•
The study of light and vision is called optics.•
Light is a form of energy which is propagated as Electromagnetic waves which produces the sensation of sight in us.
•
Geometrical optics treats propagation of light in terms of rays and is valid only if wavelength of light much lesses than the size of obstacles.i) Light does not require a medium for its propagation ii) It’s speed in free space (vaccum) is 3 x 108m/s iii) It is transverse in nature
•
In the spectrum of e.m.w. it lies between u.v. and infra-red region and has wavelength between 4000 to 7000O
A
. i.e(
0
.
4
µ
m
to
0
.
7
µ
m
)
Indigo is not distensible from blue. BASIC - DEFINATIONS
•
Source:A body which emits light is called source. Source can be a point one (or) extended one. (a) Self-luminous-source: The source which possess light of it own.
Ex:- Sun, Electric arc, Candle, etc.
(b) Non-luminous-Source: It is a source of light which does not possesses light of its own but acts as source of light by reflecting the light received by it.
Ex: Moon, object around us, Book…….etc.
•
Isotropic Source: It gives out light uniformly in all directions.•
Non-isotropic Source: It do not give out light uniformly in all direction.
•
Ray: The straight line path along with the light travels in a homogeneous medium is called a ray. A single ray cannot be propagated form a source of light.•
Beam: A bundle can bunch of rays is called beam it is called beam it is of following 3 types•
Convergent-beam: In this case diameter of beam decreases in the direction of ray
•
Divergent Beam: It is a beam is with all the rays meet at a point when produced backward and the diameter of beam goes on increasing as the rays proceed forward.
•
Parallel Beam: It is beam in which all the rays constituting the beam move parallel to each other and diameter of beam remains same
•
Object: An optical object is decided by incident rays only. It is if two kinds•
Real Object: In this case incident rays are diverging and point of divergence is the position of real object.
•
Virtual Object: In this case incident ray are converging and point of convergence is the position of virtual object. Virtual object cannot be seen by human eye be cause for an object can image to be seen by eyes, ray received by eyes must be diverging.•
Image: An optical image is decided by reflected (or) refracted rays only. It is of two types.(a) Real Image: This is formed due to real intersection of reflected (or) refracted rays, Real image can be obtained on screen.
•
Virtual-Image: This is formed due to apparent intersection of reflected (or) refracted light rays. Virtual image can’t be obtained on screen.(Note: Human ray can’t distinguish between real and virtual image because in both case rays are diverging)
•
REFLECTION:The phenomenon by virtue of which incident light energy is partly or completely sent back into the same medium from which it is coming after being obstructed by a surface is called reflection.
The direction of incident energy is called incident ray and the direction in which energy is thrown back is called reflected ray. It is of two types.
•
LAWS OF REFLECTION:1) First Law: The incident ray, the reflected ray and the normal to the reflecting surface at the point of incidence, all lie in one plane which is
⊥
'
rto the reflecting surface.2) The angle of incidence is equal to the angle of reflection ∠i=∠r. Note:
2) Whenever reflection takes place, the component of incident ray parallel to reflecting surface remains uncharged, while component perpendicular to reflecting surface (i.e. along normal) reverse in direction.
^ ^ ^ 1
x
i
y
j
z
k
r
=
+
+
→ , ^ ^ ^ 2x
i
y
j
z
k
r
=
+
−
→3) Vector form of laws of Reflection: ^ ^ ^ ^ ^
.
2
I
N
N
I
R
−
=
→R Unit vector along the reflected ray
→
^I
Unit vector along the Incident ray→
^N
Unit vector along the normal ray•
Image formed by a plane mirror:a) Point Source: For construction of image of a point source it is sufficient to consider any two rays falling on mirror. The point of intersection of corresponding reflected rays give the position of image as shown in figure.
OA = AI
(
∴
∆
ABD
≅
∆
ABI
)
Image I lies as much behind the mirror as the object is in front of it. b) Extended source:
•
Characteristics of the image formed by a plane mirror: 1) The image formed by a plane mirror is Virtual 2) The image formed by a plane mirror is Erect3) The image formed by a plane mirror is of same size as object.
4) The image formed by a plane mirror is at the same distance behind the mirror is the object is infront of it.
5) The image is laterally inverted (i.e.) right appear as left and vice-versa.
6) Note: If two plane mirror faring each other are inclined at an angle θ with each other, then number of images are formed due to multiple reflection. This principle is used in the toy
kaleidoscope. (a) If
θ
360
is even integer, then number of images formed is =360−1
θ η Ex: If
θ
=
60
0 then 1 6 1 5 60 360 = − = − = η (b) If θ 360is odd integer, then number of images formed is
θ η =360
Ex: If
θ
=
40
0 (which is not the complete part of 1800) then 9 40360 =
= η
•
Deviation (δ ): The angle between incident and reflected (or) refracted ray is termed as deviation. For reflectionδ
= π
−
2
i
Cases: When i = 0 (Normal incidence)
π
δ
max=
Multiple Reflection:
∑
= i
net δ
δ
δ
i = deviation due to single reflection.Note while summing up, sense of rotation is taken into account.
Q: 1) Two plane mirror are inclined to each other such that a ray of light incident on the first mirror and parallels to the second is reflected from the second mirror parallel to the first mirror. Determine the angle between the two mirror. Also determine the total deviation produced in the incident ray due to the two reflections.
Solution: From figure
3
θ
=
180
060
=
θ
i
2
1= π
−
δ
=
180
−
2
×
30
0=
120
↑
A.C.W. 0 0 2=
π
−
2
i
=
180
−
2
×
30
=
120
δ
( )
↓ ↑ = + = ∴δnet δ2 δ1 240 or 120 Or From fig. δ =180+θ =180+60=
240
0↑
( )
or
120
↓
Q: 2) Calculate deviation suffered by incident ray in situation as shown in figure, after three successive reflections?
i
2
1= π
−
δ
=
180
−
2
×
50
0=
100
↑
↓
=
×
−
=
0 0 2180
2
20
140
δ
↓
=
×
−
=
0 0 2180
2
10
160
δ
( )
↓ ↑ = ↓ + ↓ + ↑ = 0 260 100 160 140 100 or net δQ: 3) Two plane mirrors are inclined to each other at an angle θ. A ray of light is reflected first at one mirror and then at the other. Find the total deviation of the ray?
Solution: Let
α
=
Angle of incidence for M1=
β
Angle of incidence for M2=
1δ
Deviation due to M1=
2δ
Deviation due to M2 From figureα
π
δ
1=
−
2
β
π
δ
2=
−
2
Also ray is rotated in same secure (i.e.) anticlockwise
2 1
δ
δ
δ
Net=
+
Now in∆
OBC
=
π
−
2
α
+
π
−
2
β
∠
OBC
+
∠
BCO
+
∠
COB
=
180
0(
α
β
)
π
δ
Net=
2
−
2
+
(
90
0−
α
) (
+
90
0−
β
)
+
θ
=
180
0( )
θ
π
δ
=
2
−
2
∴
Netα
+
β
=
θ
θ
π
δ
Net=
2
−
2
yI/m = y-co-ordinate of image w.r.t. mirror
For plane mirror
xO/m = -xI/m
Differentiating both sides w.r.t. time (t)
(
O m)
( )
xI m dt d x dt d / / =− x m I x m OV
V
−
=
→ → / / mx Ix mx Ox V V V V→ − =− −→ Ox mx Ix V V V → → − =2 Similarly yI/m = yO/mDifferentiating both sides w.r.t. time we get
y m O y m I
V
V
=
→ → / /In nutshell, for solving numerical problems involving calculation of velocity of image of object with respect to any observer, always calculate velocity of image first with respect to mirror using following points.
11 / 11 /
=
→ → m O m IV
V
1 / 1 /
−
=
→ → m O m IV
V
1 / 11 / /
+
=
→ → → m I m I m IV
V
V
Velocity of image with respect to required observer is then calculated using basic equation for relative motion.
B A B A
V
V
V
→ → →−
=
/Note: If the velocity of the object (w.r.t mirror) is not in a direction normal to the mirror, then the velocity of the object can be resolved into two components one normal to the mirror (vn) and the other along the mirror
(vp). The image has velocities –Vn and VP, normal to and along the mirror.
Q: 1) Point object is moving with a speed V before an arrangement of two mirrors as shown in figure. Find the velocity of image in mirror M1 w.r.t. image in mirror M2?
Solution: 1/2 1 2 → → →
−
=
V
V
V
=2V
sin
θ
F.B.D Angle between 2 1 I Iand
V
V
Is2
θ
∴
their magnitude is V.Q: 2) Find the velocity of image of a moving particle in situation as shown in figure.
Solution: Analysis:
For component of velocity of image
⊥
1/2to mirror0
2
→ → →−
=
V
V
V
I m( )
V
I 1/2=
2
( )
−
2
−
6
=
−
10
m
/
s
∴
⊥For component of velocity of image parallel to the mirror
( )
V
I 11=
8
m
/
s
A point object is approaching the intersection point of mirror with a speed of 100cm/s. The velocity of the image of object formed by M2 w.r.t. velocity of image of object formed by M1 is:
Solution: The components of various velocities are as shown in the figure below
2
IM
V
→
is given by the vector sum of components of velocity of image w.r.t. M2 along the normal and
⊥
1rto thenormal.
−
+
+
+
=
→ ^ 0 0 ^ 0 2 ^ 0 0 ^ 0 237
cos
37
sin
100
37
cos
100
37
cos
37
sin
100
37
sin
100
2i
j
i
j
V
IMs
cm
j
i
48
/
28
^ ^
−
+
=
1 2 1 2,IM IM IM IMV
V
V
→ → →−
=
sec
/
48
128
^ ^cm
j
i
−
+
=
Q: 4) In the situation show in figure, find the velocity of image? Solution: Along x – direction, applying
(
m)
m iV
V
V
V
=
=
−
0−
(
0)
[
0(
0)
]
30 cos 5 60 cos 10 30 cos 5 =− − − − − i V(
)
im s Vi 51 3 / ^ + − = ∴ Along y-direction V0 = Vi s m j Vi 10sin60 5 / ^ 0 = = ∴∴ Velocity of the image
5
(
1
3
)
i
5
j
m
/
s
^ ^+
+
−
=
Q: 5) An object moves with 5m/s towards right while the mirror moves with 1m/s towards the left as shown. Find the velocity of image.
Solution: Take
→
as +ve direction.0
V
V
V
V
i−
m=
m−
( ) ( )
−
1
=
−
1
−
5
−
iV
( )
6
,
0
'
=
A
( )
0
,
6
'
=
M
( )
0
,
9
'
=
N
Q: 7) An object moves towards a plane mirror with a speed v at an angle 600 to the
⊥
1r to the plane of the mirror. What is the relative velocity between the object and the emage?a) V b)
V
2
3
c)2
V
d) 2 V Solution:V
OIV
OV
I → → →−
=
+
+
−
^ 0 ^ 0 ^ 0 ^ 060
sin
60
cos
60
sin
60
cos
i
V
j
V
i
V
j
V
Q: 8) A ray of light making angle 200 with the horizontal is incident on a plane mirror with itself inclined to the horizontal at angle 100, with normal away from the incident ray. What is the angle made by the reflected ray with the horizontal?
Solution: AO = Incident ray
OB = Reflected ray
The reflected ray goes along the horizontal. Hence angle made by the reflected ray with the horizontal is zero. Q: 9) A ray of light making angle 100 with the horizontal is incident on a plane mirror making angle θ with the
horizontal. What should be the value of θ, so that the reflected ray goes vertically upwards?
a) 300 b) 400 c) 500 d) 600
Solution:
•
Number of Images Formed by two Inclined-Plane Mirrors:
b) When mirror are perpendicular: In this case, three images are formed. The ray diagram is shown.
Note that the third image is formed due to rays undergoing two successive reflection. Also, object and its images lie on a circle whose equation is given by
x
2+
y
2=
a
2+
b
2.When an object is placed in front of arrangement of three mutually perpendicular mirror, then total seven images are formed.
Further, object and its image lie on a sphere whose equation is given by
x
2+
y
2+
z
2=
a
2+
b
2+
c
2, where a, b and c are co-ordinates of object.•
Minimum size of Mirror to see Full-Image:AB is the person with E as his eyes,
M1 M2 = plane mirror infront of him.
For the length of the mirror to be minimum, the rays coming from the extreme top and bottom portions of his body. (i.e.) A and B, Should after reflection, be able to just enter his eyes.
The light ray AM, is incident ray and M1E the reflected ray.
So
∠
AM
1N
1=
∠
EM
1N
1As
∆
'
s
AM
1N
1and
EM
1N
1 are similar.( )
AE E N Say x E M 2 1 1 1 1 = = = ∴ ……..(1)Similarly the light rays BM2 and M2E are incident and reflection rays respectively
So
∠
BM
2N
2=
∠
EM
2N
2 2 2 2 2N
and
EM
N
BM
s
S∆
∴
are similar( )
Say N E( )
BE y E M 2 1 2 1 2 = = = ∴ ………(2)Adding equation (1) and (2) yield
= +y x length of mirror
(
) ( )
2 1 2 1 2 1 + = = = AE BC AB (Height of person)Note:- Minimum size is independent of distance between man and mirror.
Q: 1) A plane mirror is inclined at an angle θ with the horizontal surface. A particle is projected from point P (see fig.) at t = 0 with a velocity v at angle
α
with the horizontal. The image of the particle is observed from the frame of the particle projected. Assuming the particle does not collide the mirror, find the (a) time when the image will come momentarily at rest w.r.t. the particle (b) path of the image as seen by the particle.Solution:
(a) The image will appear to be at rest w.r.t. the particle at the instant, the velocity of the particle is parallel to the mirror.
θ
tan
=
x yV
V
θ α α tan cos sin = − V gt V(
)
g
V
t
=
cos
α
tan
α
−
tan
θ
(b) St. line
⊥
1r to mirrorQ: 2) An a oblong object PQ of height ‘h’ stands erect on a flat horizontal mirror. Sun rays fall on the object at a certain angle. Find the length of the shadow on screen placed beyond the shadow on the mirror.
Solution: PS = Shadow on the mirror
P’ Q’ = Inversed shadow of PQ on the screen
Let
α
=
angle of incidence Then PS = h tanα
and QS = h secα
From the properly of image P’ Q’ =
2
(
h
sec
α
)
cos
α
=
2
h
Q: 3) A plane mirror is placed at parallel of y-axis, facing the positive x-axis. An object starts form (2m, 0, 0) with a velocity of
2
i
2
j
m
/
s
^ ^
+
. The relative velocity of image with respect to object is alongSolution: 0
=
=
( ) ( )
2
2+
2
2 → → IV
V
s
m
V
0=
2
2
/
→Relative velocity of image with respect to object is in negative x-direction as shown in figure.
Q: 4) A reflection surface is represented by the equation
x
2+
y
2=
a
2. A ray traveling in negative x-direction is directed towards positive y-direction after reflection from the surface at some point ‘P’. Then co-ordinates of point ‘P’ areSolution: From figure
2
9
,
2
9
=
=
y
x
=
∴
2
,
2
q
q
P
Q: 5) A ray is traveling along x-axis in negative x-direction. A plane mirror is placed at origin facing the ray. What should be the angle of plane mirror with the x-axis so that the ray of light offer reflecting from the plane mirror passes through point (1m, 3m)?
Solution:
Q: 6) Two plane mirror A and B are aligned parallel to each other as shown in the figure. A light ray is incident at an angle 300 at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflection
(including the first one) before it emerges out is____ Solution:
3
2
.
0
30
tan
2
.
0
0=
=
d
30
3
/
2
.
0
3
2
.
.
=
=
∴
Max
No
of
reflection
REFLECTIN FROM CURVED SURFACE: (Spherical – Surface only)
A curved mirror is a smooth reflecting part of any geometry. The nomenclature of curved mirror depends on the geometry of reflecting surface. There are different types of curved mirror like paraboloidal, ellipsoidal, cylindrical, spherical ….etc.
•
Rules for Ray-Diagrams1) A ray of light parallel to principal axis passes (or) appears to pass through four after reflection.
2) A ray of light passing through focus (or) appears to pass through focus becomes parallel to principal-axis after reflection.
3) A ray of light passing through (or) appears to pass through centre of curvature is reflected back.
4) A ray of light hitting pole is reflected making equal angle with principal oxis
Note: 1) Focal length and radius of curvature of plane mirror =
α
2) Concave mirror = Convergent mirrorConvex mirror = Divergent mirror
f
=
u
+
v
•
Relation between the speeds of object and image formed by a spherical mirror We know that, mirror formula is given byf
v
u
1
1
1
=
+
……….(1)Differentiating both sides w.r.t. time (t), we get 0 . 1 . 1 2 2 − = − dt dv v dt du u 0 . 1 . 1 2 2 =− = dt dv u dt dv v
dt
dy
u
v
dt
dv
.
2 2−
=
…………(2)Since v speed of time
dt dv i = = object of speed v dt du = = 0 0 2 .v u v vi = ∴ ………….(3) From equation (1),
( )
f
x
f
u
v
or
f
u
f
u
v
−
=
−
=
Hence equation (2) become
0
.v
f
u
f
v
i
−
−
=
•
Linear magnification: It is defined as the ratio of the size (or) height of the image to the size (or) height of the object.image
of
height
image
of
height
image
of
size
image
of
size
m
=
=
=
∴
o
I
M
=
'
' B
A
image of object ABABp
S∆
andA
'
B
'
p
are similarPA PA AB B A' '= ' ∴ …………(1)
Applying sign conversion
u
PA
AB
=
+
0
=
−
v
PA
B
A
'
'
=
−
1
'
=
−
∴ Equation (1) can be rewritten as
u v O I − − = + − u v O I =−
=
−
∴
u
v
m
same for convex-mirror also.•
Magnification in terms of u, v and f a) As we know thatf
v
u
1
1
1
=
+
Multiplying both sides by ‘u’ we get
f
u
v
u
u
u
=
+
f
u
v
u
=
+
1
f
f
u
f
u
v
u
−
=
−
=
1
f
u
f
m
v
−
=
∴
Since u v m=−( )
−
=
−
−
=
∴
u
f
f
m
or
f
u
f
m
b) As we know thatf
v
u
1
1
1
=
+
f
f
v
f
v
u
v
−
=
−
=
1
Since u v m=−( )
f
v
f
m
or
f
f
v
−
=
−
−
=
Note: a) +ve magnification mean both object and image are upright
b) –ve magnification means, object and image have different orientation (i.e.) if object is upright, then image is inverted.
•
LATERAL-MAGNIFICATION (mL) 0L
L
object
of
length
image
of
length
m
i L=
=
For extended objects the lateral magnification can be obtained by independently imaging the two end points and calculating the length of the image. There is no direct formula to obtain the magnification.
However, if the length of the object is small, them the lateral magnification can be directly obtained from equation
f
v
u
1
1
1
=
+
Differentiating both sides, we get 0 2 2 − = − v dv u du =− = L m u v du dv 2 2
Q: 1) What do we do if the size of the object is large as compared to the distance u? Analysis:
For extended object
B A B A
u
u
V
V
m
−
−
=
2 For tip A(
x
L
)
u
=
−
+
2 R f =− B
V
V
=
f
u
v
1
1
1
=
+
R
l
x
v
B2
1
1
−
=
+
−
from which VB can be obtained∴
Subtracting VB from VA, we can calculate the length of the image.•
Combinations of mirrors:What do we do if we have a combination of mirror? If an object is placed between the mirrors, how do we find the final position of he image?
Analysis: In such situations, we need to simply solve for the reflection at each of the mirror keeping in mind that the image formed by the first mirror is the object of the second mirror and so on.
Case must be taken to correctly apply the sign conversion at each of the mirror. Q: 1) Find the velocity of image in situation as shown in figure?
Solution:
V
9
i
2
i
11
i
m
/
s
^ ^ ^ 0
=
+
=
→ ^ 2 i V→m =− m/s( )
2
30
20
20
−
=
−
−
−
−
=
−
=
∴
u
f
f
m
(
)
11 / 2 11 /
−
=
∴
V
I Mm
V
→O M = -(-2)2 11 ^i
= -44 ^i
m/s.( )
I m I n m I m IV
V
V
+
=
∴
/ − / − /=
−
( )
−
2
12
x
=
−
24
j
m
/
s
I m I n m I n m IV
V
V
+
=
∴
− / − / − /44
i
24
j
m
/
s
^ ^
−
−
=
^ ^ /V
44
i
24
j
2
i
V
V
I I m m
=
−
−
=
=
=
− → →
−
−
=
46
i 24
→j
^Q: 2) A thin rod of length 3
f
is placed along the principal axis of a concave mirror of focal-length ‘f’ such that its image just touches the rod, calculate magnification?
Solution: Since image touches the rod, the rod must be placed with one end at centre of curvature. Case –I Case –II
3
5
3
2
f
f
f
u
=
−
−
−
=
f
f
=
−
( )
( )
2
5
3
5
3
5
f
f
f
f
f
f
u
f
u
v
=
−
−
−
−
−
=
−
=
(
)
(
)
2
3
2
3
5
2
3
5
f
f
f
f
f
u
u
V
V
m
C A C A=
−
−
−
−
−
−
−
=
−
−
=
∴
3
7
3
2
f
f
f
x
=
−
+
−
=
f
f
=
−
( )
( )
4
7
3
7
3
7
f
f
f
f
f
f
u
f
u
V
=
−
−
−
−
−
−
=
−
=
(
)
(
)
4
3
2
3
7
2
4
7
−
=
−
−
−
−
−
−
=
−
−
=
∴
f
f
f
f
u
u
V
V
M
C A C A CONCEPTUAL POINTS•
It a hole is formed at the center of mirror, the image position and size will not change. The intensity will reduce depending on the size of the hole.•
For all object positions a convex-mirror forms a virtual and erect image PROBLEMS OF MIRRORSQ: 1) A short linear object of length ‘b’ lies along the axis of a concave mirror of focal-length f, at a distance u from the mirror. The size of image approximately is
Solution: 2 2
−
=
=
u
f
f
u
V
M
axial 2
−
=
u
f
f
O
I
− = ⇒ − = 2 2 u f f b I u f f b IQ: 2) Two spherical mirrors M1 and M2 one convex and other concave having same radius of curvature R are
arranged coaxially at a distance 2R (consider their pole separation to be 2R). A bead of radius a is placed at the pole of the convex mirror as shown. The ratio of the sizes of the first three images of the bead is
(
R
)
R
V
+
−
=
−
2
2
1
1
1R
R
V
2
4
2
1
1
1−
=
R
V
2
3
1
1−
=
3 1 2 3 2 3 2 1 / 1 − = − − = ⇒ − R R m R V . ⇒ object distance 3 4 3 2 2R− R = R = = − + 2 2 3 4 1 1 2 R R VR
R
V
4
2
2
1
2+
=
11 4 2 R V =11
3
3
4
11
4
2 2 2−
=
−
=
−
=
R
R
u
V
m
11 3 2 = ∴mSo radius of second image
11 3 . 11 3 2 a a a = = ⇒
Similarly radius of third image is
41 3 a a = 41 1 : 11 1 : 3 1 ∴ Answer
Q: 3) When an object is placed at a distance of 60cm from a convex spherical mirror, the magnification produced is 1/2. where should the object be placed to get a magnification of 1/3?
Solution: u=−60cm u V m=− 60 2 1 − − = V (or) V =+30cm
60
1
30
1
60
1
1
1
1
=
+
−
=
+
=
∴
v
u
f
cm
f
=
+
60
∴
In second case 3 100 3 1 u V u V m= =− =− Asf
v
u
1
1
1
=
+
60 1 3 1− = u ucm
u
=
−
120
Q: 4) Two objects A and B when placed one after another in front of a concave mirror of focal-length 10cm, form images if same size. Size of object A is 4 times that of B. If object A is placed at a distance of 50cm from the mirror, what should be the distance of B from the mirror?
Solution: For object A For object B
1 1 2
u
f
f
h
h
m
−
=
=
1 1 2'
'
'
u
f
f
h
h
m
−
=
=
1 2 2 2 1 1 1 2'
f
u
u
f
h
h
h
h
m
m
−
−
=
×
=
∴
Ash
1=
4h
11 andh
2=
h
21,f
=
−
10
cm
cm
u
1=
−
50
50 10 10 4 1 2 + − − − = ∴ ucm
u
2=
−
20
Q: 5) A concave mirror of focal length 10cm is placed at a distance of 35cm form a wal. How far from the wall should an object be placed to get in image on the wall?
Distance of the object form wall = 35 – 14 = 21 cm
Q: 6) An object is placed at a distance of 36cm form a convex mirror. A plane mirror is placed in between so that the two virtual images so formed coincide. If the plane mirror is at a distance if 24cm from the object, find the radius of curvature of the convex mirror.
Solution:
OP
=
u
=
−
36
cm
cm
PI
V
=
=
+
12
18
1
36
3
1
12
1
36
1
1
1
1
=
+
−
=
+
=
+
=
∴
V
u
f
cm
f
=
18
∴
cm
f
R
=
2
=
2
×
18
=
36
∴
Q: 7) A convex mirror of focal length ‘f’ forms an image which is
n
1
times the object. The distance of the object which is
n
1
times the object. The distance of the object from the mirror is Solution:
u
V
−
=
+
=
η
η
1
η
u
V
=
−
u
V
f
1
1
1
+
=
∴
η
η
1
1
1
+
−
=
u
f
(
)
f
u
= η
−
−
1
Q: 8) An object of size 7.5cm is placed in front of a convex mirror of radius of curvature 25cm at a distance of 40cm. The size of the image should be
Solution:
u
f
f
O
I
−
=
=
η
u=−40(
)
(
)
(
25
/
(
2
) (
)
40
)
2
/
25
2
/
2
/
5
.
7
=
R
−
u
=
−
−
R
I
cm
I
=
+
1
.
78
Q: 9) The image formed by a convex mirror of focal length 30cm is a quarter of the size of the object. The distance of the object from the mirror is
Solution:
u
f
f
m
−
=
u
−
+
+
=
+
30
30
4
1
cm
u
=
−
90
Q: 10) A concave mirror of focal length f (in air) is immersed in water
(
µ
=
4
/
3
)
. The focal length of the mirror in water will be a)f
b) f 3 4 c) f 4 3 d) f 3 7 Solution: On immersing a mirror in water, focal length of the mirror remains uncharged.Q: 11) An object is 20cm away form a concave mirror with focal-length 15cm. If the object moves with a speed of 5m/s along the axis, then the speed of the image will be
Solution: 15 1 20 1 1 − = − V
cm
V
=
−
60
0 2 . V u V Vi − =( )
5 . 20 60 2 =s
m /
45
=
Q: 12) A concave mirror is placed at the bottom of an empty tank with face upwards and axis vertical when Sun-light falls normally in the mirror, it is focused at distance of 32cm form the mirror. If the tank filled with water
=
3
4
µ
upto a height of 20cm, then the Sunlight will now get focused at Ans: 9cm above water levelSolution: f 5cm 2 2 = = = For part PQ 0 1
L
u
f
f
L
−
=
(
20
)
3
5
5
0 0L
L
=
−
×
−
−
−
−
=
For part QR 0 2 2L
u
f
f
L
−
=
(
20
)
9
5
5
0 0 2L
L
=
×
−
−
−
−
=
1
3
2 1=
∴
L
L
•
CONCEPT OF NEWTON’S FORMULA (FOR A MIRROR)In this formula, the object and image distance are expressed w.r.t. focus. Consider an object O kept beyond ‘C’ of a concave mirror, and whose image is formed at I with in C.
Let OF = x and IF = y From triangle OMC
(
π
α
)
α
θ
sin
sin
sin
OM
OM
OC
=
−
=
…………(1)α θ sin sin IM IC = ………….(2) Dividing equation (1) and (2) yields
IP OP IM OM IC OC = = (since M is close to P) (or)
y
f
f
x
y
f
f
x
+
+
=
−
−
y
f
f
y
x
f
x
y
f
y
x
f
f
x
−
2+
−
=
−
+
2−
(or)x
y
=
f
2 y x f = ∴ 1) Asx
y
=
f
2 (or)y
x
α
1
(i.e.) The distance of object and image form the focus are inversely proportional to each other. In other words, the more the object distance (from the focus), the less will be the image distance (from the focus) and vice versa
2) If
x
→
0
;
y
→
∞
and ifx
→
∞
;
y
→
0
. If the object is at focus the image is a far off distance and vice-versa.3) From
xy
=
f
2;
if
x
=
f
, theny
=
f
.Thus, if the object be at C, then image will also be at C (for a concave mirror) and if the object is at P, then the image will also be at P (for a convex mirror)
4) Since
f
2 is necessarily +ve for both types of mirror, so x and y bear the same sign, which implies that both the object and the image always lie an the same side of focus.A) GRAPH OF |x| Versus |y| :
Since
xy
=
f
2 represents a rectangular hyperbola, existing in the first and third quadrant (f
2 being positive).∴
The graph of |y| vs |x| will be a rectangular hyperbola existing only in the first quadrant.B) GRAPH OF U Versus V : Since
xy
=
f
2(
)(
)
2f
f
v
f
u
−
−
=
∴
f
f
and
y
v
x
u
=
=
=
−
∴
,
We have, form(
)(
)
2f
f
v
f
u
−
−
=
(
)(
)
2f
f
y
f
u
+
+
=
Or[
x
−
( )
−
f
]
[
y
−
( )
−
f
]
=
f
2Evidently it is again an equation of a rectangular hyperbola with origin of coordinates shifted to the point (-f, -f) (see figure
↓
)3) GRAPH OF v 1 VERSUS u 1 From mirror formula
f
u
v
1
1
1
=
+
Putting x u = 1 and y v = 1 , we havey
y
x
+
=
1
(or) opposite signs. The intercepts on x and y axis are each
( )
f
or
f
1
1
−
+
according as theobject and image are to the right (or) left of the mirror. For a concave mirror:
u is always –ve
v can be positive (or) negative and f is –ve.
For a convex mirror u is always negative v is always positive and f is always positive.
•
CONCEPT OF CRITICLA ANGLEWhen a ray of light is traveling form denser medium to rarer medium, it get refracted and the ray derivates away form the normal.
If we keep increasing angle of incidence then at an angle, the angle of refraction becomes 900 . This is known as Critical –Angle (c).
When angle of incidence is increased, further the ray gets reflected back in the same medium. This phenomena is known as T.I.R.
According to Snell’s Law
C a b
i
sin
=
µ
=
C a bi
sin
1
µ
⇒
C depends on colour and and temp⇒
CRed > Cviolet∴
CRed < Cviolet⇒
If temp↑
C
↓
Q: 1) The sum (diameter d) subtends an angle
θ
radius at the pole of a concave mirror of focal length f. Find te diameter of the image of sun formed by mirror?Solution:
f
u
v
1
1
1
+
=
we getf
v
=
−
1
1
(∴
u is very large so1
≈
0
u
Orv
=
−
f
It means image is formed at focus Taking
' f
'
as radius and usingf
r
and
d
when
r
=
=
=
l
l
θ
f
d
or
f
d
θ
θ
=
=
∴
From
∆
les
OBC and IBC We havei
=
α +
β
And
β
=
r
+
r
( )
or
r
=
β
−
r
From Snell’s law
r
i
sin
sin
1 2=
µ
µ
µ
1sin
i
=
µ
2sin
r
For small angle of incidence I, we can write
sin
i
≈
i
and
sin
r
≈
r
∴
µ =
1i
µ
2r
µ
1[
α
+
β
]
=
µ
2[
β
−
r
]
As ‘i’ is small, and so
α
,
β
and
r
are also small. Thus(
α
+
β
)
=
tan
α
+
tan
β
R
h
u
h
+
+
−
=
And(
)
V
h
R
h
r
=
−
−
β
=
−
+
+
−
∴
v
h
R
h
R
h
u
h
2 1µ
µ
After simplifying we get
R
u
v
1 2 1 2µ
µ
µ
µ
−
=
−
R
v
1
1
1 2 1 2−
=
−
µ
µ
µ
µ
µ
−
=
−
→
R
u
v
1
1
2 1 2 1µ
µ
This formula is derived for convex surface and for real Image
Analysis: Here two cases a rise. Refractive index is varying either as function of y (or) function of x. Case-I:
µ
=
f
( ) ( )
y
i
.e
.
Refractive index varies with yAt some height h angle of incidence is
θ
y and refractive index isf
( )
y
from Snell’s Law
=
θ
µ
sin
constant( )
y
y
f
θ
α
sin
sin
1
×
=
∴
……..(1) Slope of curve at A(
y
)
dx
dy
θ
−
=
tan
90
dx
dy
y
=
⇒
cot
θ
From equation (1)( )
{
}
α
α
sin
sin
2 2−
=
f
y
dx
dy
Case-I:
µ
=
f
( ) ( )
y
i
.e
.
Refractive index varies as function of x. According to Snell’s Law=
θ
µ
sin
constant(
0)
0sin
90
sin
1
×
α
=
µ
−
θ
(
0)
0sin
90
sin
α
=
µ
−
θ
0 0cos
sin
α =
µ
θ
∴
Here
θ
0angle of refracting ray at point A with OXSo 0 0
sin
cos
µ
α
θ =
And 2 0 2 0sin
1
sin
µ
α
θ
=
−
Now Snell’s Law at M gives
( )
x
sin
θ =
xµ
0sin
θ
0f
Or( )
2 0 2 0sin
1
sin
µ
α
µ
θ
=
−
x
f
x( )
x
f
xα
µ
θ
2 2 0sin
sin
=
−
∴
Now slope of tangent at M is given as x
dx
dy
θ
tan
=
( )
{ }
µ
α
α
µ
2 2 0 2 2 2 0sin
sin
+
−
−
=
x
f
dx
dy
Q: 1) If
µ
=
1
+
y
and ray of light is incident at grazing incidence at origin, then find equation of path of refracted ray.Solution: We can use result derived above for which
( )
090
1
+
=
=
y
and
α
y
f
2 / 1y
dx
dy
=
∴
So4
2x
y
=
Q: 2) An object is at a distance 25cm form the curved surface of a glass hemisphere of radius 10cm. Find the position of the image and draw the ray diagram
(
µ
g=
1
.
5
)
R
u
v
1 2 1 2µ
µ
µ
µ
−
=
−
cm
25
−
=
µ
1
1=
µ
2
/
3
2=
µ
R = 140cmcm
V
=
150
∴
∴
The rays are converging beyond of at 140cm form Q. Again refraction takes place at the plane surface.For refraction at second face
1
5
.
1
,
2=
1=
∞
=
µ
µ
R
cm
140
+
=
µ
?
=
v
UsingR
u
v
2 1 2 1µ
µ
µ
µ
−
=
−
∞
−
=
+
−
1
1
.
5
140
5
.
1
1
v
3
.
93
=
v
∴
The ray meet axis at 93.3cm form point Q.•
PROBLEMS ON REFRACTION1) A light ray is incident at an angle of incidence double that of refraction on one face of a parallel sided transparent slab of refractive index
'
µ
'
and thickness ‘t’. Find the lateral displacement of the ray? Solution:i
=
2
r
( )
(
)
t
r
r
r
t
r
r
r
t
r
r
i
t
D
tan
cos
sin
cos
2
sin
cos
sin
−
=
−
=
=
=
As
r
i
sin
sin
=
µ
r
r
r
r
r
sin
cos
.
sin
2
sin
2
sin
=
=
µ
r
cos
2
=
µ
2
cos
µ =
µ
1
2
tan
2−
=
∴
µ
r
1
2
2−
=
∴
µ
t
D
µ
µ
24
−
=
∴
D
t
Q: 2) A light ray is incident on a transparent slab of R. I.
µ
=
2
, at an angle of incidenceπ
/
4
. Find the ratio of the lateral displacement suffered by the light ray to the maximum value which it could have suffered?Solution:
,
2
4
=
=
π
µ
i
6
/
2
sin
4
/
sin
sin
sin
π
π
µ
⇒
=
∴
=
=
r
r
r
i
( )
6
cos
6
4
sin
cos
sin
π
π
π
−
=
−
=
t
r
r
i
t
D
−
=
6
tan
.
4
cos
4
sin
π
π
π
t
−
×
=
3
1
2
1
2
1
t
(
3
1
)
6
−
=
t
D
(
)
6
6
2
3
−
=
∴
t
D
Q: 3) Light of a certain colour has 2000 waves in one millimeter of air. Find the number of waves of that light in one millimeter length of water and glass respectively?
Solution:
λ
a=
wavelength in air=
m
λ
wavelength in mediumm m a
n
n
µ
λ
λ =
=
∴
1 2 1 2n
n
=
m×
∴
µ
2660
2000
33
.
1
2=
×
=
n
in water3000
2000
50
.
1
2=
×
=
n
in glassQ: 4) Light from a sodium-lamp
(
λ
=
58
nm
)
traverses a distance of 60m in a chloroform(
µ
=
1
.
45
)
in a certain time. Determine the time difference occurring when light happens to traverse the same length in ethyl ether(
µ
=
1
.
35
)
Solution: mc
d
t
=
( )
s
m
m
t
chloroform/
10
3
45
.
1
60
8 1=
×
For ethyl ether
( )
s
m
m
t
er/
10
3
35
.
1
60
8 .2=
×
2 1t
t
t
=
−
∆
(
1
.
45
1
.
30
)
10
30
60
8−
×
=
m
sec
10
0
.
2
×
−8=
Q: 5) A rectangular glass slab of thickness 12cm is placed over a small coin kept on a table. A thin transparent beaker filled with wager to a height 6cm & placed over the block. Find the apparent shift of the position of the coin, when viewed from a point directly above it?
Solution:
−
=
1 1 11
1
µ
t
S
−
=
3
2
1
12
cm
4
=
−
=
2 2 21
1
µ
t
S
−
=
4
3
1
6
= 1.5 cmcm
S
S
S
=
1+
2=
4
+
1
.
5
=
5
.
5
∴
Q: 6) The time taken by light to cover a distance of 9mm in water is____
Solution:
C
w10
m
/
s
4
9
3
/
4
10
3
8 8×
=
×
=
sec
10
4
10
9
4
10
9
11 8 3 − −×
=
×
×
×
=
=
wC
d
t
=
4
×
10
−11×
10
9ns
=
0
.
04
ns
Q: 7) A ray incident at an angle of incident 600 enters a glass – sphere of R.I
µ
=
3
. The ray is reflected and reflected and refracted at the farther surface of the sphere. The angle between reflected and refracted rays at this surface is_____Solution:
