(Optics)
• •
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 obs tacles. tacles. i) Light does not require require a medium medium for its propagation propagation ii) It’s speed in in free space (vaccum) (vaccum) is 3 x 10 8m/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 O
4000 to 7000 A . i.e (0.4µ 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 p oint one (or) extended one. (a)
Self-luminous-source: Self-luminous -source: The source which possess light of it own. Ex:- Sun, Electric arc, Candle, etc.
(b)
Non-luminous-Source: Non-luminous-Source: It is a source source of light which does not possesses light of its its own own but acts as source of light by reflecting the light received by it. Ex: Moon, object around us, Book…….etc.
• • •
Non-isotropic Source: 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.
Isotropic Source: It gives out light uniformly in all directions.
Medium: Substance through which light propagates is called medium
A single ray cannot b e propagated form a source of light.
• •
Convergent-beam: 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
Beam: A bundle can bunch of rays is called beam it is called beam it is of following 3 types
diameter diameter of b eam goes on increasing as the rays proceed for ward.
•
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 received by eyes eyes must be diverging. 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 reflected ray. It is of two types.
R EFLECTION: ON: • LAWS OF REFLECTI 1) First Law: The T he incident incident ray, the reflected reflected ray and the normal to the reflecting reflecting surface at the
⊥' r to the reflecting surface. 2) The angle of incidence is equal to the angle of reflection ∠i = ∠r . point of incidence, incidence, all lie in in one plane which which is
Note: 1) The laws of reflection reflection are valid va lid for any smooth reflecting surface irrespective of geometry.
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. →
r1
^
^
^
= xi + y j + z k ,
→
r2
^
^
= xi + y j − zk
3) Vector form of laws of Reflection: ^
^
^
^
^
R = I − 2 I . N N 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 p lane 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 Erect 3) The image formed by a plane mirror is of same size as object. 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 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. reflection. This principle is used in the toy kaleidoscope. (a) If
360
θ
Ex: If θ (b) If
θ
Ex: If θ
•
= 600 then η =
360
η=
is even integer, then number of images formed is
360 60
360
θ
−1
−1 = 6 −1 = 5
is odd integer, then number of images images formed is
η=
360
= 40 0 (which is not the complete part of 180 0) then η =
θ 360 40
=9
Deviation ( δ ): The angle between incident and reflected (or) refracted ray is termed as deviation. For reflection
δ = π − 2i
Cases: When i = 0 (Normal incidence)
δ max max = π When i
=
π 2
(Grazing incidence)
δ min min = 0
Multiple Reflection:
δ net = ∑ δ i
δi
= deviation due to s ingle reflection. reflection.
Note while while summing up, sense of of rotation is taken into account.
Q: 1)
Two plane mirror mirror are inclined to each other such that a ray of light incident on the first first mirror and parallels 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
3θ
= 180
θ = 600 δ1 = π − 2 i = 180 − 2 × 30 0 = 120 ↑ A.C.W. δ 2 = π − 2 i = 180 − 2 × 300 = 120 0 ∴δ net = δ 2 + δ 1 = 240 ↑ (or ) 120 ↓ Or From fig.
δ = 180 + θ = 180 + 60
= 2400 ↑ Q: 2)
(or ) 120 ↓
Calculate deviation suffered by incident ray in situatio situation n as shown in figure, figure, after three successive reflections?
Solution:
F,B.D
δ1 = π − 2 i = 180 − 2 × 50 0 = 100 ↑ δ 2 = 180 − 2 × 200 = 1400 ↓ δ 2 = 180 − 2 ×100 = 1600 ↓ δ net = 100 ↑ +140 ↓ +160 ↓ = 100 ↑ (or ) Q: 3)
260 0 ↓
Two plane mirrors are inclined inclined to each 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 t he ray? Solution:
Let
α = Angle of incidence for M 1 β = Angle of incidence for M 2 δ1 =
Deviation due to M 1
δ 2 = Deviation due to M 2
From figure
δ1 = π − 2 α
δ2 = π − 2β Also ray is rotated in same secure (i.e.) anticlockwise anticlockwise
δ Net = δ 1 + δ 2
Now in
= π − 2α + π − 2β
COB = 180 0 ∠OBC + ∠BCO + ∠COB
δ Net = 2π − 2(α + β )
(90 − α ) + (90 − β ) + θ = 180
∴ δ Net = 2π − 2(θ )
α + β =θ
δ Net = 2π − 2θ
•
∆ OBC
Velocity of Image: Let xO/m = x-co-ordinate of object w.r.t. mirror xI/m = x-co-ordinate of image w.r.t. mirror yO/m = y-co-ordinate of object w.r.t. mirror yI/m = y-co-ordinate of image w.r.t. mirror For plane mirror
xO/m = -xI/m Differentiating Differentiating both sides w.r.t. time (t)
d d ( xO / m) = − (xI / m) d d
V→ = − V→ O / m x I / m x →
→
V Ox− Vmx = −VIx − V mx VIx
→
→
= 2 V mx − V Ox
Similarly y I/m = yO/m Differentiating Differentiating both sides w.r.t. time we g et
0
0
0
V→ = V→ I / m y O/ m y In nutshell, for solving numerical problems involving calculation calculation of velocity of image of object with respect to any observer, always calculate velocity velocity of image first with respect to mirror using following points.
V→ = V→ I m O m / 11 / 11 V→ = − V→ I m O m / 1 / 1 → → V I / m = VI / m + V I / m 11 1 →
Velocity of image with respect to required observer is then calculated using basic equation for relative motion. →
V A/ B
→
→
= V A −V B
Note: If the velocity velocity of the object object (w.r.t mirror) is is not in a direction direction normal to the mirror, mirror, then the velocity velocity of the t he object can be resolved into two t wo components one normal to the mirror (v n) and the other along the mirror (v p). The image has velocities –V n and VP , normal to and along the mirror.
Q: 1)
Point object object is moving with with a speed speed V before an arrangement arrangement of two two mirrors as shown in in figure. Find the velocity of image in mirror M 1 w.r.t. image in mirror M 2? →
Solution:
F.B.D
V 1/ 2
→
→
= V1 − V 2
= 2 V sin θ
Angle between Is 2 θ
Q: 2)
VI and VI 1
2
∴ their magnitude is V.
Find the velocity velocity of image image of of a moving moving particle in situation as shown in figure.
Solution:
Analysis:
For component of velocity of image →
1/ 2
→
= 2 V m− V 0 = 2(− 2) − 6 = −10 m/ s VI
∴ (VI )⊥
→
⊥1/ 2 to mirror
For component of velocity of image parallel to the mirror
(VI )11 = 8m/ s
∴
Velocity of time VI
= (VI )12 + (VI )2n
=
100 + 64
=
164 m
4 ∴ θ = tan −1 5 Q: 3)
Two plane mirror mirror are placed as shown shown in the figure figure below:
A point object is approaching approaching the intersection point of mirror with a speed of 100cm/s. The velocity velocity of the t he image of object formed by M 2 w.r.t. velocity of image of object formed by M 1 is:
Solution:
The components components of various velocities velocities are as shown in the figure below
→
V IM is given by the vector sum of components of velocity of image w.r.t. M 2 along the normal 2
and
⊥1r to the normal.
→
V IM
2
^ ^ ^ ^ 2 0 0 0 2 0 0 0 = 100 sin 37 i + 100 sin 37 cos 37 j + − 100 cos 37 i + 100 sin 37 cos 37 j
^ ^ cm/ s i j 28 48 = − +
→
V IM
→
2,
IM 1
→
= V IM − V IM 2
1
^ ^ = − 128 i + 48 j cm/ sec
Q: 4)
In the the situation situation show in in figure, find the velocity velocity of image? image?
Solution:
Along x – direction, applying
Vi
= Vm = −(V0 − Vm)
Vi
− (− 5 cos 30 0 ) = − 10 cos 600 − (− 5 cos 300 )
∴ Vi = −5(1 +
^
)
3 i m/ s
Along y-direction V0 = Vi
^
∴ Vi = 10 sin 600 = 5 j m/ s ∴ Q: 5)
Velocity of the image
= −5(1 +
^
)
^
3 i + 5 j m/ s
An object object moves moves with 5m/s towards towards right while while the mirror moves moves with 1m/s towards towards the left left as shown. Find the velocity of image.
Solution:
Take
→ as +ve direction.
Vi − Vm = Vm − V0
Q: 6)
Vi
− (− 1) = (− 1) − 5
Vi
= −7 m/ s ⇒ 7m/ s and direction towards left.
Find the region on y-axis in which which reflected reflected rays are present object object is at A(2, A(2, 0) and MN is a plane mirror, mirror, as shown shown
Solution:
A' = (6,0 ) M ' = (0,6 ) N ' = (0,9)
Q: 7)
0
An object object moves moves towards a plane mirror mirror with with a speed v at an angle 60 to the
⊥1r
to the plane of
the mirror. What is the relative velocity velocity between the object and the emage? a) V
b) →
Solution:
V OI
→
3 2
V
c)
V 2
d)
V 2
→
= VO− V I
V cos 600 i^ − V sin 60 0 j^ + V cos 60 0 i^ + V sin 60 0 ^j
Q: 8)
0
A ray of light making angle 20 with the horizontal is incident on a plane mirror with itself 0
inclined to the horizontal at angle 10 , 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 reflected ray goes go es along the horizontal. Hence angle made by the reflected ray with the horizontal is zero.
Q: 9)
A ray of light making angle 10 0 with the horizontal is incident on a plane mirror making angle with the horizontal. What should be the value of θ , so that tha t the reflected reflected ray goes vertically upwards? a) 30 0
b) 400
c) 50 0
d) 600
Solution:
•
Number of Images Formed by two Inclined-Plane Inclined-Plane Mirrors: a) When mirror are parallel: In this case, infinite images are formed due to multiple reflections.
θ
b) When mirror are perpendicular: In In this case, case, three images are formed. The ray ray diagram is shown.
Note that the third image image is is formed due to to rays undergoing undergoing two successive successive reflection. Also, Also, object object and its images lie on a circle whose equation is given by x2
+ y2 = a2 + b2 .
When an object is placed in front of arrangement of three mutually perpendicular perpendicular mirror, then total seven images are formed. Further, object object and its image lie on a sphere whose equation is given by
x2
•
co-ordinates of object. + y2 + z2 = a2 + b2 + c2 , where a, b and c ar e co-ordinates
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.) (i.e.) A and B, Should Should after reflection, reflection, be able to just enter enter his eyes. eyes. The light ray AM, is incident ray and M 1E the reflected ray.
∠AM1 N1 = ∠EM1 N1
So As
∆' s
AM 1 N1 and EM1 N1 are similar.
∴ M 1 E1 = x
1
Say = N1 E = ( AE )
……..(1)
2
Similarly the light rays BM 2 and M2E are incident incident and reflec r eflection tion rays respectively So
∠BM 2 N2 = ∠EM 2 N2
∴ ∆S s
BM 2 N2 and EM 2 N2 are similar
∴
= y(Say) = N2 E = 1 (BE )
M 2 E1
………(2)
2
Adding equation (1) and (2) yield
1
1
1
2
2
2
+ = length of mirror = ( AE + BC ) = ( AB) =
(Height of person)
Note:- Minimum Minimum size is is independent independent of distance distance between between man man and mirror.
Q: 1)
A plane mirror is inclined at an an angle
θ
with with the horizontal surface. A particle is projected from
point P (see fig.) f ig.) at t = 0 with 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 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 as seen by by the particle. 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 is parallel to the mirror.
Vy Vx
= tan θ
V sin α − gt = tan θ V cos α
t= (b)
Q: 2)
V cos α (tan α g
St. line
⊥1 r
− tan θ )
to mirror
An a oblong object object PQ of height height ‘h’ stands erect erect on a flat horizontal horizontal mirror. 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 o n the screen
Let
α=
Then
PS = h tan α and QS = h sec α
angle of incidence
From the properly of image P’ Q’ = 2(h sec α ) cos α
Q: 3)
= 2h
A plane mirror mirror is placed at parallel of y-axis, facing the positive x-axis. An object starts start s form
^
^
(2m, 0, 0) with a velocity of 2 i+ 2 j m/ s . The relative velocity of image with respect to object is along →
Solution:
V0 →
V0
→
= VI = (2)2 + (2 )2 =2
2 m/ s
Relative velocity velocity of image with respect to object is in n egative x-direction x-direction as shown in figure.
2
Q: 4) A reflection surface is represented by the equation x
+ y2 = a2 . 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’ are Solution: From figure
x=
∴ Q: 5)
9
, y=
9
q , 2
q
2
P=
2
2
A ray is traveling along x-axis in negative x-direction. A plane mirror is 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 reflecting from the plane mirror passes through point (1m,
3 m)?
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 30 0 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 emerges out is____
Solution:
d = 0.2 tan 30 0
∴
=
0.2 3
Max Max. No. of reflection =
2 3 0.2 / 3
= 30
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, paraboloidal, ellipsoidal, cylindrical, cylindrical, spherical spherical ….etc.
•
Sign-Conversion:
•
Rules for Ray-Diagram Ra y-Diagramss 1) A ray of light parallel to 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 becomes parallel to principal-axis principal-axis after reflection. reflection.
3) A ray of light passing pa ssing through (or) appears to pass through centre centre of curvature is reflected reflected back.
4) A ray of light hitting pole is reflected making equal angle with principal oxis
Note:
1) Focal length length and radius radius of curvature curvature of plane plane mirror mirror =
α
2) Concave mirror = Convergent mirror Convex mirror = Divergent mirror
•
Relation between between focal-length and radius of curvature:
f = R 2
•
Both for concave and convex mirror.
Mirror formula (or) Mirror Equation: The relation between u, v and f of a mirror is known as mirror formula
1 1 1 f = u + v
•
Relation between the speeds of object and image formed by a spherical mirror We know that, mirror formula is given by
1
+1=
u v
1
……….(1)
f
Differentiating Differentiating both sides w.r.t. time (t), we get
−
1 du 1 dv . − 2. =0 2
u
d
d
dv 1 dv . = − 12 . = 0 2 dt
u dt
dv v2 dy =− 2. dt u dt
…………(2)
Since
dv = vi dt
= speed of tim time
du = v0 dt
= speed of object
2
v ∴ vi = .v0 u
………….(3)
From equation (1),
v=
u f u − f
(or )
v f = u x − f
Hence equation (2) become
vi
•
= − f .v0 u − f
Linear magnification: It is defined as the ratio of the size (or) height of the image to the size (or) height of the object.
size of image height of image = size of image height of image
m=
I ∴ M = o
•
Magnification produced by concave mirror:
A ' B ' = image of object AB
∆S ABp and ∴
A' B ' AB
=
A' B ' p are similar
PA ' PA
Applying sign conversion
AB = +0
PA = −u
A' B' = −1
PA' = −v
…………(1)
∴
Equation (1) can be rewritten as
−I = −v +O −u I v =− O
∴
•
m= − v u
same for convex-mirror also.
Magnification Magnification in terms of u, v and f a) As we know that
1
+1=
u v
1
f
Multiplying Multiplying both sides by ‘u’ we get
u u u + = u v f 1+
u u = v f
u u = − 1 = u − f v f f
∴
v = f m u − f
Since m= −
∴ m= −
v u
f u − f
b) As we know that
1
(or ) m=
1
1
u v
f
+ =
f f − u
Multiplying both sides by V, we get
v v v + = u v f
v v +1 = u f
v v v− f = −1 = u f f Since
v u
m= −
− f − v ( ) o r m = − v f = f f Note:
a) +ve magnificati magnification on mean both object object and image image are upright b) –ve magnificati magnification on means, object and image have different orientation (i.e.) if object object is upright, then image is inverted.
•
LATERAL-MAGNIFICATION LATERAL-MAGNIFICATION (mL)
mL
=
length of image Li = length of object L0
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
1
u
1
1
v
f
+ =
Differentiating both sides, we get
− du2 − dv2 = 0 u
dv v2 du = − u2 = mL Q: 1)
What do we do if the size of the object object is large as compared compared to the distance distance u? Analysis: For extended object
m2
= VA − VB uA − uB
For tip A
u = −( x + L ) f = −
R 2
V
= VB
1
+ =
1
1
v u
f
1
vB
−
1
x+ l
=
−2 R
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 correctly apply the sign conversion at each of the mirror.
Q: 1)
Find the velocity velocity of image in situation as shown in figure?
→
Solution:
^
^
^
V0 = 9 i + 2 i = 11i m/ s
→
^
Vm = −2 i m/s
∴ m=
f = f − u
− 20 = −2 − 20 − (− 30)
→ ∴ (VI / M )11 = −m V O / M 11 2
2
^
= -(-2) 11 i ^
= -44 i m/s.
− − ∴ (VI / m) = V I / m + V I / m n I
= −(− 2 ) 12 x = −24 j
m/ s
− − − ∴ = + V V V I / m I / m I / m n n I ^ ^ = − 44 i − 24 j m/ s
VI
− = V→ m = − 44 i^ − 24 j→ = 2 i^ V = I m /
→ ^ = − − i j 46 24
Q: 2)
A thin rod of length
f
is placed along the principal axis of a concave mirror of focal-length ‘f’
3
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
f = − f
u = − 2 f −
Case –II
f
= 3
− 5 f 3
− 5 f (− f ) 5 u f 3 v= = = f u − f − 5 f − (− f ) 2 3
∴
m=
VA − VC uA − uC
5 f
− − (− 2 f ) 3 f 3 = =− − 5 f ( f ) 2 − −2 3
f = − f
x = − 2 f +
f
= − 3
7 f 3
− 7 f (− f ) 7 f u f 3 V= = =− u − f − 7 f 4 − (− f ) 3
∴M =
VA − VC uA − uC
− 7 f ( − − 2 f ) 3 4 = =− − 7 f ( − − 2 f ) 4 3
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 v irtual and erect image
PROBLEMS OF MIRRORS Q: 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 2
Solution:
2 V f M axial = = − u f u 2
I f = O f − u 2
I f = b f − u
Q: 2)
2 f ⇒ I = b − f u
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 images of the bead is
Solution:
The first image is formed due to the reflection reflection from concave mirror M 2
1
+
V1 1
=
V1 1
=
V1
⇒ 1
V2 1
V2 V2
m2
1
=
(− 2 R) − R 1 2R
−
= =
R
2R
2R
1
− 2R
V1
3
− 2 R 1 3 = − . ⇒ m/ = − 3 2R = 2 R − 2 R = 4R 3
3
1 2 = 4 R R
− 2
4
−3
object distance
+
2
3
+
2
2 4R
4R 11
=−
V2 u2
− 4R 3 = 11 = − 4R 11 3
∴ m2 =
3 11
So radius of second image
⇒ a2 =
3 a a . = 11 3 11
Similarly radius of third image is a3
∴ Q: 3)
=
a 41
1 1 1 Answer : : 3 11 41
When an object is placed at a distance of 60cm from a convex spherical mirror, the magnification produced is 1/2. where where should the object be placed placed to get a magnification magnification of 1/3?
Solution:
u = −60cm m= − 1 2
=−
V u V
(or) V
− 60
1
1
1
1
f u v
60
∴ = + =−
= +30cm +
1 30
=
1 60
∴ f = +60 cm In second case
m= As
1
1 3
= −V
u
+1=
1
u v
f
1
−3=
1
u u
100V
=−u 3
60
u = −120 cm Q: 4)
Two objects objects A and B when placed one after another another in front of a concave concave mirror mirror of focal-length 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
m=
h2 h1
=
For object B
f f − u1
m' =
∴ As
h1
= 4h11 and
u1
= −50cm
∴
1
u2
= −20 cm
4
=
h2
= h21 ,
− 10 − u2 − 10 + 50
f = −10cm
h' 2 h'1
=
m h2 = m' h1
f f − u1 h11 × 2 h2
=
f − u2 f − u1
Q: 5)
A concave mirror of focal length 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?
Solution:
f = −10 cm,
∴
1
u
=−
=
1 10
1
V
= −35 cm
−1
f V
+
1 35
=−
1 14
∴ u = −14 cm Distance of the object form wall = 35 – 14 = 21 cm
Q: 6)
An object object is placed at a distance distance of 36cm form a convex mirror. mirror. A plane mirror is placed placed in between so that t he 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 V = PI
= +12 cm
Q: 7)
∴
1
∴
f = 18 cm
∴
R = 2 f = 2 ×18 = 36 cm
f
1
= +
1
u V
=
1 36
+
1 12
=
−1+ 3 36
=
1 18
A convex convex mirror mirror of focal length length ‘f’ forms an image which is is
the object which is
Solution:
1
n
1
times the object. The distance of
times the object. The distance of the object from the mirror is
η = + = −V u η 1
=−
V
u
η
∴1 = 1 +1 f V
1
u
+1 f − u η η u = −(η − 1) f Q: 8)
1
=
An object object of size 7.5cm is placed in in front of a convex mirror mirror of radius radius of curvature 25cm at a distance of 40cm. The size of the image should be
Solution:
η=
I f = O f − u
u = −40
(R / 2) (25 / 2 ) = 7.5 (R / 2 ) − u (25 / 2 ) − (− 40) I
I Q: 9)
=
= +1.78 cm
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:
m=
f f − u
+ 1 = + 30 4 + 30 − u u = −90 cm 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 Solution:
b)
4 3
f
c)
3 4
f
d)
7 3
f
On immersing immersing a mirror in water, focal length of the mirror remains uncharged.
Q: 11) An object is 20cm away away form a concave mirror mirror with focal-length fo cal-length 15cm. If the object moves with a speed of 5m/s along the axis, then the speed of the image will be Solution:
1
V
−
1 20
=
1
− 15
V = −60 cm 2
V Vi = − . V0 u 2
60 = .(5) 20 = 45 m/ s
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
µ = 4 3
upto a height of 20cm, then the Sunlight will now get
focused at Ans:
9cm above water level
Q: 13) A small piece of wire bent into an ‘L’ shape with upright and horizontal portions of equallengths, is placed with the horizontal portion along the axis of the concave mirror whose radius of
curvature is 10cm. If the bend is 20cm from the p ole of the mirror, then the ratio of the lengths of the images of the upright and horizontal portions of the wire is Solution:
f =
R 10 2
=
2
= 5 cm
For part PQ
f L0 f u −
L 1 =
− 5 L = × L0 = − 0 3 − 5 − (− 20) For part QR 2
f L 2 = L0 f − u 2
− 5 L0 × = L = 0 9 − 5 − (− 20) ∴
•
L1 L2
=
3 1
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
OC sin θ
=
OM sin(π
−α)
=
OM sin α
…………(1)
And from triangle ICM
IC sin θ
=
IM
………….(2)
sin α
Dividing equation (1) and (2) yields y ields
OC IC (or)
=
OM IM
=
OP (since M is close to P) IP
x − f x + f = f − y f + y x f − f 2 + x y − f y = x f − x y + f 2 − f y (or) x y = f 2
∴
1)
f =
xy
As x y = f 2 (or) xα
1
y
(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 if x → ∞; y → 0 . If the object is at focus the image is a far off distance and vice-versa. vice-versa.
3)
From xy = f 2 ; if x = f , then y = 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)
2
Since f is necessarily +ve for both types of mirror, so x and y bear the same sign, which implies implies that both b oth the object and the image always always lie an a n the same side of focus.
A) GRAPH OF |x| Versus |y| : Since xy = f 2 represents a rectangular hyperbola, hyperbola, existing in the first and third qua drant ( f 2 being positive). rectangular hyperbola existing only in the first quadrant. ∴ The graph of |y| vs |x| will be a rectangular
B) GRAPH OF U Versus V : Since xy = f
2
∴ (u − f )(v − f ) =
f 2
For a convex mirror, u is always negative and V is always a lways positive. positive. Further ‘f’ is also positive.
∴
Putting u =
and V
=
y we have
(x − f ) ( y − f ) = f 2 This is the equation of a rectangular rectangular hyperbola with its origin shifted to
( f , f ) and ’x’ being
always negative while y lies between O and f . (see figure) for a concave mirror, u is always negative, v can be positive po sitive (or) negative, ‘ f ’ is negative
∴ u = x, v = y
and f = − f
We have, form
(u − f )(v− f ) = f 2 (u + f )( y + f ) = f 2
[x − (− f )] [y − (− f )] = f 2
Or
Evidently it is again an equation of a rectangular hyperbola with origin of coordinates shifted to the point (-f, -f) (see figure ↓ )
3) GRAPH OF
1
1
VERSUS
u
From mirror formula
1
v
+1=
1
u
f
1
Putting
x+ y =
= x and
1
= y , we have
1
y
It is the equation of a straight line having a slope +1 (or) -1 according according as u and v bear the same (or) opposite signs. The intercepts on x and y axis are each object and image 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.
+
1
f
(or ) −
1
f
according as the
For a convex mirror u is always negative v is always positive and f is always positive.
•
CONCEPT OF CRITICLA ANGLE When a ray of light is traveling form denser medium to rarer medium, it get refracted and the ray derivates derivates away form the normal. If we keep increasing angle of incidence then at an angle, the angle of refraction becomes 90
0
.
This is known as C ritical –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 According to Snell’s Law
b
b
b
µa = µa =
sin i sin r sin iC sin 900
µ a = sin iC
b 1 µ = a sin i C
⇒ ⇒ ⇒ Q: 1)
C depends on colour and and temp
∴C
CRed > Cviolet If temp
Red
< Cviolet
↑C ↓
The sum (diameter d) subtends subtends an angle
θ
radius at the pole po le of a concave mirror of focal length
f. Find te diameter of the image of sun formed by mirror? Solution:
1
1
+ =
v u
1
1
=
1
f
(
we get
∴ u is very large so
− f Or v = − f v
1
≈0
It means image is formed at focus Taking '
θ=
l
r
∴θ=
f '
as radius and using
when
l
=d
and r = f
d or d = θ f f
REFRACTION AT SPHERICAL SURFACE: • REFRACTION
∆le s OBC and IBC We have i = α + β From
And
β =r+r
(or ) r = β − r
From Snell’s law
µ 2 sin i = µ1 sin r µ1 sin i = µ 2 sin r For small angle of incidence I, we can write
sin i
≈i
and sin r ≈ r
∴ µ1i = µ 2 r µ1 [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
= And
(β − r ) =
h
− h
+ −
h
+
h V
h h h h ∴ µ1 + = µ 2 − − u + R R v After simplifying we get
µ 2 − µ1 = µ 2 − µ1 u
µ2 µ1 v 1
µ2
−
1
R
=
µ 1
− =
µ2 −1 µ1 R 1
µ2 − 1
u
→ This
formula is derived for convex surface and for
real Image From denser to rarer medium
µ1 µ 2 µ1 − µ 2 − = u
Q:
R
How can we derive a mathematical expression for the equation of a ray in the medium? The medium is of variable refractive index. Ray of light is incident at an angle
α
at air medium
interface? Analysis: Here two cases 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 y
At some height h angle of incidence incidence is
θ y and refractive index is f ( y) from Snell’s Law
µ sin θ = constant
∴ 1 × sin α = f ( y)sin θy Slope of curve at A
……..(1)
dy = tan(90 − θy) d
⇒ cot θy =
dy d
From equation (1)
{f ( y) − sin α } 2
dy = d
Case-I:
2
sin α
µ = f ( y) (i.e.) Refractive index varies as function of x. According According to Snell’s Law
µ sin θ = constant For initial refraction at the air medium interface
1 × sin α sin α
= µ0 sin(90 − θ 0 )
= µ0 sin(90 − θ0 )
∴ sin α = µ0 cosθ 0 Here So
And
θ 0 angle of refracting ray at point A with OX cos θ 0
sin θ 0
= =
sin α
µ0 1 − sin 2 α
µ 02
Now Snell’s Snell’s Law at M gives
f (x)sin θ x = µ0 sin θ 0 Or
sin θ x =
µ0 f ( x)
∴ sin θ x =
1−
sin 2 α
µ 02
µ 02 − sin 2 α f (x)
Now slope of tangent at M is given given as
dy = tan θ x d 2 2 µ dy 0 − sin α = dx { f ( x)}2 − µ 02 + sin 2 α
Q: 1)
If
µ=
1 + y and ray of light is incident at grazing incidence at origin, then find equation of
path of refracted refracted ray. ray. Solution:
We can use result derived above for which
f ( y) = 1 + y and
α = 900 1/ 2
dy ∴ =y d So Q: 2)
y=
x2 4
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
Solution:
For refraction at first face
µ g = 1.5
µ 2 − µ1 = µ 2 − µ1 u
R
µ = −25cm
µ1 = 1 µ2 = 3 / 2 R = 140cm
∴ V = 150 cm ∴ The rays are converging beyond of at 140cm form Q. Again refraction takes place at the plane surface. For refraction at second face
R = ∞,
µ 2 = 1.5 µ1 = 1
µ = +140 cm v= ? Using
µ1 µ 2 µ1 − µ 2 − = u
1
−
R
= 1 − 1.5 + 140 ∞ 1.5
v = 93.3
∴ The ray meet axis at 93.3cm form point Q. •
PROBLEMS ON REFRACTION REFRACTION
1)
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 refractive index ' µ '
and
thickness
‘t’.
Find
the
lateral
displacement of the ray? Solution:
i = 2r
D=
As
µ= µ=
sin r t sin (i − r ) t sin (2r − r ) = =t cos r cos r cos r
= t tan r
sin i sin r sin 2r sin r
=
2 sin r. cos r sin r
µ = 2 cos r cos µ
=
µ 2 2
2 ∴ tan r = − 1 µ 2
2 ∴ D = t − 1 µ ∴
Q: 2)
D=
t 4 − µ2
µ
A light light ray is incident incident on a transparent transparent slab of R. I.
µ=
2 , at an angle of incidence
π /4.
Find the ratio of the lateral displacement suffered by the light ray to the maximum value which it could have suffered? suff ered? Solution:
i=
π 4
,
µ=
2
sin i
=µ ⇒
sin r
D=
sin π / 4 sin r
=
∴r = π / 6
2
π − π 4 6 π cos
t sin
t sin (i − r ) = cos r
6
π π π = t sin − cos . tan 4 6 4 = t
1 2
t
D=
Q: 3)
6
−
1 2
×
3
1
( 3 − 1)
∴
D
=(
3 2− 6
)
6
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:
λa =
wavelength in air
λm =
wavelength in medium
The number of waves of that light in a length of ‘d’ will be
And
d
n1
=
n2
=
∴
n2 n1
=
∴
n2
= µm × n1
n2
= 1.33 × 2000 = 2660 in water
n2
= 1.50 × 2000 = 3000
λa d
λm λa =µ λm m
in glass
Q: 4)
(λ = 58 nm)
Light from a sodium-lamp
(µ = 1.45)
in a certain time. Determine the time difference occurring when light happens to
traverse the same length in ethyl ether Solution:
t=
traverses a distance of 60m in a chloroform
d cm
(µ = 1.35) 60m (1.45)
tchloroform1 =
3 × 108
/s
For ethyl ether
te.r
2
=
60m(1.35) 3 × 108
/s
∆ t = t1 − t2 =
60m 30 × 108
(1.45 − 1.30)
= 2.0 × 10−8 sec Q: 5)
A rectangular glass slab of thickness 12cm is 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:
S1
1 = t1 1 − µ1
2 = 12 1 − 3 = 4 cm
S2
= t2 1 − 1 µ 2
3 = 6 1 − 4 = 1.5 cm
∴ S = S1 + S2 = 4 + 1.5 = 5.5 cm Q: 6)
The time time taken by light to cover a distance distan ce of 9mm in in water is____
Solution:
Cw =
3 × 108
d t= Cw
4/3
=
9
= × 108 4
9 × 10 −3 × 4 9 × 10
8
m/ s
= 4 × 10 −11 sec = 4 × 10 −11 × 109 ns = 0.04 ns
Q: 7)
A ray incident at an angle of incident incident 60 0 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 reflected and refracted rays at this t his surface is_____
Solution:
µ=
sin i sin r 3
sin r =
sin 60
∴ r = 300
µ
0
=
2 3
=
1 2
PC = QC
∴ ∠CPQ = ∠PQC PQC = ∠r = 300 Angle between between reflected ray QR and refracted ray QS at the other face
= 180 − r − 60 0 = 180 − 300 − 60 0 = 900
•
REFRACTION AT SPERICAL SURFACES
Q: 1)
Sunshine recorder globe of 30cm diameter is made of glass of refractive index
µ = 1.5 . A ray
enters the globe parallel to the axis. Find the position from the centre of the sphere where the ray crosses the principal axis? Solution:
For first refraction (Rares to deuces)
u = −∞
µ1 = 1 ∴
µ 2 = 1.5 R = +15cm
µ 2 µ1 µ 2 − µ1 − = 2
R
µ 2 µ 2 − µ1 µ1 = + V R u
1.5
V
1.5 − 1
=
1
1
= + − ∞ 30 ( )
15
V = 45 cm For second refraction (douses to rarer)
R = −15cm,
µ1 µ 2 µ1 − µ 2 − =
Using
V'
u'
1
−
V'
∴ Q: 2)
u' = (45 − 30) = 15cm R
1.5 15
=
1 − 1.5
−15
=
1 30
(or ) V ' =
30 4
= 7.5 cm
Distance of image from from centre of globe is (15 (1 5 + 7.5) = 22.5cm
A particle executes a simple harmonic motion of amplitude 1.0cm along the principal axis of a convex lens of focal length 12cm. The mean position of oscillation is at 20cm from the lens. Find the amplitude of oscillation of the image of the particle?
Solution:
When the particle is
At R
u = −19cm V1
=?
f = 12 cm 1
V1 1
V1
−1 =
1
u
=
1 12
−
f 1 19
1 = V
=
+ 1 f u
1
7 12 × 19
V1
= 12 × 19 7
When the particle is at left extreme position
u = −21 cm, V2 1
V2
−1= u
1
=?
f = 12 cm 1
(or )
f
V2
=
1
+1=
f u
1 12
−
1 21
=
9 12 × 21
∴ Amplitude of oscillation of image = V1 − V2 2
= 1 12 × 19 − 12 × 2 2 7 9 = 2.2857 cm Q: 3)
cm
A point object is is moving moving with velocity velocity 0.01m/s on principal axis towards a convex lens of locallocallength 0.3m when object is a distance distance of 0.4m 0 .4m form the lens, find a) Rate of charge of position of the image and b) Rate of charge of lateral magnification magnification of image
Solution:
Differentiating Differentiating the
1
a) Equation
0=−
f 1
V2
.
1
= −
1
v u
w.r.t. time
dv 1 du + . d u2 d
dv V 2 du = . dt u2 dt 1 30
=
1
V
−
1
− 40
u = −40cm,
⇒ V = 120cm du = 0.01 m/ s d
∴ dv = 120 × 120 × 0.01 m/ s d 40 × 40 = 0.09 m/ s b) M
=
dv V 2 = du u2
2
V = 1 − f
dm V d V = 21 − dt 1 − f dt f
V 1 dv = −21 − f . dt f dv 2 2 120 = − 1 − V = − 1 − 0.09 s−1 f f dt 30 30 = 0.018 / sec Q: 4)
Find the position position of the image formed by the lens lens combination combination given given in figure? figure?
Solution:
Image formed by first lens
1
V1 1
V1
− +
1
u1 1 30
=
1
=
1
(or )
f 1 10
1
V1
−
1
− 30
=
1 10
(or ) V1 = 15 cm
The image formed by the first f irst lens server as the object ob ject for the second. This is at a distance of (15 – 5)cm i.e. 10cm to the right of the second lens. It is a v irtual object Now
1
V2
−
1 10
=
1
− 10
1
V2 V2
=−
1
+
10
1 10
=0
=∞
The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object for the third lens.
1
V3 1
V3 V3
− =
1
−∞
=
1 30
1 30
= 30cm
The final image is formed 30cm to the fight of the third lens.
Q: 5)
Two Plano-concave lenses lenses of glass of refractive index 1.5 1.5 have radii of curvature 20cm and 30cm respectively. They are placed in contact with the curved surfaces towards each other and the space between them is filled with a liquid of refractive index 5/2. Find the focal length of the combination.
Solution:
For first plano concave lens
f 1 =
− R2 = − 20 = − 20 = −40 cm µ1 − 1 1.5 − 1 0.5
For second plano concave lens
f 2
=
− R2 − 30 − 30 = = = −60 cm µ1 − 1 1.5 − 1 0.5
The focal length of the liquid lens is given by
1 1 = (µ2 − 1) + f 3 R1 R2
1
R 1 = 20cm,
R 2 = 30cm
µ2 = 5 / 2
∴
f 3
= 8 cm
∴1= f
1
f 1
+
1
f 2
+
1
f 3
=−
1 40
−
1 60
+1 = 8
1 12
cm
∴ f = 12 cm Q: 6)
Given the the object object image and principal principal axis find the positions and nature of of the lens
Solution:
First join the object and image
If the one point is above the optical axis and the other below it, then the lens is always a convex lens.
If object and image points are Above the principal axis and image point is higher, then the lens is convex and is present between the image and object points. Other wise the lens is concave.
Q: 7)
For the given positions of the objects and the image in figure determine determine the location and the nature of the lens used?
Solution:
Q: 8)
A ray of light passes through a medium medium whose refractive refractive index varies with distance distance as
x µ = µ 0 1 + . If the ray enters the medium parallel to the x-axis, what will be the trajectory of a the ray and what will be the time taken for the ray to tra vel a distance a? Solution:
The ray enters normally and proceeds proceeds along a straight line. At a distance ‘x’ in the medium consider consider a slab s lab of thickness “dx”. Velocity of the light ray at this point is
Time taken to cross the distance ‘dx’ is
dt =
∴
x = dx × µ 0 1 + x C a µ 1 + a
dx = V
Total time of travel is a
t=
dx C
∫ C µ 1 + a dx
x
0
0
=3
µ0 a
2 C
Q: 9)
A fish is rising up vertically inside inside a pond with with velocity 4cm/s and notices a bird which which is diving vertically downward and in velocity appears to be 16cm/s (to the fish). What is the velocity of the diving bird, if R. I of water is 4/3?
Solution:
− Vf )
Vb f = Vb −
16 = Vb + Vf
∴µ=
16 = Vb + 4
4
= 12
V
Vb
3
Vb V
= 12 V
3
= × 12 = 9 cm/ s 4
0
Q: 10) Solar rays are incident incident at 45 on the surface of water
(µ = 4 / 3) .
What is the length of the
shadow of a pole of length 1.2m erected at the bottom of the pond, if the pole is vertical assuming that 0.2m of the p ole is above the water surface? Solution:
Applying Applying Snell’s law at point c
1 × sin 45 sin θ
=
0
=
3 4 2
4 3
sin θ
Here AE
tan θ
=
sin θ
=
3
=
4 2
= CD = 0.2m
BC CE
=
BE 1
= BE
BE 1 + (BE )
2
BE 1 + (BE )
2
BE = 0.625 m
∴
The length of shadow = AB = AE + EB = 0 .2+0.625 = 0.825
Q: 11) In a lake, a fish rising vertically to the surface of water uniformly at the rate of 3m/s, observes a bird diving vertically towards the water water at a rate of 9m/s vertically vertically above it. The actual actual velocity velocity of the dive of the bird is_______ (µ Solution:
µ=
A.D R.D
µ=
y' y
∴ y' = µ y
∴ h = x+ y' h = x+ µ y Differentiating
dh dx dy = +µ dt d dt
= 4 / 3)
9 = 3+ µ
dy d
dy 6 = = 4.5 m/ s d t (4 / 3)
Q: 12) A convex lens of focal focal length 0.2m is cut into into two halves each of which which is displaced by 0.0005m and a point object is placed at a distance of 0.3m form the lens, as shown in figure. The position of the image is ________ Solution:
1
=1−1
f v u 1
v
1
= +
1
f u f = 0.2
u = −0.3m 1
=
1 0.2
−
1 0.3
v = 0.6m Q: 13) A pole 5m high is situated on a horizontal surface. surface. Sun rays are incident incident at an angle 30 0 with the vertical. vertical. The size of shadow in horizontal surface is______ Solution:
tan 30 0
=
BC 5
BC = 5 tan 300
=
Q: 14) The Sun subtends an an angle
5 3
m
α = 0.5 0
at the pole of a concave mirror. The radius is curvature of
concave mirror is R = 1.5m. The size of image formed by the concave mirror is_____ Solution:
1)
As Sun is at infinity image is formed at the focus of mirror
1
= × 0.50 ×108 2
= 1 × 0.5 × π × 105 2
180
= 0.654 cm
POQ = α = ∠POQ
Di DS
DS Or u
= magnification = v = u
f u
Di DS Di
•
=
R 2u
R × DS 2u
=
MAGNIFICAIOTN IN CASE OF CURVED SURFACE: Consider an extended object “QO” placed
⊥ 1r
to the principal axis at the point O. The image of
the point ‘O’ is formed at I. Image of extended object object is ‘IQ’
∴
From
tan θ1
POQ POQ and PIQ ' we can say that
OQ and tan θ 2 PO
=
But since Also
les ∆les
θ1
and
I Q' PI
θ 2 are very small, we can approximate
µ1 sin θ1 = µ 2 sin θ 2
Sign conversion
µ1θ1 = µ 2θ 2
h0
= +OQ
hi
= − IQ '
∴ µ1.
•
=
OQ PO
= µ2 .
IQ ' PI
∴ µ1
h0 (− hi ) = u2 . (− u) (v)
u = − PO
hi h0
µ1 V µ2 u
V
=
LATERAL MAGNIFICATION MAGNIFICATION
The lateral magnification is
= + PI
mL
= dv d
Which Which can be obtaine ob tained d by differentiating differentiating equation
µ 2 − µ1 = (µ2 − µ1 ) u
Thus
•
− µ22 .
dv µ1 + du u2
=0
REFRACTION REFRACTION AT PRISM A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle. a) Calculation of angle of Derivation:
In passing through the prism, ray KL suffers two refractions and has tweed through an
LQPN = δ . (with is angle of direction) In
∆ PLM ,
δ = ∠PLM + ∠PML δ = (i1 − r1 ) + (i2 − r2 ) δ = (i1 + i2 ) − (r1 + r2 ) In
……………(1)
∆ OLM ∠O + r1 + r2 = 180 0
……………(2)
In Quadrilateral ALOM, As
∠L + ∠M = 1800 ∴ A + ∠O = 1800
(∴ Sum of 4 angles of a Quad = 360)
∴ using equation (2) we can write ∠O + r1 + r2 A + ∠O r1 + r2
=A
…………….(3)
Putting (3) in in (1)
δ = (i1 + i2 ) − A
……………(4)
(for bigger refracting angles)
If ' µ ' is the refractive index of the material material of the prism, then according to Snell’s Law
µ=
sin i1 sin r1
=
i1 (when angles are small) r1
∴ i1 − µ r1 similarly
i2
= µ r2
Putting the above two in equation (4) we get
δ = (u r1 + u r2 ) − A δ = µ (r1 + r2 ) − A ∴ δ = (µ − 1)A (when refracting angle are small) This is the angle through which a ray derivate on passing t hrough a thin prism of small refracting refracting angle A.
Prism formula: In minimum derivation p osition i1 = i2 = i and r 1 = r 2 = r
∴ from equation (3) r+r = A 2r = A
And from equation (4)
δ m = (i + i ) − A δm = 2i − A ∴ A+ δm = 2i ∴
(or )
from Snell’s Law
i
=
A+ δm 2
µ = sin i = sin r
sin
A+ δ m
2 sin ( A/ 2)
Note : The deviation deviation through a prism is maximum maximum when i1 = 900. Thus
•
δ m = 90 + i2 − A
MAXIMUM DEVIATION:
For
i1
= 90 0
∴
i2
=?
At surface BD
µ1 sin i1 = µ. sin r1 1× sin 90 0
= µ. sin r1
1 ∴ sin r1 = µ r1
= sin −1 1 µ
We know that r1
or (r1
+ r2 = A r2
= ( A − r1 )
r2
= (A− θC )
For surface CD
µ sin r2 = sin i2
= θC )
sin i2
i2
= µ sin ( A − θ C )
= sin −1 [µ sin ( A − θ )]
CONCEPT: Sometimes a part of a prism is given and the keep on thinking whether how should we proceed? To solve such p roblems roblems first complete the prism then solve as the prob lems lems of prism are solved
Defects in Image formed by leuses: The defect in the lens on account of which it does not form a white point image of white point object is defined as Aberration.
Axial Chromatic Aberration: The variation of the image distance form the lens with the colour measures measures axial a xial chromatic Aberration. Lateral Chromatic Aberration: The variation in the size of the image with colour measures the lateral transverse chromatic aberration.
Concept of Dispersion of Light: L ight: Dispersion of light is the phenomenon of splitting of a beam of white light into its constituent constituent colours on passing through a prism. p rism. (accuse (accuse due to wavelength). wavelength). The band of seven colours so obtained is called the visible spectrum. The order of colours from the lower end of spectrum is “VIBGYOR”. Violet colour deviates through maxi. Value and red colour deviates through the minimum angle. Causes of Dispersion: Each colour has in own wavelength according according to Cauchy’s Ca uchy’s formula
R.I. of a material depends on wavelength
µ = A+
B
λ
2
+
C
λ4
(λ )
+ ..........
For a prism of small refracting angle of deviation is Angular dispersion: It is the angle in which all colous of white light are contained dv = deviation of violet colour dr = deviation of red colour
(θ ) ∴ As
Angular dispersion = d v - dr
dV
= (µV − 1) A
dS
= (µV − 1) A
∴ dV − dr = (µV − 1)A − (µ r − 1) A = (µV − 1 − µ r + 1) A dV
− dr = (µV − µ r ) A
Dispersive Dispersive Power (w): Ratio of angular deviation to the mean position produced produced by the prism.
ω= Note:
dv − dr d
=
(µv − µ r ) dµ = (µ − 1) (µ − 1)
Single prism produces both deviation and dispersion dispersion simultaneously. simultaneously. It cannot give deviation deviation without dispersion (or) dispersion without deviation. However a suitable combination of two prisms can can do so. Dispersion of light occurs because velocity of light in a material depends depends upon its colour There is no dispersion of light refracted through a rectangular glass slab.
Combinations of prim: I) Deviation without dispersion: Net dispersion dispersion = 0, Net deviation deviation Necessary Necessary condition condition
≠0
(δV − δ R ) + (δ V1 − δ R1 ) = 0 (µV − µ R ) A + (µV' − µ 'R ) A ' = 0
In this Situation
µ − µ R δ1 + δ ' = (µ − 1) + (µ '−1) A' = (µ − 1) A + (µ + − 1) − V' ' µ − µ V R (µ − µ R ) = (µ − 1) A − (µ '−1) V' A (µV − µ R' )
Net deviation deviation =
(µ − µ ) ( − ) = (µ − 1) A V R × µ' ' 1 ' (µ − 1) (µV − µ R ) ω δ 1 1 − ω1 Usually ω ' < ω so δ
is in the same direction as produced by the first prism.
The prism which which produces p roduces deviation deviation without dispersion dispersion is called achromatic prism. II)
Dispersion without deviation: deviation: Net deviation deviation = 0, Net dispersion dispersion Necessary Necessary condition condition
δ1 + δ ' = 0
(µ − 1) A + (µ '−1) A' = 0 ( ) ∴ A' = − µ − 1 A (µ '−1)
≠0
In this situation Net dispersion dispersion
= (µ v − µ r ) A + (µ v' − µ r' ) A'
θ = (µ v − µ r ) A + (µ v' − µ r' )
(µ − 1) A (µ '−1)
θ = (µ v − µ r ) A + µv' − µ r' ×
(µ − 1) A (µ '−1)
µ v − µ r µ v' − µ r' = (µ − 1) A − 1 µ − ( ) (µ '−1) θ = δ [ω − ω '] If ω ' > ω , the resultant dispersion is negative. (i.e.) opposite to that produced by the first prism. This prism which p roduces dispersion without derivation is called direct vision pr ism.
Q: 1)
A prism having an apex angle 4 0 and refractive index 1.5 is located in front of a vertical plane mirror. Through what total angle is the ray deviated after reflected reflected form the mirror?
Solution:
δ prism = (µ − 1) A = (1.5 − 1)4 0 = 2 0
δ total = δ prism + δ mirror
= (µ − 1) A + (180 − 2 i ) = (1.5 − 1) 4 0 + (180 − 2 × 20 )
= 20 + 1760 = 178 Q: 2)
A container container contains water water upto a height height of 20cm and there is a point source at the centre of the bottom of the container. A rubber ring of radius ‘r’ floats centrally on the water. The ceiling ceiling of the from is 2.0m 2 .0m above the wager surface. (a) Find the radius of the shadow of the ring for med on the ceiling if r = 15cm. 1 5cm. (b) Find the maximum value of ‘r’ for which the shadow of the ring is formed on the ceiling
(µ w = 4 / 3) ? Solution: a)
Using Snell’s Law
µ sin i = sin r 4 3
×
15 15 2
∴
+ 20 2
800
∴ x=
= 1×
3
x x2
+ 200 2
cm
Radius of shadow = 15 +
=
800 3
845 3
cm
= 2.81 m b)
For shadow to be formed formed angle of incidence incidence should be less than critical critical angle using Snell’s Law
rmax max
4
2 max
3 V r
+ 20
2
= 1 × sin 900
2 = 9rmax + 9 × 20 2
2 16rmax 2 7rmax
rmax max Q: 3)
= 9 × 202 9
=
7
× 20cm= 0.2268 m
0 Monochromatic Monochromatic light light falls on a right angled angled prism at an angle of incidence incidence 45 . The emergent
light is found to slide along the face AC. Find the refractive refractive index of material material of prism? Solution:
= θC
r2
r1 + r2
………..(1)
= A = 900
= 90 − r1
r2
………..(2)
∴ θ C = 90 − r1 sin θ C
= sin (90 − r1 )
sin θ C
= cos r1
1
= cos r1
µ
and sin r1
=
1 − cos 2 r1
Snell’s Snell’s law AR
1 sin 450 sin 450 1 2 1 2
− µ sin r1
= µ 1−
µ2
= µ 2 −1
= µ 2 −1
µ=
1
1.5
µ 2 = 3 = 1.5 2
=
1−
1
µ2
