Fiitjee Physics

OPTICS Definitions Light is a kind of energy which produces the sensation of vision on eyes. The light itself is invisible. Ray: A directed line segment which is used to describe the direction of propagation of light, that is, the direction in which light is traveling is called a ray. Beam: A beam of light is a bunch of light rays. It may be convergent, divergent or parallel. 1. A convergent beam is that in which the rays are directed towards a point. 2. A divergent beam is that in which the light rays are directed away from a point. 3. In a parallel beam of light, all the rays are parallel to each other. In a homogeneous medium light travels along a straight line. Reflection: Whenever light travels from one medium to the other medium, at the boundary a part of the light beam is returned back into the previous medium Returning of light rays into the previous medium is called reflection. Laws of reflection: There are two laws of reflection (i) Incident ray, reflected ray and normal to the point of incidence, all lie in same plane. (ii) Angle of incidence i, is equal to the angle of reflection r. Both angles are measured with the normal to the point of incidence.

Regular Reflection

irregular Reflection

Regular and irregular reflection If reflected beam follows a certain pattern/order reflection in known as regular. For an example if the incident beam is a combination of parallel rays and reflected rays are either parallel among them selves or converge at a point or appear to diverge from a point, these are the regular reflection. If reflected rays does not follow any fixed pattern that is path of two reflected rays can not be predicted by simple observation the reflection is known as irregular. Regular reflection lakes place at even surface and irregular reflection takes place at uneven surface. Irrespective of the situation (either regular or irregular reflection) laws of reflection are valid. Deviation due to reflection Deviation of a ray is the angle through which an incident ray changes its direction from the original direction. If for an incident ray angle of incidence is i then the angle through which the reflected ray would deviate is given by  =  - 2i. Consider the given figure. AB is an incident ray with angle of incidence i and BD is the reflected ray. Angle of reflection is also i. Had the mirror been not present the ray AB would have followed the path BC. A This shows that the ray has rotated through an D i i angle . = - 2i Along the anticlockwise sense So i + i +  =  B So  =  - 2i C

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PDT Courseware (FTRE) 10th Moving to 11th

PH-OPTICS-2

Mirror: If reflecting surface reflects maximum portion of light falling on it, then the surface is called a mirror. Plane Mirror: If reflecting surface is a plane, that is its radius of curvature is infinite and it behaves like a mirror, then it is called a plane mirror. Object and Image: Any thing from which rays are coming to the mirror is called an object. After reflection the reflected rays either actually converge or they appear to converge at a point. The point of intersection of reflected rays is called image. Real and virtual Object: If incident rays are coming from an object lying in front of the mirror, object is known as real. If incident rays appear to converge behind the mirror, the point of convergence is known as virtual object. Real and virtual image: If reflected rays actually converge the image formed is real. A real image is always formed in front of the mirror. If reflected rays are divergent that is they do not meet actually, rather they appear to come from a point, that point is called virtual image. A virtual image is always fromed behind the mirror. In the adjacent figure a point object is placed in front of a plane mirror. We consider three incident rays say 1, 2 and 3. The corresponding reflected rays 1, 2 and 3 meet at I, when extended. The I is the virtual image of O.

1 1 1 1

3 O

x 2

I

3 y

2 2

2

In the adjacent figure two incident rays 1 and 2 appears to meet at point O behind the mirror. So point O is a virtual point object for the plane mirror. The corresponding reflected rays 1 and 2 meet at I in front of the mirror. The I is the real image of O. The real image can be shown on a screen whereas virtual image can not be.

1 1 1 1 I

x 2

O y

2 2

2

Properties of plane mirror: Using simple geometry, it can be shown that, object distance = image distance; where distances are measured from the mirror.  The image is laterally inverted. i.e. if you stand in front of a plane mirror, your left hand becomes right hand of your image.  The magnification is unity, that size of the image is equal to the size of the object.  When the plane mirror is rotated through an angle , the reflected ray turns through double the angle, i.e. 2.

  object

Laterally Inverted image


PH-OPTICS-3


Illustration 1: In the figure, M is a plane mirror parallel to Y-axis. A point object P is placed in front of the mirror. Position of the object is shown. We have to locate the position of the image.

Y

45 X

O

10 cm P

Solution:

To locate the position of the image, we need at least two incident rays and their corresponding reflected rays. Consider the figure, for the sake of simplicity we take two incident rays PQ normal to the mirror and PO incident at 450. According to the laws of reflection reflected rays corresponding to PQ retraces that path as angle of incidence is zero. The reflected ray corresponding to PO makes an angle of 450 with the normal to the mirror (here with the x-axis) when we extend these two reflected rays, they meet at P behind the mirror. Now in  PQO and PQO POQ = POQ = 450 PQO = PQO = 900 and OQ is common. This implies that  PQO and PQO are congruent. Hence PQ = PQ. This shows that perpendicular distance of the object from the mirror is equal to the perpendicular distance of the image from the mirror.

Y

X

O

10 cm

P

P

Illustration 2: Two plane mirrors M1 and M2 are inclined to each other as shown in the figure. A ray of light parallel to M2 is incident on the M1. After two reflections one at each mirror it becomes parallel to M1. Determine the angle between the two mirrors. Solution:

Let  be the angle between the two mirrors M1 and M2. The incident ray AB is parallel to mirror M2 and strikes the mirror M1 at an angle of incidence equal to . It is reflected along BC; the angle of reflection being . From figure, we have M1BA = M1OM2 =  Similarly, for reflection at mirror M2, we have O M2CD = M2OM1 =   = 900 -  and x = 900 -  0 0 So x = 90 - (90 - ) =  0 0 Similarly  = 90   and y = 90 -  So y = . Now, in triangle OBC, 3 = 180o, therefore,  = 60o.

M1 B

D E

 x



Illustration 3: A point source of light O is placed at a distance L in front of the centre of a mirror of width d hung vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown. Find the greatest distance over which he can see the image of the light source in the mirror.

 N 1 

y

N2  

A 

 M2

C

Q` A

Q

d B

P 2L


P`


Solution:

PH-OPTICS-4

A large number of rays coming from B falls on the mirror at different angles. At the two extremes of the mirror, rays fall with the greatest angle of incidence and the corresponding reflected rays are the extremes amongst the reflected rays. Now consider the figure. The observer can see through the distance PQ, where extreme reflected rays meet the line, along which man is walking From geometry, CD = d = CD Since  BDO  BDP  PD = DO Also and ACQ  ACD  CQ = CO  CO = DO = d/2  PD = CQ = d/2 BDP ~ BDP PD BD   PD BD  PD = 2PD similarly CQ = 2CQ PQ = CD + PD + QC = d + 2 PD + 2CQ = d + 2 ( PD + QC) = d + 2d = 3d.

Q`

Q A

C

C`

B

O D P

D`

2L

P`

d

Illustration 4: (a) What is the minimum length of mirror required for a person to see his or her full height? Assume that the eyes are a distance ‘a’ below the top of the head and a distance b above the feet. (b) Does the person’s horizontal distance from the mirror matter? (c) Does the vertical position of the mirror matter? Solution:

(a) Rays enter the eye from the feet and the top of the a/2 i A head after reflection at the mirror, as shown in the a/2 i figure. We know that the angle of incidence is b/2 r equal to the angle of reflection. Light from the feet r B reaches the eye after reflection at point B, which is b/2 b/2 above the floor. Light from the top of the head reaches the eye after reflection at point A, located a/2 below the top of the head. The person’s total height is a + b, and the required length of mirror is a/2 + b/2, which is 50% of the full height of the person. (b) No. (c) Yes, the bottom of the mirror must be at a height b/2.

Spherical mirrors If mirror is a portion of a sphere, then it is known as spherical mirror. Type of Spherical mirrors There are two types of spherical mirror, (i) concave mirror (ii) convex mirror In concave mirror reflection takes place at the inner surface and in convex mirror reflection takes place at the outer surface.

P

F

C

C

F

P

Concave mirror

Convex Mirror



PH-OPTICS-5

Elements of Spherical mirrors Centre of curvature:

It is the centre of the sphere of which the mirror is a part.

Radius of curvature:

It is the radius of the sphere of which the mirror is a part.

Pole: It is the geometrical centre of the spherical reflecting surface. Principal axis: (For a spherical mirror) It is the straight line joining the centre of curvature to the pole. Focus: When a narrow beam of rays of light, parallel to the principal axis and close to it, is incident on the surface of a mirror, the reflected beam is found to converge to or appear to diverge from a point on the principal axis. This point is known as focus of the mirror. Focal length: (For a mirror) it is the distance between pole and the focus. Paraxial approximation In all the mirrors, an approximation is made that the aperture (size) of the mirror is small compared to its radius of curvature. Thus, most of the incident rays are nearly parallel to the principal axis. Ray tracing:

In geometrical optics, to locate the image of an object, tracing of a ray as it reflects, is very important. Concave Convex 1. A ray going through centre of curvature is reflected back along the same direction. (As it is normal to the mirror). C C

2. If incident ray is parallel to principal axis then the reflected ray either passes through the focus or appears to pass, and vice-versa. Also, parallel rays after reflection intersect on the focal plane. 3. If incident ray hits the mirror at pole, the principal axis bisects the angle between incident ray and reflected ray.

F

F

i i

F

P

i i

Sign convention There are several sign-conventions. Students should try to follow any one of these. In this package, the new Cartesian Sign Convention has been used wherever required. (i) All distances are measured from the pole P. (ii) Distances measured along the direction of incident rays are taken as positive. (iii) Distances measured along a direction opposite to the incident rays are taken as negative. (iv) Distances above the principal axis are positive. (v) Distances below the principal axis are negative.

Incident Ray +ve -ve

P

Mirror / Lens



PH-OPTICS-6

For a Concave mirror positions of image and its nature for different positions of real object are given in the following table. S. No.

Position of object

Position of image

Nature of image

Magnification

1.

At infinity

At focus

Real and inverted

Highly diminished

2.

Beyond 2f

Between f and 2f

Real and inverted

Diminished

3.

At 2f

At 2f

Real and inverted

Size of the image = size of the object

4.

Between 2f and f

Beyond 2f

Real and inverted

Magnified

5.

At f

At infinity

Real and inverted

Highly magnified

6.

Between f and P

Behind the mirror

Virtual and erect

Magnified

For a real object image formed by the Convex mirror lies behind the mirror between P and f, and it is erect and diminished Mirror formula Mirror formula gives the mathematical relation between object distance image distance and focal length of a mirror. If object distance image distance and focal length are represented by u, v and f respectively, then 1 1 1   f v u Note: In this equation u, v and f must be placed with proper sign. In our domain of syllabus f = R/2, where R is the radius of curvature of the mirror. Magnification Magnification formula describes the relation between sizes of object and its image. Transverse Magnification Transverse magnification is defined as m = - v/u =

image height object height

Note: v and u must be placed with proper sign Lateral or Axial Magnification The axial length of the image can be found by taking the difference of two image distance corresponding to two extreme lateral position of the object. Illustration 5: A line object of length 4 mm is placed in front of a concave mirror of radius of curvature 20 cm. at a distance of 15 cm from the mirror. The length of object is perpendicular to the principal axis and its one end lies on the principal axis. Calculate the height of the image. Solution:

Consider the figure u = - 15 cm, v = ? f =  R/2 = - 10 cm 1 1 1    f v u 1 1 1    10 v 15  v =  30 cm


PH-OPTICS-7


Negative sign indicates that image is formed in front of the mirror and hence real. To find the height of the image we will use the concept of transverse magnification himage v  hobject u himage

 30cm     4mm  15cm   himage =  8 mm Here negative sign indicates that image is inverted.



Illustration 6: When an object is at two different positions whose distances are u1 and u2 from the poles of a concave mirror, images of the same size are formed. Find the focal length of the mirror. Solution:

One image will be real and the other will be virtual. Since they are of the same size, one will have magnification m and the other m. v m =  1 so v 1 = mu1 u1 so from mirror formula, or

1 1 1   u1 u1m  f

1 1 1 1    u1  m f

…(1)

In the second situation, m =  and

1 1 1   u2 u2m  f

1 1 1 1    u2  m   f From equation (1) and (2) u1 u2  2 f f u  u2 u  u2  1  2 or f  1 . f 2

or

v2 u2

…(2)



PH-OPTICS-8

SOLVED PROBLEMS Problem 1:

A ray of light is incident on a plane reflecting surface at an angle of 30 to its normal. (a) Find the deviation in the incident ray. (b) What will be the deviation if the ray suffers a reflection again at a surface inclined at 60 to the first surface?

Solution:

(a) The ray AB is incident on the first mirror M1 at an angle of incidence of 30 as shown in the figure. After reflection, the reflected ray is along BC which also makes an angle of reflection 30 with the normal.  Deviation in the incident ray AB, D1 = DBO + ODC = 60 + 60 = 120 (b) The reflected ray BC is incident on the second mirror M2. It is again reflected to CD. Since M1OM2 = 60and OBC = 600 So BCO = 60  BCN2 = N2CD = 30.  Deviation D2 of the ray = 60 + 60 = 120  Total deviation due to the two mirrors = D1 + D2 = 240

M2

B

D2 C 30 N1 60 30

D

A N2 30 30

60

M1

60 O

B D

Problem 2:

A soldier directs a laser beam on an enemy by reflecting the beam from a mirror. If the mirror is rotated by an angle , by what angle will the reflected beam rotate?

Solution:

Let M1OM2 be the initial position of the mirror. ON is the normal to the initial mirror position M1OM2. The mirror is rotated through an angle  to the position M1OM2 . Now ON is the new normal to the new mirror position M1OM2. PO is the incident light. OQ was the initial reflected ray and OQ is N Q’ the reflected ray after rotating the mirror by angle . If N i = initial incidence angle, then POQ  2i and Q P  POQ'  PON' N'OQ  (i  )  (i  )  2i  2 M2 M1  The reflected beam rotates through an angle 2. O



M2

M1

Problem 3:

The sun (diameter D) subtends an angle of  radians at the pole of a concave mirror of focal length f. Calculate the diameter of the image of the sun formed by the mirror.

Solution:

Since the sun is at very large distance, p =  1 1 1     q f  q=f If the diameter of the image be d,

Image of the sun

 d q

f

d/2    d = (2) q q Putting 2  =  and q = f, we obtain d = f 

d



2


PH-OPTICS-9


Problem 4:

When an object is placed at a distance of 0.60 m from a convex spherical mirror, the 1 magnification produced is . Where should the object be placed to get magnification of 2 1 ? 3

Solution:

From mirror formula, Magnification m = 

1 1 1   v u f

v u

. . . (ii)

From both the equation, m = In the first situation, 

. . . (i)

f u f

1 f  2 f  0.60

(because u = 0.6m)

f = + 0.60 m

In the second situation 0.60 – u = 1.80

1 0.60  3 0.60  u  u =  1.20 m



PH-OPTICS-10

ASSIGNMENTS 1.

2.

(i) A ray of light falls on a plane mirror, show that if the mirror is tilted through an angle , as shown in the figure, the reflected ray tilts through an angle 2. (ii) What is the magnification produced by a plane mirror? (iii) Prove that the size of the image formed by a plane mirror is equal to that of the object.





M2

By what angle should M2 be rotated, so that the light ray after reflection from both the mirrors become horizontal.

25 40

3.

An object P is placed in front of a plane mirror at a distance of 10 cm along line PX as shown in the figure. What is the perpendicular distance of the image of p from the mirror?

M1

Y

10cm P

X

O

30 

4.

Two mirrors M1 and M2 are placed with one edge together such that angle between them is 60° a point object O is placed on the bisector of the angle between the mirrors at a distance of 10 cm from the common edge. Calculate the distance between the images formed by the two mirrors.

5.

Find the angle between two flat mirrors positioned such that a beam of light incident on one of the mirrors at an arbitrary angle with a plane that is perpendicular to the mirror surface on reflection from both mirror surface it becomes parallel to incident beam but in opposite direction.

6.







 

The adjacent figure shows two plane mirrors parallel to each other and an object O placed between them. Then, find the distance of first three images from the mirror M2 (in cm)

5 cm O 15 cm

cm M2

M1

7.

Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle 30 at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. Calculate the maximum number of times the ray undergoes reflections (including the first one) before it emerges out.

B

23m

0.2m

30

A


PH-OPTICS-11

8.


A light ray is observed to be reflected at a reflection angle r from a plane mirror. What is the angle by which the light ray has been deviated?

r 

9.

A concave mirror having a radius of curvature of 40 cm is placed in front of an illuminated point source at a distance of 30 cm from it. Find the distance of the image.

10.

An object 5 cm high is placed 20 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror, should a screen be placed to obtain a sharp image?

11.

A man uses a concave mirror for shaving. He keeps his face at a distance of 25 cm from the mirror and gets an image which is 1.4 times enlarged. Find the focal length of the mirror.

12.

A particle goes in a circle of radius 2.0 cm. A concave mirror of focal length 20 cm is placed with its principle axis passing through the centre of the circle and perpendicular to its plane. The distance between the pole of the mirror and the centre of the circle is 30 cm. Calculate the radius of the circle formed by the image.

13.

A concave mirror produces three times magnified real image of an object placed at 10cm in front of it. Find the position of image from the pole of mirror.

14.

An object is placed at 18cm in front of a spherical mirror. If the diminished image is formed at 4cm to the right of the mirror. Find focal length of mirror and nature of mirror. Also find the nature of image.

15.

In the figure shown a linear object AB of length 10cm is placed in front of a concave mirror. Whose focal length is 10cm. Calculate the length of the image of AB. Hint: Calculate the image distance for A & B and take their difference.

16.

17.

18.

20cm

A

B

P

30cm The image formed by a convex mirror of focal length 0.30 m is a quarter of the object. What is the distance of the object from the mirror?

In the figure shown a square of side 10 cm is placed in front of a concave mirror with sides AD and BC parallel to the principle axis. Focal length of the mirror is 15cm and other distances are shown. Calculate the area of the image.

B

A

C

D

P

30cm 40cm An object is moving with speed 20 cm/s along principle axis of concave mirror of radius of curvature 20 cm towards the pole. Find the speed of image in cm/s when object is at 30 cm from the pole of mirror.

19.

A convex mirror of focal length f produces an image half of the size of the object. Find the distance of the object from the mirror.

20.

An object is kept in front of a concave mirror of focal length 15 cm. the image formed is three times the size of object. Calculate two possible distance of the object from the mirror.



21.

22.

PH-OPTICS-12

A point source S is placed midway between two converging mirrors having equal focal length f as shown in the figure. Find the values of d for which only one image is formed. (Hint: image due to one mirror is the object for the other)

A converging mirror M, a point source S and diverging mirror M2 are arranged as shown in the figure. The source is placed at a distance of 30 cm from M1. the focal length of each of the mirrors is 20 cm. Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself. (a) Find the distance between the two mirrors. (b) Find the location of the image formed by the single reflection from M2.

d

S

M2

M1

S

23.

An object is placed 0.05 m in front of a concave mirror of radius of curvature 0.15 m. Determine the position of the image.

24.

A man uses a concave mirror for shaving. He keeps his face at a distance of 25 cm from the mirror and gets an image which is 1.4 times enlarged. Find the focal length of the mirror.

25

Find the distance of an object from a concave mirror of focal length 10cm so that the image size is four times the size of the object.

26.

A spherical mirror is to be used to form on a screen 5.0m from an object an image 5 times the size of the object. (a) What type of mirror is required? (b) What is its focal length? (c) What should be the separation between the mirror and the object?

27.

A point object is placed in front of spherical mirror at point P as shown figure. Consider OX and OY positive X and Y axis. Find coordinate of image of point P

Y f=15cm P O

1.2mm

X

40cm

28.

Find the co-ordinate of final image in each case after two reflection, one at each mirror (focal length of the concave mirror is 20cm and all other values are shown.) Case I: Ist reflection on spherical mirror Case II: Ist reflection on plane mirror Hint:

29.

Y

F=20cm

O 40cm

P 2mm

X

10cm

Image due to first reflection will be the object for the second reflection. For each reflection sign convention must be used.

Find the focal length of spherical mirror if both the images coincides at the same point in each of the following cases during to single reflection from both mirror.

24cm 4cm

(i)

O

P

P

30cm

15cm

(ii)



PH-OPTICS-13

30.

Find the position of final image from P2 after two reflection of object as shown in the figure if first reflection at concave. Where AB = 1mm, P1 P2 = 50cm, P1 B = 30 cm. f 1 = 20 cm f 2 = 20cm

M1

M2 A

P1

B f1

P2 f2



PH-OPTICS-14

ANSWERS TO ASSIGNMENTS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

(ii) +1 95 clockwise. OR 50 anticlockwise 5 cm 10 3 cm 900 5, 25, 35 30 180  2r 60 cm from the mirror on the side of the object 8.57 cm 87.5 cm 4.0 cm 30cm in front of the mirror. 5.14 cm, convex mirror, erect and virtual image. 5cm 0.9 m 48 cm2 5 f 20 cm, 10 cm 2 f, 4 f (a) 50 cm (b) 10 cm from the diverging mirror farther from the converging mirror 0.15 m 87.5 cm 12.5 cm (for real image), 7.5 cm (for virtual image) 25 (a) concave (b) f  m (c) 1.25m 24 X = 24cm, Y = 0.72 mm Case I: X = 76cm, Y = – 1.2 mm Case II:X = 20cm, Y = 6 mm (i) 48cm (ii) 20cm 20cm towards left of the P2.


Fiitjee Physics

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