LAQ for Light Reflection and Refraction Class 10 Science NCERT
Important Questions1
(a) Define the following terms in the context of spherical mirrors :
(i) Pole
(ii) Centre of curvature
(iii) Principal axis
(iv) Principal focus
(b) Draw ray diagrams to show the principal focus of a :
(i) Concave mirror
(ii) Convex mirror
(c) Consider the following diagram in which M is a mirror and P is an object and Q is its magnified image formed by the mirror.
State the type of the mirror M and one characteristic property of the image Q.
Answer
(a) (i) Pole – Centre of the reflecting surface of the mirror.
(ii) Centre of curvature – The centre of the hollow sphere of which the reflecting surface is a part.
(iii) Principal axis - Straight – line passing through the pole and the centre of curvature of a spherical mirror.
(b)
(c) Concave mirror
Image formed is virtual.
2
(a) To construct a ray diagram we use two rays which are so chosen that it is easy to know their directions after reflection from the mirror. List two such rays and state the path of these rays after reflection in case of concave mirrors. Use these two rays and draw ray diagram to locate the image of an object placed between pole and focus of a concave mirror.
(b) A concave mirror produces three times magnified image on a screen. If the object is placed 20 cm in front of the mirror, how far is the screen from the object?
Answer
(a) Two light rays whose path of reflection are known are :
(i) The incident ray passes through the centre of curvature: In this case, light after reflecting from the concave mirror moves back in the same path. This happens because light is incident perpendicular on the mirror surface.
(ii) The ray incident obliquely to the principal axis: In this case, the incident ray will be reflected back by the reflecting surface of the concave mirror obliquely and making equal angles with the principal axis.
Let an object is placed between the focus and pole of the concave mirror. Then using above two rays, image of the candle can be located as shown below:
The image is formed behind the mirror. The image is virtual, erect and magnified.
(b) Given, m = -3, u = -20 cm, v = ?
As we know,
m = - v/u
⇒ -3 = -(v/-20)
⇒ v = - 60 cm
The screen is placed in front of the mirror at a distance of 60 cm from the pole of the mirror. Thus, the screen is placed 40 cm away from the object.
3
Suppose you have three concave mirrors A, B and C of focal lengths 10 cm, 15 cm and 20 cm. For each concave mirror, you perform the experiment of image formation for three values of object distances of 10 cm, 20 cm and 30 cm. By giving reason, answer the following :
(i) For the three object distances, identify the mirror/mirrors which will form an image of magnification –1.
(ii) Out of the three mirrors identify the mirror which would be preferred to be used for shaving purposes/makeup.
(iii) For the mirror B draw ray diagram for image formation for object distance 10 cm and 20 cm.
Answer
(i) fa = 10 cm; fb = 15 cm; fc = 20 cm
u1 = 10 cm; u2 = 20 cm; u3 = 30 cm
(i) Mirror A will form image of m = -1 for object distance 20 cm since u2 = 2fa
(ii) Mirror B or C – distance should be less than focal length for erect and magnified image, face is generally kept at a distance more than 10 cm.
(iii) 
4
(a) If the image formed by a mirror for all positions of the object placed in front of it is always diminished, erect and virtual; state the type of the mirror and also draw a ray diagram to justify your answer. Write one use such mirrors are put to and why.
(b) Define the radius of curvature of spherical mirrors. Find the nature and focal length of a spherical mirror whose radius of curvature is + 24 cm.
Answer
(a) A convex mirror forms an erect, diminished and virtual image for all positions of the object placed in front of
Convex mirrors are commonly used as rear-view mirrors in vehicles as they always give an erect, diminished image.
(b) Radius of curvature is the radius of sphere (imaginary) of which spherical mirror is a part. It is a part. It is represented by ‘R’.
R = + 24 cm
∴ It is a convex mirror.
F = R/2 = 24/2 = + 12 cm
Thus, the focal length of a convex spherical mirror is 12 cm.
5
It is desired to obtain an erect image of an object, using concave mirror of focal length of 12 cm.
(i) What should be the range of distance of an object placed in front of the mirror ?
(ii) Will the image be smaller or larger than the object ? Draw ray diagram to show the formation of image in this case.
(iii) Where will the image of this object be, if it is placed 24 cm in front of the mirror ? Draw ray diagram for this situation also to justify your answer.
Show the positions of pole, principal focus and the centre of curvature in the above ray diagram.
Answer
(i) Range of distance between 0 cm to 12 cm
(ii) Larger than the object.
(iii) Image also at 24 cm in front of the mirror,
6
A student has focused the image of a candle flame on a white screen using a concave mirror. The situation is as given below:
Length of the flame = 1.5 cm
Focal length of the mirror = 12 cm
Distance of flame from the mirror = 12 cm
Distance of flame from the mirror = 18 cm
If the flame is perpendicular to the principal axis of the mirror, then calculate the following :
(i) Distance of the image from the mirror
(ii) Length of the image
If the distance between the mirror and the flame is reduced to 10 cm, then what would be observed on the screen? Draw ray diagram to justify your answer for this situation.
Answer
(i) h = + 1.5 cm; f = - 12 cm; u = -18 cm; v = ? h’ = ?
(a) 1/f = 1/v + 1/u
∴ 1/v = 1/f – 1/u
= 1/(-12) – 1/(-18)
= -1/12 + 1/18
= (-3 + 2)/36
= -1/36
∴ v = - 36 cm
(b) h’ = - v/u × h
= -(-36 cm/-18 cm) × 1.5 = -3 cm
(Magnified inverted image)
(ii) if, u = -10 cm
No distinct image would be formed on the screen. In this case, the image formed will be virtual (object will be between focus & pole)
(iii) 
7
A student wants to project the image of a candle flame on the walls of the school laboratory by using a mirror.
(i) Which type of mirror should he use and why ?
(ii) At what distance, in terms if focal length ‘f’ of the mirror, should he place the candle flame to get the magnified image on the wall ?
(iii) Draw a ray diagram to show the formation of the image in this case.
(iv) Can he use this mirror to project a diminished image of the candle flame on the same wall? State ‘how’ of your answer is ‘yes’ and ‘why not; if your answer is ‘no’.
Answer
(i) He should use a concave mirror as it forms real images.
(ii) He should place the candle flame between the focus and centre of curvature of the mirror to het the magnified image on the wall.
(i) The ray diagram for the formation of the image is shown below :
(ii) Yes, he can get a diminished image of the candle flame when the object is located at infinity.
8
(i) 4-5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of image and the magnification. Describe what happens as the needle is moved farther from the mirror.
(ii) What kind of mirror is used in a solar furnace? Give reason for using this mirror.
(iii) One half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Justify your answer.
Answer
(i) u = -12 cm
f = + 15 cm
Using mirror formula,
1/f = 1/u + 1/v
⇒ 1/v = 1/f – 1/u
⇒ 1/v = 1/15 – 1/-12
⇒ v = 60/9 = 6.7 cm
Magnification, m = -(-v/u) = h2/h1
= -6.7/-12
= 0.558
m1 = h2/h1
⇒ h2 = h1 × m
⇒ h2 = 0.558 × 4.5
⇒ h2 = 2.5 cm
As the needle is moved farther from the mirror, image moves to the focus and the size of image goes on decreasing.
(ii) Concave mirrors are used in solar furnaces, as they concentrate solar energy in the focal plane and help in attaining high temperatures.
(iii) When one half of a convex lens is covered with a black paper, the lens will produce a complete image of the object, but the intensity of the image is reduced because rays only from the top portion of the lens are refracted and form the image.
9
(i) An object is placed at a distance of 60 cm from a convex mirror where the magnification produced is ½. Where should the object be placed to h=get a magnification of 1/3 ?
(ii) A small electric lamp is placed at the focus of a convex lens. State the nature of beam of light produced by the lens. Draw a diagram to show this.
Answer
(i) u = - 60 cm
m = + ½
⇒ m = - v/u = ½
⇒ ½ = (-v/-60)
Using mirror formula
1/f = 1/u + 1/v = 1/60
⇒ f = 60 cm
Now, m = 1/3
∴ 1/3 = -v/u
⇒ 3v = -u
∴ 1/u + 1/v = 1/f
⇒ 1/u + (-3/u) = 1/60
∴ u = -120 cm
(ii) When a small electric lamp is placed at the focus of a convex lens, a parallel beam of light is produced by the lens.
Ray diagram:
10
(i) Between pole and focus of a concave mirror.
(ii) Between focus and centre of curvature of a concave mirror.
(iii) At the centre of curvature of a concave mirror.
(iv) Between infinity and pole of a convex mirror.
(v) At infinity from a convex mirror.
Answer
11
(i) Define optical center of a spherical lens.
(ii) A divergent lens has a focal length 20 cm. At what distance should an object of height 4 cm from the optical centre of the lens be placed so that its image is formed 10 cm away from the lens. Find the size of the image also.
(iii) Draw a ray diagram to show the formation of image in above situation.
Answer
(i) Optical centre: the central point of a lens.
(ii) h1 = 4 cm, v = -10 cm, u = ?, h2 = ?
1/f = 1/v – 1/u
⇒ 1/u = -1/10 + 1/20
= (- 2 + 1)/20
= -1/20
u = -20 cm
hi = v/u h0
= - 10 cm/ - 20 cm × 4
= 2 cm
(iii) 
12
(i) Define focal length of a spherical lens.
(ii) A divergent lens has a focal length of 30 cm. At what distance should an object of height 5 cm from the optical centre of the lens be placed so that its image is formed 15 cm away from the lens ? Find the size of the image also.
(iii)Draw a ray diagram to show the formation of image in the above situation.
Answer
(i) Distance between optical centre and focus of the lens.
(ii) f = -30 cm; u = ?; h1 = 5 cm; h2 = ?, v = -15 cm
1/f = 1/v – 1/u
⇒ 1/u = 1/v – 1/f
⇒ u = vf/(f – v)
=( -15 cm × - 30 cm)/{- 30 cm – (-15 cm)}
= -30 cm
m = v/u = h2/h1
⇒ h2 = v/u × h1
= (-15 cm/-30 cm) × 5 cm
= 2.5 cm
(iii)
13
(a) Define focal length of a divergent lens.
(b) A divergent lens of focal length30 cm forms the image of an object of size 6 cm on the same side as the object at a distance of 15 cm from its optical centre. Use lens formula to determine the distance of the object from the lens and the size of the image formed.
(c) Draw a ray diagram to show the formation of image in the above situation.
Answer
(a) The distance between the principal focus and the optical centre of the concave lens or diverging lens is called the focal length of a diverging lens.
(b) f = -30 cm
h = 6 cm
v = -15 cm, u = ?, h = ?
lens formula :
1/v – 1/u = 1/f
⇒ -1/u = 1/f – 1/v
⇒ -1/15 – 1/u = -1/30
⇒ -1/u = -1/30 – (-1/15)
⇒ -1/u = -1/30 + 1/15
⇒ -1/u = (-15 + 30)/450
⇒ - 1/u = 15/450
= 1/30
⇒ - u = 30
⇒ u = 30
The object distance is – 30 cm.
m = v/u = h’/h
⇒ -15/-30 = h’/6
⇒ 6/2 = h’
⇒ h’ = 3 cm
The height of the image is 3 cm.
(c) Position of object is between infinity and optical centre.
Image position is between focus, and optical centre.
14
(a) State the laws of refraction of light. Explain the term absolute refractive index of a medium and write an expression to relate it with the speed of light in vacuum.
(b) The absolute refractive indices of two media ‘A’ and ‘B’ are 2.0 and 1.5 respectively. If the speed of light in medium ‘B’ is 2 × 108 m/s, calculate the speed of light in :
(i) Vacuum
(ii) Medium ‘A’
Answer
(a) Laws of refraction of light :
(i) The incident ray, the normal and the refracted ray at the point of incidence all lies in the same plane for the two given transparent media.
(ii) The ratio of size of angle of incidence (i.e. sin i) to the sine of angle of refraction (i.e., sin r) is always constant for the light of given colour and for the given pair of media.
Mathematically, sini/sinr = constant = n2
The constant ‘n’ is called refractive index of the second medium with respect to the first medium.
Absolute refractive index of the medium is given by,
n = Speed of light in a vacuum (c)/Speed of light in medium (v)
(b) Given nA = 2.0 and nB = 1.5
Speed of light in medium B = 2 × 108 m/s
nB = Speed of light in vacuum (c)/Speed of light in medium (v)
1.5 = c/(2 × 108)
Speed of light in vacuum c = 2 × 108 × 1.5
= 3.0 × 108 m/s
Speed of light in medium ‘A’
nA = Speed of light in vacuum/Speed of light in medium ‘A’
2.0 = (3 × 108)/2
= 1.5 × 108 m/s
15
Analyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follow, without doing any calculation:
(a) What is the focal length of the convex lens ? Give reason in support of your answer.
(b) Write the serial number of that observation which is not correct. How did you arrive at this conclusion?
(c) Take an appropriate scale to draw ray diagram for observation at S. No. 4 and find the approximate value of magnification.
Answer
(a) f = + 15 cm
Reason : Objects at S. No. (3) indicates u = -30 cm, v = +30 cm
Thus, object is 2F (2f = 30 cm)
∴ f = 15 cm
(b) Observation at S.No. (6)
The value, u = - 10 cm, indicates that the object is in between the optical centre and the focus (i.e., less than the focal length) of the lens and hence the image should be on the same side as the object. Thus the image distance cannot be positive.
(c) u = -20 cm ; v = + 60 cm; f = + 15 cm
m = h2/h1 = -4.5 cm/+ 15 cm
= -3
Detailed Answer :
(i) From S.No. 3, we can say that the radius of curvature of the lens is 30 cm because when an object is placed at the centre of curvature of a convex lens, its image is formed on the other side of the lens at the same distance from the lens. And, we know that focal length is half of the radius of curvature. Thus, the focal length of the lens is + 15 cm.
(ii) S. No. 6 is not correct as the object distance is between focus and pole so for such cases the image formed is always virtual but in this case a real image is forming as the image distance is positive.
(iii) Approximate value of magnification for distance object - 2- cm and the image distance +60 cm is -3.
16
Analyse the following observation table showing variation of image-distance (v) with object-distance (u) in case of a convex lens and answer the questions that follow without doing any calculations:
(a) What is the focal length of the convex lens ? Give reason to justify your answer.
(b) Write the serial number of the observation which is not correct. On what basis have you arrived at this conclusion ?
(c) Select an appropriate scale and draw a ray diagram for the observation at S.No. 2. Also find the approximate value of magnification.
Answer
(a) The focal length of the convex lens can be calculated from S.No. 3 as when an object is placed at a distance from the convex lens its image is formed on the other side of the lens at the same distance from the lens. So, the focal length is + 20 cm.
(b) S.No. 6 is incorrect as the object distance between focus and pole and here the real image is formed as the image distance is positive. But in such situation virtual image should form.
(c) Approximate value magnification for object distance - 60 cm and image distance +30 cm is -0.5.
17
“A convex lens can form a magnified erect as well as magnified inverted image of an object placed in front of it.” Draw ray diagram to justify this by statement stating the position of the object with respect to the lens in each case.
An object of height 4 cm is placed at a distance of 20 cm from a concave lens of focal length 10 cm. Use lens formula to determine the position of the image formed.
Answer
Convex lens form magnified erect image when object is placed between F1 and optical centre O. Image formed is virtual, erect and enlarged and on the same side of the lens as the object.
Convex lens form magnified inverted image when object is placed at F1. Image formed is real, inverted and enlarged at infinity.
Given focal length f = - 10 cm (concave)
Object distance (u) = -20 cm
For concave lens;
1/f = 1/v – 1/u, we get
⇒ 1/f + 1/u = 1/v
⇒ 1/-10 + 1/-20 = 1/v
⇒ 1/v = -1/10 – 1/20
⇒ 1/v = (-2 -1)/20
= -3/20
∴ v = - 20/3 cm
m = h1/h0 = v/u
⇒ h1 = v/u × h0
= -20/(3 × - 20) × 4
= 4/3
= 1.3 cm
18
(a) Explain the following terms related to spherical lenses :
(i) Optical centre
(ii) Aperture
(ii) Centres of curvature
(iii) Principal Axis
(iv) Aperture
(v) Principal Focus
(vi) Focal length
(b) A converging lens has focal length of 12 cm. Calculate at what distance should the object be placed from the lens so that it forms an image at 48 cm on the other side of the lens.
Answer
(a) (i) Optical centre : The centre point of a lens is known as its optical centre. It always lies inside the lens. A light beam passing through the optical centre emerges without any deviation.
(ii) Centre of curvature : It is defined as the centre of the sphere of which the lens is originally a part of. Because the spherical lens consists of two spherical surfaces, the lens has two centres of curvature.
(iv) Aperture : This is the length or breadth of the lens through which refraction takes place.
(v) Principal focus : A light rays parallel to the principal axis of the lens meet at a point on the principal axis. This point is called the principal focus.
(vi) Focal length : The distance of the point from the centre of lens or mirror at which a parallel ray of beam converge (or diverge) is called focal-length and the point is called focus.
(b) Focal length of the converging lens, f = 12 cm
Image distance, v = 48 cm
Using the lens formula, we get :
1/f = 1/v – 1/u
⇒ 1/u = 1/v – 1/f
= 1/48 – 1/12
= (1 – 4)/48
= -3/48
∵ u = - 48/3 = - 16 cm
So, the distance of the object from the lens is 16 cm.
19
(i) Define power of a lens. Write its SI units.
(ii) You are provided with two convex lenses of focal length 15 cm and 25 cm, respectively. Which of the two is of larger power? Give reason for your answer.
(iii) A 20 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification.
Answer
(i) Ability of a lens to converge or diverge light rays is termed as power of a lens.
Power = 1/Focal length
SI unit of power is Dioptre.
(ii) Lens of focal length 15 cm is of larger power because power is inversely proportional to the focal length.
(iii) 1/v – 1/u = 1/f
1/v = 1/-15 + 1/10
= 1/30
∴ v = 30 cm
Image is real and inverted
Magnification = h’/h = v/u’ m = - 2
20
(i) Draw a ray diagram to show the formation of image by a convex lens when an object is placed in front of the lens between its optical centre and principal focus.
(ii) In the above ray diagram mark the object-distance (u)and the image-distance (v) with their proper signs (+ve or –ve as per the new Cartesian sign convention) and state how these distances are related to the focal length (f) of the convex lens in this case.
(iii) Find the power of a convex lens which forms a real, and inverted image of magnification -1 of an object placed at a distance of 20 cm from its optical centre.
Answer
(i) 
(ii) Relation: 1/f = 1/v – 1/u
(ii) m = -1; u = - 20 cm; v = ?, f = ?
Marking of u & v,
m = v/u
∴ v = + 20 cm
Thus, object is at 2F
i.e., 2f = 20 cm
∴ f = 10 cm = 0.1 m
P = 1/f = 1/0.1 = + 10 D
21
(i) Draw a ray diagram to show the formation of image by a concave lens when an object is placed in front of it.
(ii) In the above diagram mark the object-distance (u) and the image-distance (v) with their proper signs (+ve or –ve as the per the new Cartesian sign convention) and state how these distances are related to the focal length (f) of the concave lens in this case.
(iii) Find the nature and power of lens which forms a real and inverted image of magnification -1 at a distance of 40 cm from its optical centre.
Answer
f = -20 cm; h1 = 6 cm; v = - 15 cm; u = ?
Lens formula: 1/f = 1/v – 1/u
⇒ u = vf/(f – v)
= (-15 cm × - 20 cm)/{-20 cm – (-15 cm)}
= -60 cm
Object at 60 cm from the lens
h2 = v/u × h1
= -15 cm/-60 cm × 6 cm
= +1.5 cm diminished, erect
22
A student wants to project the image of a candle flame on the walls of school laboratory by using a lens :
(i) Which type of lens should he use and why ?
(ii) At what distance in terms of focal length ‘F’ of the lens should he place the candle flame so as to get : (i) A magnified, and (ii) a diminished image respectively on the wall ?
(iii) Draw ray diagram to show the formation of the image in each case ?
Answer
(i) He should use a convex lens as it forms real images.
(ii) He should place the candle flame between F and 2F (the focus and centre of curvature of the lens) to get the magnified image on the wall while the diminished image is obtained when the object is located at a distance greater than 2F.
(iii) The ray diagram for the formation of the magnified image is shown below :
The ray diagram for the formation of the diminished image is shown below :
23
(i) The refractive index of diamond is 2.42. What is the meaning of this statement?
(ii) Redraw the diagram given below in your answer book and complete the path of the ray.
(iii) What is the difference between virtual images produced by concave, plane, and convex mirrors?
(iv) What does the negative sing in the value of magnification produced by a mirror indicates about the image?
Answer
(i) This means that the ration of speed of light in air and speed of light in diamond is 2.42.
(ii) 
(iii) Virtual image produced by concave mirror is magnified, that produced by plane mirror is of the same and the virtual image produced by convex mirror is diminished.
(iv) Real.
24
(i) Two convex lenses A and B have power P1, and P2, respectively and P2 is greater than P1. Draw a ray diagram for each lens to show which one will be more converging. Give reason for your answer.
(ii) A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification.
Answer
(i) P2 > P1
∴ F2 < F1
Power = P1
Focal length = F1
(b) Power = P2
Focal length = F2
(ii) 1/v = -1/15 + 1/10
⇒ 1/v = 1/30
⇒ v = + 30 cm
The image formed is real and inverted and on the other side of optical centre.
m = h’/h = v/u = 30/-15
⇒ m = -2 inverted and 4 cm tall
25
A very thin narrow beam of white light is made incident on three glass objects shown below. Comment on the nature and behaviour of the emergent beam in all the three case.
There is a similarity between two of the emergent beams. Identify the two.
When light enters from air to glass, the angles of incidence and refraction in air and glass are 45° and 30°. Respectively. Find the refractive index of glass.
(Given that sin 45° = 1/√2; sin 30° = ½)
Answer
In (i) emergent beam is white and laterally displaced.
In (ii) emergent beam is a spectrum of seven colours bent in different angles.
In (iii) emergent beam from the second prism is white only.
Similarly between (i) and (ii) as both emergent rays are white in colour.
ang = sini/sinr = sin 45°/sin 30°
= (1/√2)/(1/2)
= √2