we can get the same image distance for various positions of objects using concepts of refraction through lenses? (AS 1) 4 Marks
A: The focal length of a lens depends on the material by which it has been made and radii of curvature of lens. For different positions of object, the image distance remain constant only when there is a change in focal length of lens. This is possible only when the eye lens is able to change its shape.
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 focussed 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.
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.
10. What is the function of iris in human eye? (AS 1) 1 Mark
A: It controls the light which enters into eye by pupil.
11. Write the differences between cones and rods of human eye. (AS 1) 2 Marks
A: Rods:
* Respond to intensity of light
* Enables to see in dim light
* can not distinguish various colours
Cones:
* Respond to colour
* Become active in bright light
* It can distinguish between various colours.
12. What is the nature of the image formed on the retina? (AS 1) 1 Mark
A: Real, inverted and diminished image is formed on the retina.
13. What kind of lens is present in an eye? (AS 1) 1 Mark
A: A convex lens made up of a transparent jelly like proteinious material is present in our eye.
14. Why sun does not appear red during noon hours? (AS 6) 1 Mark
A: During noon hours, the distance to be travelled by the sun rays in the atmosphere is less than that compared to morning and evening hours. Therefore all colours, reach our eye without much scattering. Hence the sun appears white during noon hours.
15. Explain the phenomenons (i) Dispersion of light (ii) Scattering of light. (AS 1) 2 Marks
A: i) Dispersion of light: The splitting of white light into different colours (VIBGYOR) is called dispersion.
e.g.: Formation of rainbow.
ii) Scattering of light: Atoms or molecules which are exposed to light absorbs light energy and emit some part of the energy in different directions. This is the basic process happens in scattering of light.
e.g.: This is the reason for the appearance of clear blue colour when look at the sky in a direction perpendicular to the direction of the sun rays.
PROBLEMS - SOLUTIONS
1. A boy uses spectacles of focal length -50 cm. Name the defect of vision he is suffering. Compute the power of the lens. (AS 1) 2 Marks
Sol: As the focal length is negative the lens is concave. The boy is suffering from myopia.
Here f = -50 cm
2. Doctor advised to use 2D lens. What is its focal length? (AS 1) 2 Marks
Sol: Given that power of lens P = 2D
f = 50 cm.
3. The far point of a myopic person is 100 cm in front of the eye. What is the nature and power of the lens required to correct the defect? (AS 1) 4 Marks
Sol: The defect called myopia is corrected by using a concave lens.
u = ∞
v = -100 cm
f = ?
Thus, the focal length of the required concave lens is - 100 cm
= -1 D
So, the power of concave lens is required is -1 D.
4. The power of lens is +1.5 D. What kind of lens it is? And what is its focal length? (AS 1) 2 Marks
Sol: P = +1.5 D
As the power of lens is positive, so it is a convex lens.
f = 0.66 m
5. What are the maximum and minimum focal lengths of the eye lens? How can we find them? (AS 1) 4 Marks
Sol: Maximum focal length is 2.5 cm and minimum focal length is 2.27 cm.
(i) When the object is at infinity
u = -∞
v = 2.5 cm
ii) When the object is at a distance of 25 cm from eye.
u = -25 cm
v = 2.5 cm
6. A person wears glasses of power -2.5 D. Is the person far-sighted (or) near sighted, what is the far point? (AS 1) 4 Marks
Sol: P = -2.5 D
-'ve' sign shows he is suffering myopia.
7. The near point of a person suffering from hypermetropia is 75 cm. Calculate the focal length and power of the lens required to enable him to read the news paper which is kept at 25 cm from the eye. (AS 1) 2 Marks
Sol: u = -25 cm
v = -75 cm
8. A prism with an angle A = 60° produces an angle of minimum deviation of 30°. Find the refractive index of material of the prism.
Sol: Given that A = 60°
D = 30°
SHORT QUESTIONS
1. What is the principle on which human eye functions? (AS1) 1 Mark
Ans: Principle of sensation of vision.
2. What is the least distance of distinct vision for common man? (AS1) 1 Mark
Ans: 25 cm.
3. What is the least distance of distinct vision for child below 10 years? (AS1) 1 Mark
Ans: 7 to 8 cm.
4. What is the least distance of distinct vision for old aged people? (AS1) 1 Mark
Ans: 1 to 2 m.
5. What is the angle of vision for a healthy human being? (AS1) 1 Mark
Ans: 60°
6. What are the observations from the figure regarding angle of vision?
(AS4) 1 Mark
Ans: i) A'B' is partially observed.
Only EF part is observed.
ii) AB and CD are completely observed.
7. What happens when an object is seen with angle of vision below 60°? (AS2) 1 Mark
Ans: We can see the complete part of the object.
8. What happens when an object is seen with angle of vision above 60°? (AS2) 1 Mark
Ans: We can see the partial part of the object.
9. What is a protective membrane of eye? (AS1) 1 Mark
Ans: Cornea.
10. Which part of eye control the light entering the eye? (AS1) 1 Mark
Ans: Iris.
11. Which is the coloured part that is seen in the eye? (AS1) 1 Mark
Ans: Iris.
12. Why does pupil appear black in colour? (AS1) 1 Mark
Ans: When any light falling on it goes into eye and their is almost no chance of light coming back to the outside.
13. What is the distance between the lens and retina in eye? (AS1) 1 Mark
Ans: 2.5 cm.
14. Which part of eye is used to change radii of curvature of eye lens? (AS1) 1 Mark
Ans: Ciliary muscles.
15. What is the process of adjusting the focal length of eye lens? (AS1) 1 Mark
Ans: Accomidation.
16. What is nature and type of image is obtained on retina in human eye? (AS1) 1 Mark
Ans: Real and inverted.
17. What is the role of rods and cones in retina? (AS1) 1 Mark
Ans: Rods are used to identify intensity of light. Cones are used to identify the colours.
18. What is the minimum and maximum focal length of human eye lens? (AS1) 1 Mark
Ans: Fmin = 2.27 cm
Fmax = 2.5 cm
19. What happens, if the human eye lens have maximum focal length less than 2.5 cm?
(AS2) 1 Mark
Ans: Myopia.
20. What happens, if the human eye lens have minimum focal length greater than 2.27 cm? (AS2) 1 Mark
Ans: Hypermetropia.
21. What type of eye defect is seen for the aged people? (AS1) 1 Mark
Ans: Presbyopia.
22. How can you rectify presbyopia? (AS1) 1 Mark
Ans: Bi−focal lens.
23. In finding refractive index of prism experiment at what situation 
i1 = i2? (AS1) 1 Mark
Ans: Critical angle.
24. What are the precautions we have to take while performing experiment in finding the refractive index of prism? (AS3) 1 Mark
Ans: i) Pins are to be of same height.
ii) In observation of the pins, we should observe without any parallax error.
iii) Prism should not be disturbed through out the experiment.
25. What is the phenomenon of splitting of white light into different colours? (AS 1) 1 Mark
Ans: Dispersion of light.
26. According to which principle light ray always chooses the path of least time? (AS1) 1 Mark
Ans: Fermat's principle.
27. In VIBGYOR which colour as low refractive index? (AS1) 1 Mark
Ans: Red
28. What are the various phenomenon involved in formation of rainbow with a water drops? (AS1) 1 Mark
Ans: Refraction, Dispersion of light, total internal refraction.
29. What should be the angle between the incoming and out going rays of the water droplet the violet colour is seen? (AS1) 1 Mark
Ans: 40°
30. If the angle between the incoming and outgoing rays of the water droplet is 42, which colour is seen? (AS1) 1 Mark
Ans: red.
31. What is the name of the phenomenon used in figure? (AS5) 1 Mark
Ans: Scattering of light.
32. Due to the scattering of which molecules present in atmosphere are responsible for appear sky in blue colour? (AS6) 1 Mark
Ans: N2 and O2.
33. Draw the graph between angle of incidence and angle of deviation from the experiment in finding the refractive index of the prism? (AS5) 2 Marks
Ans:
34. Draw the diagram of scattering of light. (AS5) 2 Marks
Ans:
35. Does rainbow form always opposite to sun? Why? (AS1) 2 Marks
Ans: Yes. Arainbow is always formed in the opposite direction of the Sun. As the Sun rays should fall on rain drops it happen so.
36. What happens to the image distance in the eye when we increase the distance of an object from the eye? (AS2) 2 Marks
Ans: In the eye the image distance (distance between eye lens and retina) is fixed and cannot be changed, so when we increase the distance of an object, there is no change in the image distance.
37. When a monochromatic light passes through a prism will it show dispersion?
(AS2) 1 Mark
Ans: No, it will not show any dispersion but shows deviation.
38. A short − sighted person may read a news paper without spectacles. Comment.
(AS1) 2 Marks
Ans: The statement is true, because a short sighted person has difficulty in observing far off objects.
39. Write the difference between 'Myopia' and 'Hypermetropia'? (AS1) 2 Marks
Ans: i) The eye defect in which people cannot see at long distances but can see near by objects clearly is called myopia.
ii) The eye defect in which people cannot see near distant objects but can see distant objects is called hypermetropia.
40. What will be colour of the sky in the absence of atmosphere? (AS2) 2 Marks
Ans: In the absence of any atmosphere there will be no scattering of sunlight and the sky will appear dark.
41. What is the cause of blue colour of the ocean? (AS1) 2 Marks
Ans: The water molecules of the ocean scatter blue light strongly than light of other colours. So the ocean appears bluish.
