CBSE Class 10 Science Reflection And Refraction Of Light Worksheet

DIRECTION : In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
(e) Both Assertion and Reason are false.

Question. Assertion : When the object moves with a velocity v̅, its image in the plane mirror moves with a velocity of -2v̅.
Reason : The minimum height of the mirror to be required to see the full image of man of height h is h/2 .
Answer : B 

Question. Assertion : The small object, to be seen in a microscope, is kept within the two foci of its objective.
Reason : In this case, the image formed by the objective is nearer to the eyepiece.
Answer : C

Question. Assertion : Mirror formula can be applied to a plane mirror.
Reason : A plane mirror is a spherical mirror of infinite focal length.
Answer : A

Question. Assertion : If both plane mirror and object are moved through a distance x , then the image moves through a distance 2x .
Reason : When the object is fixed and plane mirror is moved through a distance x . Then the image is also moves through the distance 2x .
Answer : C

Question. Assertion : Plane mirror may form real image.
Reason : Plane mirror forms virtual image, if objects is real.
Answer : B

Question. Assertion : The height of an object is always considered positive.
Reason : An object is always placed above the principal axis in this upward direction.
Answer : A

Question. Assertion : Red light travels faster in glass than green light.
Reason : The refractive index of glass is less for red light than for green light.
Answer : A

Question. Assertion : A convex mirror is used as a driver’s mirror.
Reason : Because convex mirror’s field of view is large and images formed are virtual, erect and diminsed.
Answer : A

Question. Assertion : When a concave mirror is held under water, its focal length will increase.
Reason : The focal length of a concave mirror is independent of the medium in which it is placed.
Answer : D

Question. “A concave mirror of focal length 15 cm can form a magnified erect as well as inverted image of an object placed in front of it.” Justify this statement stating the position of the object with respect to the pole of the mirror in both cases for obtaining the images. 
Answer: Case 1. When the object is placed between less than 15 cm from the pole, i.e., between the pole and the
focus of a concave mirror, a magnified, erect and virtual image will be formed.
Case 2. When the object is placed between 15 to 30 cm, i.e., between the focus and the centre of curvature of the concave mirror, a magnified, inverted and real image will be formed.

Question. List four characteristics of the images formed by plane mirrors. 
Answer: The characteristics of the images formed by plane mirrors are:
(i) The image formed by a plane mirror is virtual and erect. It cannot be received on a screen.
(ii) The image formed by a plane mirror is of the same size as the object.
(iii) The image formed by a plane mirror is at the same distance behind the mirror as the object is in front
of the mirror.
(iv) The image formed in a plane mirror is laterally inverted.

Question. List four specific characteristics of the images of the objects formed by convex mirrors. 
Answer: The images of the objects formed by convex mirrors are always—
(i) virtual,
(ii) erect,
(iii) diminished and
(iv) formed behind the mirror between focus and pole of the mirror.

Question. The absolute refractive indices of glass and water are 3/2 and 4/3 respectively. If the speed of light is 2 × 108 m/s, calculate the speed of light in (i) vacuum, (ii) water.
Answer: (i) Given: vg = 2 × 108 m/s (Speed of light in glass)
We know, Absolute Refractive Index of a Medium = Speed of light in Vacuum (c)/Speed of light in the Medium
ng = 3/2 , nw = 4/3 ⇒ ng = c/vg = c = ngvg ⇒ c = 3/2 × 2 × 108
∴ c = 3 × 108 m/s 
(ii) nw = 4/3 c = 3 × 108 m/s 
⇒ vw = c/nw = 3 x 108/4/3 = 3 x 3 x 108/4 = 2.25 × 108 m/s

Question. State two positions in which a concave mirror produces a magnified image of a given object. List two differences between the two images. 
Answer: The two positions are:
(i) When an object is placed between the pole (p) and focus (f) of a concave mirror, the image formed is larger than the object (or magnified).
(ii) When an object is placed between the focus (f) and centre of curvature (c) of a concave mirror, the image formed is larger than the object.

Question. An object 5 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer: h = 5 cm u = –20 cm r = 30 cm v = ? h’ = ? nature = ?
∴ f = r/2 = 30/2 = 15 cm Now , 1/f = 1/v + 1/u ∴ 1/15 = 1/v + 1/-20
⇒ 1/15 + 1/20 = 1/b ⇒ 4 + 3 /60 = 1/v ⇒ = 60/7 = 8.56 cm
Nature of image. Virtual, erect and behind the mirror.
and h'/h = -v/u ⇒ h'/5 = -8.56/-20 ∴ h' = 8.56/20 x 5 = 2.14 cm
∴ An erect and virtual image 2.14 cm high will be formed behind the mirror at a distance of 8.56 cm.

Question. What is meant by power of a lens? What does its sign (+ve or –ve) indicate? State its S.I. unit. How is this unit related to focal length of a lens? 
Answer: • The power of a lens is a measure of the degree of convergence or divergence of light rays falling on it.
• +ve sign Æ converging lens/convex lens
–ve sign Æ diverging lens/concave lens
• The SI unit of lens is dioptre. One dioptre is the power of a lens whose focal length is 1 metre.

Question. An object is placed at a distance of 30 cm in front of a convex mirror of focal length 15 cm. Write four characteristics of the image formed by the mirror. 
Ans. Four characteristics of the image formed are:
(i) Image is erect.
(ii) Image is virtual.
(iii) Image is diminished in size.
(iv) The image is formed behind the mirror between P & F points of the mirror.
(v) The image is laterally inverted. (any four)

Question. A student places a candle flame at a distance of about 60 cm from a convex lens of focal length 10 cm and focuses the image of the flame on a screen. After that he gradually moves the flame towards the lens and each time focuses the image on the screen.
(a) In which direction-toward or away from the lens, does he move the screen to focus the image?
(b) How does the size of the image change?
(c) How does the intensity of the image change as the flame moves towards the lens?
(d) Approximately for what distance between the flame and the lens, the image formed on the screen is inverted and of the same size? 
Answer: (a) He will move the screen away from the lens to focus the image.
(b) Size of the image goes on increasing.
(c) Intensity of image goes on decreasing.
(d) About 20 cms.

Question. Draw the ray diagram and also state the position, the relative size and the nature of image formed by a concave mirror when the object is placed at the centre of curvature of the mirror. 
Answer: When the object is at the centre of curvature of a concave mirror, i.e., point C:
The image formed is
(i) real,
(ii) inverted,
(iii) same size as the object at C, and
(iv) at C.

Question. Define, ‘refractive index of a transparent medium.’ What is its unit? Which has a higher refractive index—glass or water? 
Answer: • The light bending ability of a transparent medium is called the refractive index of that medium.
• The ratio of speed of light in vacuum to the speed of light in a medium is called the refractive index of that medium.
Refractive index (of a medium) = Speed of light in vacuum/Speed of light in medium
• Since refractive index is a ratio of two similar quantities therefore it has no units.
• The refractive index of glass is more than water.

Question. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards or away from the normal? Why? Draw a ray diagram to show the refraction of light in this situation. 
Answer: When a ray of light travelling in air enters obliquely into water, the light ray will bend towards the normal. Because when a ray of light travels obliquely from rarer medium to
denser medium, it will bend towards the normal.

Question. (i) “The refractive index of diamond is 2.42”. What is the meaning of this statement?
(ii) Name a liquid whose mass density is less than that of water but it is optically denser than water.
Answer: (i) The refractive index of diamond is 2.42. It means that the ratio of the speed of light in air and the speed of light in diamond is equal to 2.42.
Higher is the refractive index of a medium, lower is the speed of light in that medium. Because the refractive index of diamond is very high, therefore the speed of light in diamond is very low.
(ii) Kerosene has the mass density less than water but it is optically denser than water.

Question. “The magnification produced by a spherical mirror is –3”. List four informations you obtain from this statement about the mirror/image. 
Ans. Magnification produced by a spherical mirror, m = –3
• image is 3 times magnified than the object. • image is inverted (as m has negative sign).
• image is real. • nature of the mirror is concave.

Question. The refractive indices of glass and water with respect to air are 3/2 and 4/3 respectively. If speed of light in glass is 2 × 108 m/s, find the speed of light in water.
Answer: Refractive index of a medium = Speed of light in air/Speed of light in that medium
Given: ng = 3/2 , nw = 4/3
Speed of light in glass = 2 × 108 m/s; Speed of light in water = v = ?
ng= Speed of light in air/Speed of light in glass ⇒ 3/2 = Speed of light in air/2×108 m/s
∴ Speed of light in air = 3/2 × 2 × 108 = 3 × 108 m/s 
nw = Speed of light in air / Speed of light in water ⇒ 4/3 = 3 x 108m/s/v 
3 × 108 x 3/4 = v ⇒ v = 2.25 × 108 m/s
∴ Speed of light in water = 2.25 × 108 m/s

Question. What is the principle of reversibility of light? Show that the incident ray of light is parallel to the emergent ray of light when light falls obliquely on a side of a rectangular glass slab. 
Answer: Principle of reversibility of light. Refractive index for light going from medium 1 to medium 2 is equal to the reciprocal of refractive index for light going from medium 2 to medium 1.
1η2 = 1/2η1
According to Snell’s law
ah sin i = gh sin r1 ...(i)
Similarly, gh sin r2 = ah sin e ...(ii)
r2 = r1 because glass slab is rectangular and therefore the two normals are parallel.
r1 and r2 are alternate interior angles.
Comparing (i) and (ii), we get
ah sin i = ah sin e
sin i = sin e ⇒ i = e
∴ angle of incidence is equal to the angle of emergence when a ray of light falls obliquely on a rectangular glass slab.

Question. The image of a candle flame placed at a distance of 36 cm from a spherical lens is formed on a screen placed at a distance of 72 cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2.5 cm, find the height of the image. 
Answer: Object distance, u = –36 cm, Images distance, v = +72 cm
...[+ve sign is due to the image being formed on the screen hence it is real Focal length, f = ?, Nature of the lens = ?
Height of the object, h1 = 2.5 cm, Height of the image, h2 = ?
According to lens formula:
1/f = 1/v - 1/u ⇒ 1/f = 1/72 - 1/-36 ⇒ 1/f = 1/72 + 1/36 = 1+2/72 = 3/72 = 1/24
∴ f = +24 cm ⇒ +ve sign of f shows that the lens is convex having focal length 24 cm.
Magnification, m = v/u = 72/-36 = –2
Formula: m = h2/h1 = h2/2.5 = –2 ⇒ h2 = –2 × 2.5 = –5 cm
• –ve sign of h2 shows that the image is inverted.
• Thus an inverted, magnified, 5 cm long image is formed on the screen.

Question. To construct ray diagram we use two light rays which are so chosen that it is easy to know their directions after reflection from the mirror. List these two rays and state the path of these rays after reflection. Use these rays to locate the image of an object placed between centre of curvature and focus of a concave mirror. 
Answer: Rays used and their path after reflection: (i) A ray parallel to the principal axis after reflection this ray will pass through the principal focus in case
of a concave mirror or appear to diverge from the principal focus in case of a convex mirror.
(ii) A ray passing through the centre of curvature of a concave mirror or directed in the direction of the
centre of curvature of a convex mirror, after reflection, is reflected back along the same path.
When the object is between centre of curvature C and focus F of
mirror:
The image formed is
(i) real,
(ii) inverted,
(iii) magnified, and
(iv) beyond C.

Question. The image of a candle flame placed at a distance of 45 cm from a spherical lens is formed on a screen placed at a distance of 90 cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2 cm, find the height of its image.
Answer: Object distance, u = –45 cm, Image distance, v = +90 cm (real) Focal length, f = ?,
Nature of lens = ?, Height of the object, h1 = 2 cm, Height of the image, h2 = ?
According to lens formula:
1/f = 1/v - 1/u ⇒ 1/f = 1/90 - 1/-45 ⇒ 1/f = 1/90 + 1/45 = 1+ 2/90 = 3/90 = 1/30
∴ f = 30 cm
The positive sign of f shows that the given lens is a convex lens of focal length 30 cm.
Magnification, m = v/u = +90/-45 = -2
⇒ h2/h1 = -2 ⇒ h2/h = -2 ∴ h2 = –4
• Thus height of the image is 4 cm.
• The negative sign shows that this image is in the downward direction below the axis, i.e., image is inverted.

Question. A 2.4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 18 cm.
The distance of the object from the lens is 12 cm. Find the position, size and nature of the image formed, using the lens formula. 
Answer: h1 = +2.4 cm (upright) (Concave lens) v = ? h2 = ?
f = +18 cm (for convex lens) u = –12 cm Nature of the image = ?
According to lens formula:
1/v - 1/u = 1/f ⇒ 1/v - 1/-12 = 1/18 ⇒ 1/v + 1/12 = 1/18
⇒ 1/v = 1/18 - 1/12 = 2-3 /36 = -1/36 ⇒ v = –36 cm
• Image is formed at a distance of 36 cm from the convex lens.
• The negative (–) sign of v shows that the image is formed on the left hand side of the convex lens. Only virtual image is formed on the left hand side
m = v/u = -36/-12 = 3 ⇒ m = h2/h1 = h2/2.4 ⇒ h2/2.4 = 3
• The height of the image is 7.2 cm.
• The positive (+) sign shows that the image is formed above the axis. Thus the image is virtual and erect.

Question. A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 12 cm.
The distance of the object from the lens is 8 cm. Using the lens formula find the position, size and nature of the image formed.
Answer: f = +12 cm (for convex lens), h1 = 5 cm, u = –8 cm
v = ?, h2 = ?, Nature of the image = ?
According to lens formula:
1/v - 1/u = 1/f ⇒ 1/v - 1/-18 = 1/12 ⇒ 1/v + 1/8 = 1/12
⇒ 1/v = 1/12 - 1/12 - 1/8 = 2-3/24 = -1/24 ⇒ v = –24 cm
• Image is formed at a distance of 24 cm from the convex lens.
• The negative (–) sign of v shows that the image is formed on the left hand side of the convex lens and only virtual image is formed on the left hand side.
• A virtual image is formed at 24 cm from the lens.
∴ m = v/u = 24/8 = 3 ⇒ m = h2/h1 ⇒ h2/5 = 3
∴ h2 = 3 × 5 = 15 cm
• Thus size of the image is 15 cm.
• The positive (+) sign shows that the image is formed above the axis.
• Thus a virtual, magnified and erect image is formed.

Question. State the type of mirror preferred as (i) rear view mirror in vehicles, (ii) shaving mirror. Justify your answer giving two reasons in each case. 
Answer: (i) Convex mirror is used as rear view mirror in vehicles because the image formed in a convex mirror is highly diminished thus a convex mirror gives a wide field of view. Therefore a convex mirror enables a driver to view a much larger area of the traffic behind him.
(ii) Concave mirror is used as shaving mirror because when face is held within the focus of a concave mirror, an enlarged image of the face is seen in the concave mirror.

Question. A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position, size and nature of the image formed, using the lens formula. 
Answer: Height of the object, h1 = 4 cm
Convex lens:
Focal length, f = +24 cm, Object distance, u = –16 cm, Image distance, v = ?
Height of the image, h2 = ?, Nature of the image = ?
According to lens formula:
1/v - 1/u = 1/f ⇒ 1/v - 1/-16 = 1/24 ⇒ 1/v + 1/16 = 1/24
⇒ 1/v = 1/24 - 1/16 = 2-3/48 = -1/48 ∴ v = –48 cm
• The image is formed at a distance of 48 cm from the convex lens.
• The minus sign for image distance shows that the image is formed on the left side of the convex lens.
• Only virtual image is formed on the left hand side.
According to formula: m = h2/h1 , m = v/u
h2/h1 = v/u ⇒ h2/4 = -48/-16 = 3 ∴ h2 = 3 × 4 = 12 cm.
Thus a magnified (12 cm high), virtual and erect image is formed.

Question. The image of a candle flame placed at a distance of 30 cm from a spherical lens is formed on a screen placed at a distance of 60 cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2.4 cm, find the height of its image. 
Answer: Object distance, u = –30 cm, Image distance, v = +60 cm
...[+ve sign is due to the image formed on the screen, hence it is real f = ?, Type of lens = ?
Height of the object, h1 = 2.4 cm, Height of the image, h2 = ?
According to lens formula:
1/f = 1/v - 1/u ⇒ 1/f = 1/60 - 1/- 30 ⇒ 1/f = 1/60 + 1/30 + = 1+2/60 = 3/60 = 1/20
∴ f = +20 cm
The positive (+ve) sign of f shows that the lens is convex having focal length 20 cm.
Now Magnification, m = v/u = 60/-30 = -2
Formula: m = h2/h1 ⇒ h2/2.4 = -2 ∴ h2 = –2 × 2.4 = – 4.8 cm
The negative (–ve) sign of h2 shows that the image is inverted.

Question. Mention the types of mirrors used as (i) rear view mirrors, (ii) shaving mirrors. List two reasons to justify your answers in each case.
Answer: (i) Convex mirror is used as rear view mirror in vehicles because
• it always produces an erect image of the objects;
• the image formed in a convex mirror is highly diminished thus it gives a wide field of view.
(ii) Concave mirrors are used as shaving mirrors because
• when the face is held within the focus of a concave mirror, then an enlarged image of the face is seen in the concave mirror. This helps in making a smooth shave.

Question. An object of height 6 cm is placed perpendicular to the principal axis of a concave lens of focal length 5 cm. Use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 10 cm. 
Answer: Height of the object, h1 = 6 cm Focal length of the concave mirror, f = –5 cm
Position of the image, v = ? Size of the image, h2 = ?
Object distance, u = –10 cm
According to lens formula:
1/v - 1/u = 1/f ⇒ 1/v - 1/-10 = 1/-5 ⇒ 1/v + 1/10 = -1/5
⇒ 1/v = -1/5 - 1/10 = -2 - 1/10 = -3/10 ∴ v = -10/3 = -3.3cm 
h2/h1 = v/u ⇒ h2/6 = -10/3 / -10 ⇒ h2/6 = 10/3 x 10 = 1/3
⇒ h2/= 6/3 ∴ h2 = +2 cm
Thus the image is formed at a distance of 3.3 cm from the concave lens. The negative (–) sign for image distance shows that the image is formed on the left side of the concave lens (i.e., virtual). The size of the image is 2 cm and the positive (+) sign for hand image shows that the image is erect.
Thus a virtual, erect, diminished image is formed on the same side of the object (i.e., left side).

Question. An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. Use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 20 cm. 
Answer: Height of the object, h1 = 5 cm Focal length of the concave lens, f = –10 cm
Position of the object, v = ? Size of the object, h2 = ?
Object distance, u = –20 cm
According to lens formula:
1/v - 1/u = 1/f ⇒ 1/v - 1/-20 = 1/-10 ⇒ 1/v + 1/20 = -1/20
⇒ 1/v = -1/10 - 1/20 = -2 -1/20 = -3/20 ∴ v = -20/3 = – 6.67
The negative (–) sign for image distance shows that image is formed on the left side of the concave lens. So the image is virtual.
Magnification, m = v/u , m = h2/h1 ∴ m = -20/3/10 = 20/10 x 3 = + 2/3 = 0.67
⇒ h2/h1 = v/u ∴ h2 = h1 x v /u = 5x - 20 /3x - 20 = 5/3 ⇒ m = +1.66 cm
Since h2 < h1 therefore image is diminished. The positive (+) sign for the magnification shows that image is erect and virtual.

Question. An object of height 4 cm is kept at a distance of 30 cm from a concave lens. Use lens formula to determine the image distance, nature and size of the image formed if focal length of the lens is 15 cm. 
Answer: Height of the object, h1 = 4 cm Object distance, u = –30 cm (It is to the left of the lens)
Focal length of concave lens, f = –15 cm Image distance, v = ?
Nature of the image = ? Image height, h2 = ?
According to lens formula:
1/v - 1/u = 1/f ⇒ 1/v - 1/-30 = 1/-15 ⇒ 1/v + 1/30 = -1/15
 ⇒ 1/v = -1/u 15 - 1/30 = -2 -1//30 = -3/30 = 1/10 ∴ v = –10 cm
Thus the image is formed at a distance of 10 cm from the concave lens.
The negative (–) sign for image distance shows the image is formed on the left side of the concave lens, i.e., it is virtual.
Magnification, m = v/u , m = h2/h1 
∴ h2/h1 = v/u ⇒ h2/4 = -10/-30 = 1/3 ⇒ h2 = 1/3 x 4 = 4/3 = +1.33 cm
Thus a 1.33 cm high image is formed and positive (+) sign of h2 shows that image is erect.
Thus image distance = 10 cm, image height = 1.33 cm.
Nature of the image is virtual and erect.

Question. Name the type of mirror used (i) by dentists and (ii) in solar furnaces. Give two reasons why such mirrors are used in each case. 
Answer: (i) Concave mirrors are used by dentists to see the large images of the teeth of patients because when a tooth is within the focus of a concave mirror, then an enlarged image of the tooth is seen in the concave mirror. Thus it becomes easier to locate the defect in the tooth.
(ii) Large concave mirrors are used in solar furnaces as reflectors. Solar furnace is placed at the focus of the concave reflector which focusses the Sun’s heat rays on the furnace due to which the solar furnace gets very hot. Even steel can be melted in this solar furnace.

Question. A student focussed the image of an object on a white screen using a converging lens. He noted down the positions of the object, screen and the lens on a scale as given below:
Position of object = 10.0 cm; Position of lens = 50.0 cm; Position of screen = 90.0 cm
(a) Find the focal length of the converging lens.
(b) Find the position of the image if the object is shifted towards the lens at a position of 30.0 cm.
(c) State the nature of the image formed if the object is further shifted towards the lens.
Answer: Object distance, u = –40 cm Image distance, v = (90 – 50) = +40 cm
(a) Focal length of converging lens, f = ?
According to lens formula: 
1/f = 1/v - 1/u ⇒ 1/f = 1/40 - 1/-40 = 1/40 + 1/40 ⇒ 1/f = 2/40 = 1/20
∴ f = 20 cm
∴ Focal length of the converging lens, f = +20 cm
(b) If u = –20 cm (Object is shifted from 10 cm to 30 cm with respect to 50 cm) f = +20 cm
1/v - 1/u = 1/f ⇒ 1/ v = 1/f + 1/u = 1/20 + 1/-20 = 1/20 = -1/20 = 0 ⇒ 1/v = 0
⇒ v = 1/0 ∴ v = ∞ (∴ Reciprocal of 0 is ∞)
So image is formed at infinity.
(c) If the object is shifted from 30 cm towards 50 cm, the object lies between the focus and optical centre of the lens. Then the image formed will be behind the object, virtual, erect and enlarged (larger than the object).

Question. A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the positions of the candle flame, screen and the lens as given below: 
Position of the candle flame = 12.0 cm; Position of the lens = 50.0 cm; Position of the screen = 88.0 cm (i) Find the focal length of the convex lens.
(ii) Find the position of the image of the candle flame if it is shifted towards the lens at a position of 31.0 cm.
(iii) State the nature of the image formed if the candle flame is further shifted towards the lens.
Answer: Position of the candle flame = 12 cm, Position of the lens = 50 cm
Position of the screen = 88 cm, Object distance, u = 50 – 12 = –38 cm
Image distance, v = 88 – 50 = +38 cm (+ sign for real image as the image is formed on the screen)
(i) Focal length of convex lens, f = ?
According to lens formula:
1/f = 1/v - 1/u

Long Answer Questions

Question. A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the positions of the candle flame, screen and the lens as given below: 

Position of the candle flame = 12.0 cm; Position of the lens = 50.0 cm; Position of the screen = 88.0 cm (i) Find the focal length of the convex lens.
(ii) Find the position of the image of the candle flame if it is shifted towards the lens at a position of 31.0 cm.
(iii) State the nature of the image formed if the candle flame is further shifted towards the lens.
Answer: Position of the candle flame = 12 cm, Position of the lens = 50 cm
Position of the screen = 88 cm, Object distance, u = 50 – 12 = –38 cm
Image distance, v = 88 – 50 = +38 cm (+ sign for real image as the image is formed on the screen)
(i) Focal length of convex lens, f = ?
According to lens formula:
1/f = 1/v - 1/u ⇒ 1/f = 1/38 - 1/-38
⇒ 1/f = 1/38 + 1/38 = 2/38 = 1/19 ⇒ f = 19
∴ Focal length of the convex lens, f = +19 cm
+ve sign shows the converging lens, i.e., convex lens.
(ii) v = ?, f = +19 cm, u = (50 – 31) = –19 cm
1/v - 1/u = 1/f ⇒ 1/v = 1/f ⇒ 1/v = 1/f + 1/u = 1/19 + 1/-19 = /19 - 1/19 = 0
⇒ 1/v = 0 ∴ v = ∞ (∴ Reciprocal of 0 is ∞)
So image is formed at infinity.
(iii) When the candle flame lies between 50 cm to 31 cm, i.e., less than 19 cm, the image will be virtual, erect, magnified and behind the object.

Question. A spherical mirror ‘A’ always forms an erect image of an object and another spherical mirror ‘B’ forms erect as well as inverted image of an object. State with reasons the type of spherical mirrors ‘A’ and ‘B’ and draw ray diagrams showing formation of images to justify your answer.
Answer: Mirror ‘A’ is a convex mirror as this mirror always forms an erect and virtual image.
When an incident ray is parallel to the principal axis, the reflected ray appears to be coming from the focus and when incident ray moves towards the centre of curvature,
the reflected ray retraces the path.
These two rules show that the reflected rays are diverging and when produced back-wards, the rays appear to meet. Thus, this mirror always produces a virtual image which is always
erect. Mirror ‘B’ is a concave mirror. The nature, position and size of the image formed by a concave mirror depends on the position of the object in relation to points P(Pole), F(focus) and C(centre of curvature) of the mirror.
Case 1. When the object is placed between P and F—the two reflected rays (one passes through the focus and other retraces through centre of curvature) are diverging and appear to meet in backward direction. This produces a virtual and erect image.

Question. A spherical mirror produces an image of magnification –1 on a screen placed at a distance of 50 cm from the mirror. 
(a) Write the type of mirror.
(b) Find the distance of the image from the object.
(c) What is the focal length of the mirror?
(d) Draw the ray diagram to show the image formation in this case.
Answer: If magnification, m= –1; v = 50 cm
If the magnification has minus sign, then the image is real and inverted.
∴ v = –50 (for real image) ∴ m = -v /u ⇒ -1 = -(-50)/u
u = –50 cm
(a) Since image is formed on the screen therefore the mirror formed real image which is formed by concave mirror only.
(b) Image distance = 50 cm in front of the mirror.
(c) 1/f = 1/v + 1/u ⇒ 1/f = 1/-50 + 1/-50 ⇒ 1/f = -1/50 - 1/50 ⇒ -2/50 = -1/5

Question. A spherical mirror produces an image of magnification –1 on a screen placed at a distance of 40 cm from the mirror: 
(i) Write the type of mirror.
(ii) What is the nature of the image formed?
(iii) How far is the object located from the mirror?
(iv) Draw the ray diagram to show the image formation in this case.
Answer: • Spherical mirror • m = –1
• image is formed on a screen • image distance, v = 40 m
(i) Concave mirror
(ii) Real image (as it is formed on the screen)
(iii) m = –1
m = -v/u ⇒ –1 = -(-40)/u ⇒ –u = +40 ⇒ u = –40 cm
∴ Object is placed at 40 cm from the mirror.

Question. A student wants to project the image of a candle flame on a screen 60 cm in front of a mirror by keeping the flame at a distance of 15 cm from its pole.
(i) Write the type of mirror he should use.
(ii) Find the linear magnification of the image produced.
(iii) What is the distance between the object and its image?
(iv) Draw a ray diagram to show the image formation in this case. 
Answer: (i) Concave mirror
(ii) Linear magnification = - Image distance/Object distance ⇒ m = -v/u
Object distance, u = –15 (u is always negative)
Image distance, v = –60 (v is negative for real image)
m = -(-60)/(-15) ∴ m = –4
The minus sign in magnification shows that the image formed is real and inverted.
(iii) Distance between object and its image = 45 cm
(iv) Ray diagram:

Question. A student wants to project the image of a candle flame on a screen 48 cm in front of a mirror by keeping the flame at a distance of 12 cm from its pole.
(a) Suggest the type of mirror he should use.
(b) Find the linear magnification of the image produced.
(c) How far is the image from its object?
(d) Draw ray diagram to show the image formation in this case. 
Answer: (a) Concave mirror as only concave mirror produces the real image.
(b) Linear magnification = - Image distance / Object distance ⇒ m = -v/u
Object distance, u = –12 (u is always negative)
Image distance, v = –60 (v is negative for real image)
m = -(-48)/(-12) ∴ m = –4
∴ The minus sign in magnification shows that the image formed is real and inverted.
(c) The image is formed at a distance of 36 cm from the object.

Question. To construct a ray diagram we use two rays of light which are so chosen that it is easy to determine their directions after reflection from the mirror. Choose these two rays and state the path of these rays after reflection from a concave mirror. Use these two rays to find the nature and position of the image of an object placed at a distance of 15 cm from a concave mirror of focal length 10 cm. 
Answer: Ray 1. When an incident ray of light is parallel to the principal axis of a concave mirror, its reflected ray must pass through the principal focus of the concave mirror.
Ray 2. A ray passing through the ‘C’ point of a concave mirror after reflection will be reflected back on the same path.
• Focal length = 10 cm; Then centre of curvature, C = 20 cm
Object is placed at 15 cm, i.e., between F & C
When the object is between F and C (centre of curvature):
The image formed is real, inverted and magnified.
It is formed beyond C.

Question. If the image formed by a mirror for all positions of the object placed in front of it is always erect and diminished, what type of mirror is it? Draw a ray diagram to justify your answer. Where and why do we generally use this type of mirror? 
Answer: Convex mirror always forms an erect, virtual and diminished image for all positions of the object placed in front of it.
Uses:
(i) Convex mirrors are used as rear-view mirrors in vehicles to see the traffic at the rear side (or back side) because—(i) a convex mirror always produces an erect image of the object; (ii) the image formed in a convex mirror is highly diminished due to which a convex mirror gives a wide field of view.
(ii) Big convex mirrors are used as ‘shop security mirrors’. By installing a big convex mirror at a strategic point in the shop, the shop owner can keep an eye on the customer to look for thieves and shoplifters among them as convex mirrors always form a virtual, diminished and erect image.

Question. The image formed by a spherical mirror is real, inverted and is of magnification –2. If the image is at a distance of 30 cm from the mirror, where is the object placed? Find the focal length of the mirror. List two characteristics of the image formed if the object is moved 10 cm towards the mirror. 
Answer: Nature of image formed by spherical mirror = Real, inverted Magnification, m = –2 Object distance, u = ?
Image distance, v = –30 cm (–ve sign is for real image) f = ?
m = -v/u ⇒ -v/u = –2 ⇒ -30/u = 2
2u = –30 ∴ u = –15 cm
According to the mirror formula:
1/f = 1/-30 + 1/-15 ⇒ 1/f = -1/30 - 1/15 = -1-2/30 = -3/30
1/f = -1/10 ∴ f = –10 cm
If the object is shifted 10 cm towards the mirror then, u = 15 – 10 = 5 cm.
In this case the object is placed between pole and focus of the spherical mirror.
Therefore, the image formed is virtual, erect, magnified and behind the mirror.

Question. An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. If the distance of the object from the optical centre is 20 cm, determine the position, nature and size of the image formed using the lens formula. 
Answer: Object height, h1 = +5 cm Focal length, f1 = –10 cm Object distance, u = –20 cm
Image distance, v = ? Nature of the image = ? Image height, h2 = ?
According to the lens formula:
1/f = 1/v - 1/u ⇒ 1/v = 1/f + 1/u = 1/(-10) + 1/(-20) = -1/10 - 1/20
⇒ 1/v = -2 -1/20 = -3/20 ∴ v = -20/3 = –6.67 cm
Negative sign of v shows that image is formed on the same side of the object, i.e., virtual.
h2/h1 = v/u ⇒ h2/5 = -20/3 x (-20) ∴ h2 = + 5/3 = +1.67 cm
Positive sign of h2 shows that image is erect.
Thus a virtual, erect, diminished image is formed at a distance of 6.67 cm away from the lens on the same side of the object.