1. LECTURE ON NUMERICAL PROBLEMS IN OPTICSBY
KAVITA MONGA(LECTURER-PHYSICS)
GOVT POLYTECHNIC COLLEGE
KHUNIMAJRA,
MOHALI
DATE- 18/4/2013

2. Light :- i) Light is a form of energy which helps us to see objects.
ii) When light falls on objects, it reflects the light and when the
reflected light reaches our eyes then we see the objects.
iii) Light travels in straight line.
iv) The common phenomena of light are formation of shadows,
formation of images by mirrors and lenses, bending of light by a
medium, twinkling of stars, formation of rainbow etc.

4. The overall study of how light behaves is called optics.
The branch of optics that focuses on the creation of images is called geometric optics, because it is based on relationships between angles and lines that describe light rays.

5. Some Definitions Absorption
When light passes through an object the intensity is reduced depending upon the color absorbed. Thus the selective absorption of white light produces colored light.
Refraction
Direction change of a ray of light passing from one transparent medium to another with different optical density. A ray from less to more dense medium is bent perpendicular to the surface, with greater deviation for shorter wavelengths
Diffraction
Light rays bend around edges - new wavefronts are generated at sharp edges - the smaller the aperture the lower the definition
Dispersion
Separation of light into its constituent wavelengths when entering a transparent medium - the change of refractive index with wavelength, such as the spectrum produced by a prism or a rainbow

6. Absorption
Control
Absorption
No blue/green light
red filter

7. A lens is an optical device that is used to bend light in a specific way.
A converging lens bends light so that the light rays come together to a point.
A diverging lens bends light so it spreads light apart instead of coming together.

8. Mirrors reflect light and allow us to see ourselves.
A prism is another optical device that can cause light to change directions.
A prism is a solid piece of glass with flat polished surfaces.

9. Optical Systems
An optical system is a collection of mirrors, lenses, prisms, or other optical elements that performs a useful function with light.
Characteristics of optical systems are:
The location, type, and magnification of the image.
The amount of light that is collected.
The accuracy of the image in terms of sharpness, color, and distortion.
The ability to change the image, like a telephoto lens on a camera.
The ability to record the image on film or electronically.

10. Reflection
Images appear in mirrors because of how light is reflected by mirrors.
The incident ray follows the light falling onto the mirror.
The reflected ray follows the light bouncing off the mirror.

11. Reflection
In specular reflection each incident ray bounces off in a single direction.
A surface that is not shiny creates diffuse reflection.
In diffuse reflection, a single ray of light scatters into many directions.

12. Law of Reflection The incident ray strikes the mirror.
The reflected ray bounces off.
The angle of incidence equals the angle of reflection.

13. Law of reflection
30o
30o
1) You are asked for a ray diagram and the angle of reflection.
2) You are given the angle of incidence.
3) The law of reflection states the angle of reflection equals the angle of incidence.
4) The angle of reflection is 30°.
A light ray is incident on a plane mirror with a 30 degree angle of incidence.
Sketch the incident and reflected rays and determine the angle of reflection.

14. A dentist uses a mirror to look at the back of a second molar (A)
A dentist uses a mirror to look at the back of a second molar (A). Next, she wishes to look at the back of a lateral incisor (B), which is 90° away. By what angle should she rotate her mirror?
Mirror A B
A. 90°
B. 45°
C. 180°

15. Specular vs. Diffuse Reflection
Specular Reflection
The surface is flat at distance scales near or above the wavelength of light
It looks “shiny”, like a mirror.

16. Specular vs. Diffuse Reflection
The surface is rough at distance scales near or above the wavelength of light
Almost all surfaces reflect in this way!

17. Two plane mirrors form a right angle
Two plane mirrors form a right angle. How many images of the ball can you see in the mirrors? 1 2 3 4
STT23.2

18. Refraction
Light rays may bend as they cross a boundary from one material to another, like from air to water.
This bending of light rays is known as refraction.
The light rays from the straw are refracted (or bent) when they cross from water back into air before reaching your eyes.

19. Refraction
When a ray of light crosses from one material to another, the amount it bends depends on the difference in index of refraction between the two materials.

20. Refraction
Refraction is the bending of light as it passes from one medium into another.
Water
Air N qw qA
refraction
Note: the angle of incidence qA in air and the angle of refraction qA in water are each measured with the normal N.
The incident and refracted rays lie in the same plane and are reversible.

22. Calculate the angle of refraction
A ray of light traveling through air is incident on a smooth surface of water at an angle of 30° to the normal.
Calculate the angle of refraction for the ray as it enters the water.
1) You are asked for the angle of refraction.
2) You are told the ray goes from air into water at 30 degrees.
3) Snell’s law:
ni sin(θi) = nr sin(θr)
ni = 1.00 (air), nr = 1.33 (water)
4) Apply Snell’s law to find θr.
1.00sin(30°) = 1.33 sin(θr)
sin(θr) = 0.5 ÷ 1.33 = 0.376
Use the inverse sine function to find the angle that has a
sine of
θr = sin-1(0.376) = 22°

23. solution 1) You are asked for the angle of refraction.
2) You are told the ray goes from air into water at
30 degrees.
3) Snell’s law:
ni sin(θi) = nr sin(θr)
ni = 1.00 (air), nr = 1.33 (water)
4) Apply Snell’s law to find θr.
1.00sin(30°) = 1.33 sin(θr)
sin(θr) = 0.5 ÷ 1.33 = 0.376
Use the inverse sine function to find the angle that has a sine of
θr = sin-1(0.376) = 22°

24. Refraction Distorts Vision
Water
Air
Water
Air
The eye, believing that light travels in straight lines, sees objects closer to the surface due to refraction. Such distortions are common.

25. Refraction
He sees the fish here….
But it is really here!!

26. Depth perception

27. Apparent Depth

28. A. a greater depth than it really is. B. its true depth.
Virtual Image of Fish
air
water
A fish swims below the surface of the water. An observer sees the fish at:
A. a greater depth than it really is.
B. its true depth.
C. a smaller depth than it really is.

29. air
water
A fish swims directly below the surface of the water. An observer sees the fish at:
A. a greater depth than it really is.
B. its true depth.
C. a smaller depth than it really is.

30. The Index of Refraction
The index of refraction for a material is the ratio of the velocity of light in a vacuum (3 x 108 m/s) to the velocity through the material.
Index of refraction c v
Examples: Air n= 1; glass n = 1.5; Water n = 1.33

31. Index of Refraction
vmedium is the speed of light in a transparent medium.
c is the speed of light in a vacuum (c=3.00×108 m/s)
n is a dimensionless constant: n≥1
n=1 in a vacuum

32. Index of refraction
The ability of a material to bend rays of light is described by the index of refraction (n).

33. Example 1. Light travels from air (n = 1) into glass, where its velocity reduces to only 2 x 108 m/s. What is the index of refraction for glass?
vair = c
vG = 2 x 108 m/s
Glass
Air
For glass:
n = 1.50
If the medium were water: nW = Then you should show that the velocity in water would be reduced from c to 2.26 x 108 m/s.

34. Analogy for Refraction
Sand
Pavement
Air
Glass
3 x 108 m/s
2 x 108 m/s
vs < vp
3 x 108 m/s
Light bends into glass then returns along original path much as a rolling axle would when encountering a strip of mud.

35. Snell's law of refraction
Snell’s law is the relationship between the angles of incidence and refraction and the index of refraction of both materials.
Angle of refraction
(degrees)
Angle of incidence
(degrees)
ni sin Qi = nr sin Qr
Index of refraction of incident material
Index of refraction of refractive material

36. Snell’s Law of Refraction

37. Snell’s Law Snell’s Law: q1 q2 Medium 1 Medium 2
The ratio of the sine of the angle of incidence q1 to the sine of the angle of refraction q2 is equal to the ratio of the incident velocity v1 to the refracted velocity v2 . v1 v2
Snell’s Law:

38. Example 2: A laser beam in a darkened room strikes the surface of water at an angle of 300. The velocity in water is 2.26 x 108 m/s. What is the angle of refraction?
The incident angle is:
Air
H2O qA
300
qA = 900 – 300 = 600 qW
qW = 35.30

39. Snell’s Law and Refractive Index
Another form of Snell’s law can be derived from the definition of the index of refraction:
Medium 1 q1 q2
Medium 2
Snell’s law for velocities and indices:

40. A Simplified Form of the Law
Since the indices of refraction for many common substances are usually available, Snell’s law is often written in the following manner:
The product of the index of refraction and the sine of the angle is the same in the refracted medium as for the incident medium.

41. Example 3. Light travels through a block of glass, then remerges into air. Find angle of emergence for given information.
First find qG inside glass:
qG = 30.70
Glass
Air
n=1.5
Apply to each interface: q qG qG
500
qe = 500
Same as entrance angle!
From geometry, note angle qG same for next interface.

42. Wavelength and Refraction
The energy of light is determined by the frequency of the EM waves, which remains constant as light passes into and out of a medium. (Recall v = fl.)
Glass
Air
n=1
n=1.5 lA lG
fA= fG
lG < lA

43. The Many Forms of Snell’s Law:
Refraction is affected by the index of refraction, the velocity, and the wavelength. In general:
Snell’s Law:
All the ratios are equal. It is helpful to recognize that only the index n differs in the ratio order.

44. Example 4: A helium neon laser emits a beam of wavelength 632 nm in air (nA = 1). What is the wavelength inside a slab of glass (nG = 1.5)?
Glass
Air
n=1.5 q qG
nG = 1.5; lA = 632 nm
Note that the light, if seen inside the glass, would be blue. Of course it still appears red because it returns to air before striking the eye.

45. Total Internal Reflection
When light passes at an angle from a medium of higher index to one of lower index, the emerging ray bends away from the normal.
When the angle reaches a certain maximum, it will be reflected internally.
Water
Air
light
900
The critical angle qc is the limiting angle of incidence in a denser medium that results in an angle of refraction equal to 900.
Critical angle qc
i = r

46. The angle at which light begins reflecting back into a refractive material is called
the critical angle, and it depends on the index of refraction.
The critical angle for water is about 49 degrees.

47. Total Internal Reflection
Occurs when n2<n1
θc = critical angle.
When θ1 ≥ θc, no light is transmitted through the boundary; 100% reflection

48. Example 5. Find the critical angle of incidence from water to air.
For critical angle, qA = 900
Water
Air qc
900
Critical angle
nA = 1.0; nW = 1.33
Critical angle:
qc = 48.80
In general, for media where n1 > n2 we find that:

49. Refraction & Dispersion
Short wavelengths are “bent” more than long wavelengths
dispersion
Light is “bent” and the resultant colors separate (dispersion).
Red is least refracted, violet most refracted.

50. Dispersion by a Prism
Red
Orange
Yellow
Green
Blue
Indigo
Violet
Dispersion is the separation of white light into its various spectral components. The colors are refracted at different angles due to the different indexes of refraction.

51. Dispersion and prisms
When white light passes through a glass prism, blue is bent more than red.
Colors between blue and red are bent proportional to their position in the spectrum.

52. Dispersion and prisms
The variation in refractive index with color is called dispersion.
A rainbow is an example of dispersion in nature.
Tiny rain droplets act as prisms separating the colors in the white light rays from the sun.

53. a) Lens formula for spherical lenses :-
The lens formula for spherical lenses is the relationship between the object distance (u), image distance (v) and focal length (f).
The lens formula is expressed as :- =
v u f
b) Magnification produced by spherical lenses :-
Magnification for spherical lens is the ratio of the height of the image to the height of the object.
Height of the image hi
Magnification = m =
Height of the object ho
The magnification is also related to the object distance and image distance. It can be expressed as :-
hi v
Magnification m = =
ho u 53

54. Power of a lens :- I 1 P = or f = f (m) P
The power of a lens is the reciprocal of its focal length (in metres). I
P = or f =
f (m) P
The SI unit of power is dioptre (D).
1 dioptre is the power of a lens whose focal length is 1 metre.
The power of a convex lens is positive ( + ve ) and the power of a concave lens is negative ( - ve ).

55. Thin Lens Equation
We can mathematically determine where an image will be formed in relation to the optical centre using the thin lens equation.

56. The Thin Lens Equation 1 = 1 + 1 f di do Where:
1 =
f di do F F’ 2F
2F’
Focal length f do di
Where:
f represents the focal length
do represents from the object to the lens
di represents from the image to the lens

57. The Thin Lens Equation – Example 1
A converging lens has a focal length of 17 cm. A candle is located 48 cm from the lens. What type of image will be formed, and where will it be located?

58. example 1 (cont'd)
G: f =
d0 =
U: di = ?
17 cm
48 cm

59. example 1
1 =
f di do E:
Rearrange equation:
1 =
di f d0

60. example 1…almost there! S: S: S: di 17 cm 48 cm 1 = 0.03799cm-1 di
1 =
di cm cm
1 = cm-1 di
di = 1 / cm-1 S:
di = cm = 26 cm S:
The image of the candle is real and will be about 26 cm from the lens (opposite the object). 60

61. example – last step!
Sketch the ray diagram for this problem.

62. Example 2: Thin Lens Equation and Diverging Lenses
A diverging Lens has a focal length of 29 cm. A virtual image of a marble is located 13 cm in front of the lens. Where is the marble (the object) located?
Follow the same steps as before, but be careful with the signs! Refer to your RULES for sign conventions.

63. Example 2: Ray Diagram Sketch

64. Example 2: f = - 29 cm di = -13 cm d0 = ? G: 1 = 1 - 1 d0 -29cm -13cm
1 =
d cm cm S: U:
1 = cm-1 d0 S:
1 =
f do di
d0 = cm = 24 cm E: S:
The marble is located 24 cm from the lens on the same side as the image.
1 =
d f di

65. Magnification Equation
M = hi = - di
ho do F F’ 2F
2F’ ho hi
Where:
M stands for magnification
hi stands for height of the image
ho stands for height of the object

66. Magnification Signs and Measurements
Magnification (M) has no units
M is POSITIVE for an UPRIGHT image
M is NEGATIVE for an INVERTED image
If M is… Then the image is…
GREATER THAN 1 LARGER than the
object
BETWEEN 0 AND 1 SMALLER than the
EQUAL TO 1 SAME SIZE as the object

67. Example 3: Finding the Magnification of a CONVERGING lens
A toy of height 8.4 cm is balanced in front of a converging lens. An inverted, real image of height 23 cm is noticed on the other side of the lens. What is the magnification of the lens?
Follow the same steps as ex 1 & 2 to solve the problem (just use the magnification equation)

68. Sketch

69. Example 3: M = -23 cm 8.4 cm M = -2.738095238 = -2.7 E: M = hi ho
h0 = 8.4 cm
hi = -23 cm
M = ? G: U:
M = -23 cm
8.4 cm S: S:
M = = -2.7 E:
M = hi ho
The lens has a magnification of -2.7, which means the image is inverted. S:

70. Example 4: Magnification of a Diverging Lens
A coin of height 2.4 cm is placed in front of a diverging lens. An upright, virtual image of height 1.7 cm is noticed on the same side of the lens as the coin. What is the magnification of the lens?

71. Sketch

72. Example 4: M = 1.7 cm h0 = 2.4 cm 2.4 cm hi = 1.7 cm M = ?S: G:
M = = 0.71 U: S: E:
M = hi ho
The lens has a magnification of 0.71, which means the image is upright. S:

73. Drawing ray diagrams
A ray diagram is the best way to understand what type of image is formed by a lens, and whether the image is magnified or inverted.
These three rays follow the rules for how light rays are bent by the lens:
A light ray passing through the center of the lens is not deflected at all (A).
A light ray parallel to the axis passes through the far focal point (B).
A light ray passing through the near focal point emerges parallel to the axis (C).

75. Thanks

76. Laws of reflection of light :-
When light falls on a highly polished surface like a mirror most of
the light is sent back into the same medium. This process is called
reflection of light.
Laws of reflection of light :-
i) The angle of incidence is equal to the angle of reflection.
ii) The incident ray, the reflected ray and the normal to the mirror at
the point of incidence all lie in the same plane.

77. Image formed by a plane mirror :-
i) The image is erect.
ii) The image is same size as the object.
iii) The image is at the same distance from the mirror as the object is in
front of it.
iv) The image is virtual (cannot be obtained on a screen).
v) The image is laterally inverted.

78. Spherical mirrors :-
Spherical mirror is a curved mirror which is a part of a hollow sphere. Spherical mirrors are of two types. They are concave mirror and convex mirror.
i) Concave mirror :- is a spherical mirror whose reflecting surface is curved inwards. Rays of light parallel to the principal axis after reflection from a concave mirror meet at a point (converge) on the principal axis.
ii) Convex mirror :- is a spherical mirror whose reflecting surface is curved inwards. Rays of light parallel to the principal axis after reflection from a convex mirror get diverged and appear to come from a point behind the mirror. F F

79. Terms used in the study of spherical mirrors :-
i) Center of curvature :- is the centre of the sphere of which the mirror
is a part (C).
ii) Radius of curvature :- is the radius of the sphere of which the mirror
is a part (CP).
iii) Pole :- is the centre of the spherical mirror (P).
iv) Principal axis :- is the straight line passing through the centre of
curvature and the pole (X-Y).
v) Principal focus :-
In a concave mirror, rays of light parallel to the principal axis after
reflection meet at a point on the principal axis called principal
focus(F).
In a convex mirror, rays of light parallel to the principal axis after
reflection get diverged and appear to come from a point on the
principal axis behind the mirror called principal focus (F).
vi) Focal length :- is the distance between the pole and principal focus
(f). In a spherical mirror the radius of curvature is twice the focal
length.
R = 2f or f = R 2

80. C – centre of curvature CP – radius of curvature
X C F P Y
C – centre of curvature CP – radius of curvature
P – pole XY – principal axis
F – principal focus PF – focal length

81. Reflection by spherical mirrors :-
i) In a concave mirror a ray of light parallel to the principal axis after reflection passes through the focus.
In a convex mirror a ray of light parallel to the principal axis after reflection appears to diverge from the focus.
C F P P F C

82. ii) In a concave mirror a ray of light passing through the focus after reflection goes parallel to the principal axis.
In a convex mirror a ray of light directed towards the focus after reflection goes parallel to the principal axis.
C F P P F C

83. iii) In a concave mirror a ray of light passing through the centre of curvature after reflection is reflected back along the same direction.
In a convex mirror a ray of light directed towards the centre of curvature after reflection is reflected back along the same direction.
C F P P F C

84. iv) In a concave or a convex mirror a ray of light directed obliquely at the pole is reflected obliquely making equal angles with the principal axis.
C F i P i P F C
r r

85. Images formed by concave mirror :-
i) When the object is at infinity the image is formed at the focus, it is highly diminished, real and inverted.
C F P

86. ii) When the object is beyond C, the image is formed between C and F, it is diminished, real and inverted.
C F P

87. iii) When the object is at C, the image is formed at C, it is same size as the object, real and inverted.
C F P

88. iv) When the object is between C and F, the image is formed beyond C, it is enlarged, real and inverted.
C F P

89. v) When the object is at F, the image is formed at infinity, it is highly enlarged, real and inverted.
C F P

90. vi) When the object is between F and P, the image is formed behind the mirror, it is enlarged, virtual and erect.
C F P

91. Images formed by convex mirror :-
i) When the object is at infinity, the image is formed at F behind the mirror, it is highly diminished, virtual and erect.
P F

92. ii) When the object is between infinity and pole, the image is formed behind the mirror, it is diminished, virtual and erect.
P F C

93. Uses of spherical mirrors :-
a) Concave mirrors :-
Concave mirrors are used in torches, search lights and head lights of vehicles to get parallel beams of light.
They are used as shaving mirrors to see larger image of the face.
They are used by dentists to see larger images of the teeth.
Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.

94. b) Convex mirrors :-
Convex mirrors are used as rear-view mirrors in vehicles. Convex mirrors give erect diminished images of objects. They also have a wider field of view than plane mirrors.

95. New Cartesian sign convention for spherical mirrors :-
i) The object is always placed on the left of the mirror and light from the object falls from the left to the right.
ii) All distances parallel to the principal axis are measured from the pole.
iii) All distances measured to the right of the pole are taken as + ve.
iv) All distances measured to the left of the pole are taken as – ve.
v) The height measured upwards perpendicular to the principal axis is taken as + ve.
vi) The height measured downwards perpendicular to the principal axis is taken as – ve.
Object
Direction of incident light
Height
upwards ( + ve )
Distance towards the left ( - ve )
Distance towards the right ( + ve )
Height
downwards ( - ve )
Image
Concave mirror 95

96. Mirror formula for spherical mirrors :-
The mirror formula for spherical mirrors is the relationship between the object distance (u), image distance (v) and focal length (f).
The mirror formula is expressed as :- =
v u f
Magnification for spherical mirrors :-
Magnification for spherical mirrors is the ratio of the height of the image to the height of the object.
Height of the image hi
Magnification = m =
Height of the object ho
The magnification is also related to the object distance and image distance. It is expressed as :-
hi v
Magnification m = =
ho u 96

97. Refraction of light :-
When light travels obliquely from one transparent medium into another it gets bent. This bending of light is called refraction of light.
When light travels from a rarer medium to a denser medium, it bends towards the normal.
When light travels from a denser medium to a rarer medium to a rarer medium, it bends away from the normal.
Normal
Normal
mmmmm
Rarer medium
Denser medium
Denser medium
Rarer medium 97

98. Refraction of light through a rectangular glass slab :-
When a ray of light passes through a rectangular glass slab, it gets bent twice at the air- glass interface and at the glass- air interface.
The emergent ray is parallel to the incident ray and is displaced through a distance.
Normal
Incident ray
Angle of incidence i
Air
Glass
Angle of refraction r
Refracted ray
Rectangular glass slab
Glass
Air
Angle of emergence e
Emergent ray
displacement
Normal 98

99. c) Laws of refraction of light :-
i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
II) The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media.( This law is also known as Snell`s law of refraction.) sine i
= constant
sine r
d) Refractive index :-
The absolute refractive index of a medium is the ratio of the speed light in air or vacuum to the speed of light in medium.
Speed of light in air or vacuum c
Refractive index = n =
Speed of light in the medium v
The relative refractive index of a medium 2 with respect to a medium 1 is the ratio of the speed of light in medium 1 to the speed of light in medium 2.
n21 = Speed of light in medium n 21 = v1 / v2
Speed of light in medium 2 99

100. Spherical lenses :-
A spherical lens is a transparent material bounded by two surfaces one or both of which are spherical.
Spherical lenses are of two main types. They are convex and concave lenses.
i) Convex lens :- is thicker in the middle and thinner at the edges. Rays of light parallel to the principal axis after refraction through a convex lens meet at a point (converge) on the principal axis.
ii) Concave lens :- is thinner in the middle and thicker at the edges. Rays of light parallel to the principal axis after refraction get diverged and appear o come from a point on the principal axis on the same side of the lens.
F F
100

101. Refraction by spherical lenses :-
i) In a convex lens a ray of light parallel to the principal axis after refraction passes through the focus on the other side of the lens. In a concave lens it appears to diverge from the focus on the same side of the lens.
2F F O F F F F O F F2

102. ii) In a convex lens a ray of light passing through the focus after refraction goes parallel to the principal axis. In a concave lens a ray of light directed towards the focus after refraction goes parallel to the principal axis.
2F F O F F F F O F F2

103. iii) In a convex lens and concave lens a ray of light passing through the optical centre goes without any deviation.
2F F O F F F F O F F2

104. Images formed by convex lens :-
i) When the object is at infinity the image is formed at the focus F2, it is highly diminished, real and inverted.
2F F O F F2

105. ii) When the object is beyond 2F1, the image is formed between F2 and 2F2, it if diminished, real and inverted.
2F F O F F2

106. iii) When the object is at 2F1, the image is formed at 2F2, it is the same size as the object, real and inverted.
2F F O F F2

107. iv) When the object is between 2F1 and F1, the image is formed beyond 2F2, it is enlarged, real and inverted.
2F F O F F2

108. v) When the object is at F1 the image is formed at infinity, it is highly enlarged, real and inverted.
2F F O F F2

109. vi) When the object is between F1 and O, the image is formed on the same side of the lens, it is enlarged, virtual and erect.
2F F O F F2

110. Images formed by concave lens :-
i) When the object is at infinity, the image is formed at the focus F1 on the same side of the lens, it is highly diminished, virtual and erect.
F O

111. ii) When the object is between infinity and F1, the image is formed between F1 and O on the same side of the lens, it is diminished, virtual and erect.
FI O

112. Sign convention for spherical lenses :-
The sign convention for spherical lenses is the same as in spherical mirrors except that the distances are measured from the optical centre (O).
The focal length of a convex lens is positive ( + ve ) and the focal length of a concave lens is negative ( - ve ). O
Object
Direction of incident light
Height
upwards ( + ve )
Distance towards the left (- ve )
Distance towards the right ( + ve )
Height
downwards ( - ve )
Image
Convex lens

113. Thanks