Chapter 17 Ray Optics Part 1 (Reflection of Light) free study material by TEACHING CARE online tuition and coaching classes
Chapter 17 Ray Optics Part 1 (Reflection of Light) free study material by TEACHING CARE online tuition and coaching classes
When a ray of light after incidenting on a boundary separating two media comes back into the same media, then this phenomenon, is called reflection of light.
Þ ∠i = ∠r
Þ After reflection, velocity, wave length and frequency
of light remains same but intensity decreases
Þ There is a phase change of p if reflection takes place from denser medium
Note : @ After reflection velocity, wavelength and frequency of light remains same but intensity decreases.
@ If light ray incident normally on a surface, after reflection it retraces the path.
If light rays, after reflection or refraction, actually meets at a point then real image is formed and if they appears to meet virtual image is formed.
Plane Mirror.
The image formed by a plane mirror is virtual, erect, laterally inverted, equal in size that of the object and at a distance equal to the distance of the object in front of the mirror.
- Deviation : Deviation produced by a plane mirror and by two inclined plane
Note : @ If two plane mirrors are inclined to each other at 90o, the emergent ray is anti-parallel to incident ray, if it suffers one reflection from each. Whatever be the angle to incidence.
- Rotation : If a plane mirror is rotated in the plane of incidence through angle q, by keeping the incident ray fixed, the reflected ray turned through an angle 2q.
- Images by two inclined plane mirrors : When two plane mirrors are inclined to each other at an angle
q, then number of images (n) formed of an object which is kept between them.
(i)
n = æ 360 – 1ö ; If 360 = even integer
ç ÷
|
|
è ø
(ii) If
360 =
q
odd integer then there are two possibilities
(a) Object is placed symmetrically (b) Object is placed asymmetrically
n = æ 360 – 1ö n = 360
|
|
ç ÷
è ø
Note : @ If θ = 0o i.e. mirrors are parallel to each other so n = ¥
|
@ If 90o, n = 360 – 1 = 3
90
i.e. infinite images will be formed.
|
@If 72o, n = 360 – 1 = 4
72
(If nothing is said object is supposed to be symmetrically placed).
- Other important informations
- When the object moves with speed u towards (or away) from the plane mirror then image also moves toward (or away) with speed u. But relative speed of image r.t. object is 2u.
- When mirror moves towards the stationary object with speed u, the image will move with speed 2u.
- A man of height h requires a mirror of length at least equal to h/2, to see his own complete
- To see complete wall behind himself a person requires a plane mirror of at least one third the height of It should be noted that person is standing in the middle of the room.
Example: 1 A plane mirror makes an angle of 30o with horizontal. If a vertical ray strikes the mirror, find the angle between mirror and reflected ray [RPET 1997]
(a) 30o (b) 45o (c) 60o (d) 90o
Solution : (c) Since angle between mirror and normal is 90o and reflected ray (RR) makes an angle of 30o with the normal so required angle will be q = 60o .
Example: 2 Two vertical plane mirrors are inclined at an angle of 60o with each other. A ray of light travelling horizontally is reflected first from one mirror and then from the other. The resultant deviation is
(a) 60o (b) 120o (c) 180o (d) 240o
Solution : (d) By using d = (360 – 2q ) Þ d = 360 – 2 ´ 60 = 240o
Example: 3 A person is in a room whose ceiling and two adjacent walls are mirrors. How many images are formed
[AFMC 2002]
(a) 5 (b) 6 (c) 7 (d) 8
Solution : (c) The walls will act as two mirrors inclined to each other at 90o and so sill form
360 – 1 = 3 images of the person.
90
Now these images with object (Person) will act as objects for the ceiling mirror and so ceiling will form 4 images as shown. Therefore total number of images formed = 3 + 4 = 7
I1
Four images by ceiling
I2
Three images by walls
|
|
|
Note : @ The person will see only six images of himself (I1 , I 2 , I 3 , I ‘ , I ‘ , I ‘ )
Example: 4 A ray of light makes an angle of 10o with the horizontal above it and strikes a plane mirror which is inclined at an angle q to the horizontal. The angle q for which the reflected ray becomes vertical is
(a) 40o (b) 50o (c) 80o (d) 100o
Solution : (a) From figure
q + q + 10 = 90
Þ q = 40 o
Vertical RR
q IR
q
10o
q
Horizontal line Plane mirror
Example: 5 A ray of light incident on the first mirror parallel to the second and is reflected from the second mirror parallel to first mirror. The angle between two mirrors is
(a) 30o (b) 60o (c) 75o (d) 90o
Solution : (b) From geometry of figure
q + q + q = 180 o
Þ q = 60 o
Example: 6 A point object is placed mid-way between two plane mirrors distance ‘a‘ apart. The plane mirror forms an infinite number of images due to multiple reflection. The distance between the nth order image formed in the two mirrors is
- na (b) 2na (c) na/2 (d) n2 a
Solution : (b)
From above figure it can be proved that seperation between nth order image formed in the two mirrors = 2na
Example: 7 Two plane mirrors P and Q are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle of q 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 reflections (including the first one) before it emerges out is
(a)
l
d tan q
d l tan q
ld tan q
- None of these
Solution : (a) Suppose n = Total number of reflection light ray undergoes before exist out.
x = Horizontal distance travelled by light ray in one reflection.
x
So nx = l also
tanq =
d
Þ n = l
d tan q
Example: 8 A plane mirror and a person are moving towards each other with same velocity v. Then the velocity of the image is
(a) v (b) 2v (c) 3v (d) 4v
Solution : (c) If mirror would be at rest, then velocity of image should be 2v. but due to the motion of mirror, velocity of image will be 2v + v = 3v.
Example: 9 A ray reflected successively from two plane mirrors inclined at a certain angle undergoes a deviation of 300o.
The number of images observable are
(a) 10 (b) 11 (c) 12 (d) 13
Solution : (b) By using
d = (360 – 2q )
Þ 300 = 360 – 2q
Þ q = 30 o . Hence number of images = 360 – 1 = 11
|
30
| \ Speed of light on screen v = wR = 2p (2n)R = 4pnR |
| Tricky example: 2 |
| A watch shows time as 3 : 25 when seen through a mirror, time appeared will be
[RPMT 1997; JIPMER 2001, 2002] (a) 8 : 35 (b) 9 : 35 (c) 7 : 35 (d) 8 : 25 Solution : (a) For solving this type of problems remember Actual time = 11 : 60 – given time So here Actual time = 11 : 60 – 3 : 25 = 8 : 35 |
| Tricky example: 3 |
| When a plane mirror is placed horizontally on a level ground at a distance of 60 m from the foot of a tower, the top of the tower and its image in the mirror subtend an angle of 90o at the eye. The height of the tower will be [CPMT 1984]
(a) 30 m (b) 60 m (c) 90 m (d) 120 m Solution : (b) Form the figure it is clear that h = tan 45 o 60 Þ h = 60 m Tower h
45o 45o 60 m Image |
Curved Mirror.
It is a part of a transparent hollow sphere whose one surface is polished.
- Some definitions :
- Pole (P) : Mid point of the mirror
- Centre of curvature (C) : Centre of the sphere of which the mirror is a
- Radius of curvature (R) : Distance between pole and centre of curvature.
(Rconcave = –ve , Rconvex = +ve , Rplane = ¥)
- Principle axis : A line passing through P and C.
- Focus (F) : An image point on principle axis for which object is at ¥
- Focal length (f) : Distance between P and F.
- Relation between f and R :
f = R
2
(fconcare = –ve , fconvex = + ve , fplane = ¥ )
- Power : The converging or diverging ability of mirror
- Aperture : Effective diameter of light reflecting Intensity of image µ Area µ (Aperture)2
- Focal plane : A plane passing from focus and perpendicular to principle
(2) Rules of image formation and sign convention :
Rule (i) Rule (ii) Rule (iii)
(3) Sign conventions :
- All distances are measured from the
- Distances measured in the direction of incident rays are taken as positive while in the direction opposite of incident rays are taken
- Distances above the principle axis are taken positive and below the principle axis are taken
Note : @ Same sign convention are also valid for lenses.
Use following sign while solving the problem :
Incident ray +
–
Mirror or Lens –
+
Principle axis
| Concave mirror | Convex mirror | |||
| Real image (u ≥ f) | Virtual image (u< f) | |||
| Distance of object u ® – | u ® – | u | ® | – |
| Distance of image v ® – | v ® + | v | ® | + |
| Focal length f ® – | f ® – | f | ® | + |
| Height of object O ® + | O ® + | O | ® | + |
| Height of image I ® – | I ® + | I | ® | + |
| Radius of curvature R ® – | R ® – | R | ® | + |
| Magnification m ® – | m ® + | m | ® | + |
(4) Position, size and nature of image formed by the spherical mirror
| Mirror
(a) Concave
¥ C F
(b) Convex
¥ P F |
P
C |
Location of the object
At infinity i.e. u = ∞ Away from centre of curvature (u > 2f) At centre of curvature u = 2f Between centre of curvature and focus : F < u < 2f At focus i.e. u = f Between pole and focus u < f At infinity i.e. u = ∞
Anywhere between infinity and pole |
Location of the image
At focus i.e. v = f
Between f and 2f i.e. f < v < 2f At centre of curvature i.e. v = 2f Away from the centre of curvature v > 2f At infinity i.e. v = ∞ v > u
At focus i.e., v = f
Between pole and focus |
Magnification, Size of the image
m << 1, diminished
m < 1, diminished
m = 1, same size as that of the object m > 1, magnified
m = ∞, magnified m > 1 magnified m < 1, diminished m < 1, diminished |
Nature | |
| Real virtual Real
Real Real Real
Real Virtual Virtual Virtual |
Erect inverted inverted
inverted inverted inverted
inverted erect erect erect |
|||||
Note : @In case of convex mirrors, as the object moves away from the mirror, the image becomes smaller and moves closer to the focus.
@ Images formed by mirrors do not show chromatic aberration.
@For convex mirror maximum image distance is it’s focal length.
@In concave mirror, minimum distance between a real object and it’s real image is zero. (i.e. when u = v = 2f)
Mirror formula and magnification.
For a spherical mirror if u = Distance of object from pole, v = distance of image from pole, f = Focal length,
R = Radius of curvature, O = Size of object, I = size of image, m = magnification (or linear magnification ), ms =
Areal magnification, Ao
= Area of object,
Ai = Area of image
- Mirror formula :
1 = 1 + 1
; (use sign convention while solving the problems).
f v u
Note : @ Newton’s formula : If object distance (x1) and image distance (x2) are measured from focus
instead of pole then
f 2 = x1 x 2
(2) Magnification : m =
Size of object Size of image
Note : @ Don’t put the sign of quantity which is to be determined.
@ If a spherical mirror produces an image ‘m’ times the size of the object (m = magnification) then
u, v and f are given by the followings
u = æ m – 1 ö f,
v = -(m – 1) f
and
f = æ m ö u
(use sign convention)
ç m ÷
ç m – 1) ÷
è ø è ø
(3) Uses of mirrors
- Concave mirror : Used as a shaving mirror, In search light, in cinema projector, in telescope, by E.N.T. specialists
- Convex mirror : In road lamps, side mirror in vehicles etc.
Note : @ Field of view of convex mirror is more than that of concave mirror.
Different graphs
Example: 10 A convex mirror of focal length f forms an image which is 1/n times the object. The distance of the object from the mirror is
(a) (n – 1) f (b)
æ n – 1 ö f
(c)
æ n + 1 ö f
(d) (n + 1) f
Solution : (a) By using
ç ÷ ç ÷
|
|
è ø è ø
m = f
f – u
Here
m = + 1 ,
n
f ® + f
So,
+ 1 =
n
+ f Þ
- f – u
u = -(n – 1)f
Example: 11 An object 5 cm tall is placed 1 m from a concave spherical mirror which has a radius of curvature of 20 cm.
The size of the image is [MP PET 1993]
(a) 0.11 cm (b) 0.50 cm (c) 0.55 cm (d) 0.60 cm
Solution : (c) By using
I = f
O f – u
Here
O + 5 cm,
f = – R = -10 cm ,
2
u = -1m = – 100 cm
So,
I = -10
+ 5 – 10 – (-100)
Þ I = – 0.55 cm.
Example: 12 An object of length 2.5 cm is placed at a distance of 1.5 f from a concave mirror where f is the magnitude of the focal length of the mirror. The length of the object is perpendicular to the principle axis. The length of the image is
- 5 cm, erect (b) 10 cm, erect (c) 15 cm, erect (d) 5 cm, inverted
Solution : (d) By using
I = f
O f – u
; where I = ? , O = + 2.5 cm.
f ® – f , u = – 1.5 f
\ I =
+ 2.5
- f
– f – (-1.5 f )
Þ I = -5 cm. (Negative sign indicates that image is inverted.)
Example: 13 A convex mirror has a focal length f. A real object is placed at a distance f in front of it from the pole produces an image at
- Infinity (b) f (c) f / 2 (d) 2f
Solution : (c) By using
1 = 1 + 1 Þ
1 = 1 + 1
Þ v = f
f v u
- f v
(- f ) 2
Example: 14 Two objects A and B when placed one after another infront of a concave mirror of focal length 10 cm from images of same size. Size of object A is four times that of B. If object A is placed at a distance of 50 cm from the mirror, what should be the distance of B from the mirror
(a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm
Solution : (b) By using
I = f Þ I A ´ OB
= f – uB
Þ 1 ´ 1 =
-10 – uB
Þ u = -20cm .
O f – u
I B OA
f – uA
1 4 – 10 – (- 50) B
Example: 15 A square of side 3 cm is placed at a distance of 25 cm from a concave mirror of focal length 10 cm. The centre of the square is at the axis of the mirror and the plane is normal to the axis. The area enclosed by the image of the wire is
(a) 4 cm2 (b) 6 cm2 (c) 16 cm2 (d) 36 cm2
Solution : (a) By using
m2 =
Ai ; where
Ao
m = f
f – u
-10 -2
2 æ – 2 ö 2 2
Hence from given values
m = – 10 – (- 25) = 3
and
Ao = 9 cm
\ Ai = ç ÷
è 3 ø
´ 9 = 4cm
Example: 16 A convex mirror of focal length 10 cm is placed in water. The refractive index of water is 4/3. What will be the focal length of the mirror in water
(a) 10 cm (b) 40/3 cm (c) 30/4 cm (d) None of these
Solution : (a) No change in focal length, because f depends only upon radius of curvature R.
Example: 17 A candle flame 3 cm is placed at distance of 3 m from a wall. How far from wall must a concave mirror be placed in order that it may form an image of flame 9 cm high on the wall
(a) 225 cm (b) 300 cm (c) 450 cm (d) 650 cm
Solution : (c) Let the mirror be placed at a distance x from wall By using
I = – v Þ – 9 = – (– x)
Þ x = -4.5m = – 450cm.
O u + 3
– (x – 3)
Example: 18 A concave mirror of focal length 100 cm is used to obtain the image of the sun which subtends an angle of 30′.
The diameter of the image of the sun will be
(a) 1.74 cm (b) 0.87 cm (c) 0.435 cm (d) 100 cm
Solution : (b) Diameter of image of sun d = fq
æ 30 ö o
æ 30 ö p
q = 30′ = ç ÷
Þ d = 100 ´ ç 60 ÷ ´ 180
è 60 ø
è ø
Þ d = 0.87 cm .
Example: 19 A thin rod of length f / 3 lies along the axis of a concave mirror of focal length f. One end of its magnified image touches an end of the rod. The length of the image is [MP PET 1995]
(a) f (b) 1 f
2
(c) 2 f (d) 1 f
4
Solution : (b) If end A of rod acts an object for mirror then it’s image will be A‘ and if
u = 2 f – f
3
= 5 f
3
So by using
1 = 1 + 1
Þ 1 = 1 + 1
Þ v = – 5 f
f v u
- f v
– 5 f 2
3
\ Length of image =
5 f – 2 f = f
2 2
Example: 20 A concave mirror is placed on a horizontal table with its axis directed vertically upwards. Let O be the pole of the mirror and C its centre of curvature. A point object is placed at C. It has a real image, also located at C. If the mirror is now filled with water, the image will be [IIT-JEE 1998]
(a) Real, and will remain at C (b) Real, and located at a point between C and ¥
(c) Virtual and located at a point between C and O (d) Real, and located at a point between C and O
Solution : (d)
O
Initially
O
Finally
Tricky example: 4
An object is placed infront of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and plane mirror is 30 cm, it is found that there is no parallel between the images formed by two mirrors. Radius of curvature of mirror will be
(a) 12.5 cm (b) 25 cm (c)
50 cm
3
(d)
18 cm
Solution : (b) Since there is no parallel, it means that both images (By plane mirror and convex mirror) coinciding each other.
According to property of plane mirror it will form image at a distance of 30 cm behind it. Hence for convex mirror u = – 50 cm, v = + 10 cm
By using
1 = 1 – 1
Þ 1 = 1 + 1 = 4
Object
A
f v u
f + 10 – 50 50
30 cm 20 cm
Þ
Tricky example: 5
f = 25 cm
2
Þ R = 2 f = 25cm.
50 cm
10 cm
A convergent beam of light is incident on a convex mirror so as to converge to a distance 12 cm from the pole of the mirror. An inverted image of the same size is formed coincident with the virtual object.
What is the focal length of the mirror
- 24 cm (b) 12 cm (c) 6 cm (d) 3 cm Solution : (c) Here object and image are at the same position so this position must be centre of curvature
\ R = 12 cm
Þ f = R
2
C
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