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Page 1 1. A point object is kept between a plane mirror and a concave mirror facing each other. The distance between the mirrors is 22.5 cm. The radius of curvature of the concave mirror is 20 cm. What should be the distance of the object from the concave mirror so that after two successive reflections the final image is formed on the object itself ? [Consider first reflection from concave mirror] : (A) 5 cm (B*) 15 cm (C) 10 cm (D) 7.5 cm Sol. Distance between object and cancave mirror O x 22.5 cm Plane Concave R = 20 cm f = 10 cm First reflection from concave mirror : 1 1 1 ( ) v x = 1 ( 10) 1 1 1 () v x = 1 10 1 1 v = 1 1 10 x = 10 10 x x v 1 = 10 10 x x it is negative Image of concave mirror becomes object for plane mirror. Second reflection from plane mirror image is formed at same distance from plane as is the distance of object. Now according to question, second image is formed at object itself. So : 10 10 x x – 22.5 = 22.5 – x x 10 (10 ) x x = 22.5 + 22.5 = 45 2 10 10 10 x x x x = 45 x 2 = 45 (10 – x) x 2 – 45x + 450 = 0 x = 2 45 45 4(450) 2 = 45 2025 1800 2 = 45 225 2 = 45 15 2 = 60 2 or 30 2 x = 30 or 15 x = 30 is not possible, so : x = 15 Ans. 2. A point object at 15 cm from a concave mirror of radius of curvature 20 cm is made to oscillate along the principal axis with amplitude 2 mm. The amplitude of its image will be : (A) 2 mm (B) 4 mm (C*) 8 mm (D) 16 mm Sol. Transverse magnification = m T = v u CPP-1 Class - XI Batches - PHONON REFLECTION

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Page 1: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 1

1. A point object is kept between a plane mirror and a concave mirror facing each other. The distance between the mirrorsis 22.5 cm. The radius of curvature of the concave mirror is 20 cm. What should be the distance of the object from theconcave mirror so that after two successive reflections the final image is formed on the object itself ?[Consider first reflection from concave mirror] :(A) 5 cm (B*) 15 cm (C) 10 cm (D) 7.5 cm

Sol. Distance between object and cancave mirror

Ox

22.5 cm

Plane Concave

R = 20 cmf = 10 cm

First reflection from concave mirror :

1

1 1( )v x

=

1( 10)

1

1 1( )v x

= 1

10

1

1v =

1 110x

= 1010

xx

v1 = 10

10xx

it is negativeImage of concave mirror becomes object for plane mirror. Second reflection from plane mirror image is formed at samedistance from plane as is the distance of object. Now according to question, second image is formed at object itself.So :

1010

xx

– 22.5 = 22.5 – x

x – 10

(10 )xx

= 22.5 + 22.5 = 45

210 10

10x x x

x

= 45

–x2 = 45 (10 – x) x2 – 45x + 450 = 0

x = 245 45 4(450)2

= 45 2025 1800

2

= 45 225

2

= 45 15

2

= 602

or 302

x = 30 or 15x = 30 is not possible, so :x = 15 Ans.

2. A point object at 15 cm from a concave mirror of radius of curvature 20 cm is made to oscillate along the principal axiswith amplitude 2 mm. The amplitude of its image will be :(A) 2 mm (B) 4 mm (C*) 8 mm (D) 16 mm

Sol. Transverse magnification = mT = vu

CPP-1 Class - XI Batches - PHONONREFLECTION

Page 2: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 2

Longitudinal magnification = mL = dvdu

= mT2

here : du = 2mm; u = 15 cm; f = 10 cm; dv = ?

1 1v u =

1f

1 1

15v = –

110

v = –30 cm

mT = 1v

= ( 30)( 15)

= –2

mL = –mT2 = –(–2)2 = –4

dvdu

= –4

dv = 8 mm Amplitude of image oscillation is 8 mm. Ans.

3. A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it. Whenits distance from the mirror is 20 cm its velocity is 4 cm/s. The velocity of the image in cm/s at that instant is :(A) 6, towards the mirror (B) 6, away from the mirror (C*) 9, away from the mirror (D) 9, towards the mirror

Sol.1 1v u =

1f

Differentiate w.r.t. time t

1 1d ddt v dt u

= 1d

dt f

2 21 1dv du

dt dtv u = 0

dvdt =

2

2v du

dtu

here :dudt

= + 4 cm/sec; u = –20 cm; dvdt

= ?

Mirror formula :1 1v u =

1f

1 120v

= 1

12

v = –30 cm

= dvdt

= 2

2( 30)( 20)

(+4) = – 9 cm/sec

So image is moving at 9 cm/sec away from mirror

4. In the figure shown consider the first reflection at the plane mirror and second atthe convex mirror. AB is object :(A) The second image is real, inverted of 1/5th magnification.

velocity

10cm 10cm50 cm

120 cm

A B C(B*) The second image is virtual and erect with magnification 1/2.(C*) The second image moves towards the convex mirror.(D) The second image moves away from the convex mirror.

Sol. First reflection at plane mirror :

50 cm10 cm

30 cm10 cm

O

R = 120f = 60

A BVelocity

PImage of A = A' 10 cm behind P.Image of B = B' 40 cm behind P.

Second reflection at convex mirrorA'B' becomes object for convex mirror

Page 3: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 3

for image of A' = A" u = –60, f = +60

1 1v u =

1f

1 1

60v =

160

v = +30 cm for image of B' = B" u = –90, f = +60

1 1v u =

1f

1 1

90v =

160

v = +45 cmSo second image A"B" Behind convex mirror

Virtual Size = 15 cm Magnification is 1/2

More over :dvdt =

2

2v du

dtu

Since : first image moves towards convex mirror so dudt positive. So

dvdt must be negative i.e. second image also

moves towards convex mirror.

5. The distance of an object from the focus of a convex mirror of radius of curvature 'a' is 'b'. Then the distance of theimage from the focus is :(A) b2 / 4a (B) a / b2 (C*) a2 / 4b (D) 4b/ a2

Sol. u = 2ab

; f =

2a

1 1v u =

1f

1 1

2av b

= 2a

1v =

2 2(2 )

a b a

= 2[2 ]

(2 )

b a aa b a

= 4

(2 )b

a b a

v = 22

4

ab ab

Distance of image from focus

= 2

av

= 22

4 2

ab a a

b = 2

4

ab

= + 2

4a

bAns.

6. I is the image of a point object O formed by spherical mirror, then which of the following statement is incorrect :(A) If O and I are on same side of the principal axis, then they have to be on opposite sides of the mirror.(B) If O and I are on opposite sides of the principal axis, then they have to be on same side of the mirror.(C*) If O and I are on opposite side of the principal axis, then they have to be on opposite side of the mirror.(D) If O is on principal axis then I has to lie on principal axis only.

Page 4: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 4

Sol. For the spherical mirror

m = I

O

hh =

vu

v = I

O

huh

O and I are on same side of principle axis means I

O

hh is positive, so v and u have opposite sign.

O and I are on opposite side of principle axis means I

O

hh is negative, so v and u have same sign.

Option (A) is correct. Option (B) is correct. Option (C) is incorrect

7. Two plane mirrors are placed as shown in the figure and a point object 'O' isplaced at the origin.(a) How many images will be formed.

(0, 0) O (2, 0)

(1, 1.25)P

(2, 2)

(2, 3)

(2, 4)

object

(b) Find the position(s) of image(s).(c) Will the incident ray passing through a point 'P' (1, 1.25) take part in imageformation.Ans. [(a) 1; (b) (4, 0); (c) No]

Sol. (a) Only one image will be formed rays after reflection converge at same position.(b) Object distance = image distance for plane mirror so image is at (4, 0).(c) Ray OP does not strike the mirros so it does not take part in image formation.

8. A converging beam of light rays is incident on a concave spherical mirror whose radius of curvature is 0.8 m.Determine the position of the point on the optical axis of the mirror where the reflected rays intersect, if the extensionsof the incident rays intersect the optical axis 40 cm from the mirror's pole.Ans. [0.2 m from the mirror]

Sol. u = +40 cm

f = 802

= –40 cm

1 1v u =

1f

1 1

( 40)v

=

1( 40)

v = –20 cm Ans.

9. A point object is placed on the principal axis at 60 cm in front of a concave mirror of focal length 40 cm on the principalaxis. If the object is moved with a velocity of 10 cm/s (a) along the principal axis, find the velocity of image(b) perpendicular to the principal axis, find the velocity of image at that moment.Ans. [(a) 40 cm/s opposite to the velocity of object; (b) 20 cm/s opposite along the velocity of object]

Sol. u = –60 cmf= –40 cm

(a)1 1v u =

1f

1 + uv =

uf

vu =

fu f

differentiate mirror formula w.r.t. time :

2 21 1dv du

dt dtv u = 0

Page 5: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 5

dvdt =

2v duv dt

dvdt =

2f du

u f dt

= 240

60 40

(+10)

= – (4) (+10)

dvdt = – 40 cm/sec Ans.

(b) Magnification formula :

m= i

o

hh =

vu

= f

u f

hi = of h

u f

idhdt =

0dhfu f dt

= 40

60 40

(+10)

idhdt = –20 cm/sec Ans.

10. Find the angle of deviation (both clockwise and anticlockwise) suffered by a rayincident on a plane mirror, at an angle of incidence 30º.

30º

MAns. [120º anticlockwise and 240º clockwise]Sol. ACW = 180° – 60°

30°

30°= 120° Anticlockwise

CW = 180° + 60°= 240° Clockwise

Page 6: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 6

1. A particle is moving towards a fixed spherical mirror. The image :(A) Must move away from the mirror (B) Must move towards the mirror(C*) May move towards the mirror (D) Will move towards the mirror, only if the mirror is convex.

Sol.1 1v u =

1f

2 21 1dv du

dt dtv u = 0

dvdt =

2

2v du

dtu

For spherical mirror : u is negative and particle is moving towards mirror i.e. u is increasing and dudt is positive,

therefore dvdt must be negative or v should decrease.

Case-1 Image real v negative decreasing v means image moving away from mirrorCase-2 Image virtual v positive decreasing v means image moving towards mirrorSo, image may move towards the mirror

2. A point source is at a distance 35 cm on the optical axis from a spherical concave mirror having a focal length 25 cm.At what distance measured along the optical axis from the concave mirror should a plane mirror (perpendicular toprincipal axis) be placed for the image it forms (due to rays falling on it after reflection from the concave mirror) tocoincide with the point source ?

Ans. [ 2454 cm = 61.25 cm]

Sol.1 1V U =

1F

35

Vx

V–35____2

1V =

1(–25) +

135

V =35 25

10

cm

Distance of mirror (x) =– 352

V

+ 35

= 61.25 cm

3. Two mirrors are inclined at an angle as shown in the figure. Light rays is incident parallelto one of the mirrors. Light will start retracing its path after third reflection if :(A) = 45º(B*) = 30º(C) = 60º(D) All three

CPP-2 Class - XI Batches - PHONONREFLECTION

Page 7: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 7

Sol. (90 – ) + = 90ºor

=

4. A light ray is incident on a plane mirror, which after getting reflected strikesanother plane mirror, as shown in figure. The angle between the two mirrors is60º. Find the angle '' shown in figure.

60ºAns. [60º]Sol. If light is incident on first mirro at angle then

in triangle ABC2.+ 2[90–(30+)] + = 180º

60º

B

C

90–

30+

A

2 + 120 – 2 + = 180º = 60º

5. A boy of height 1 m stands in front of a convex mirror. His distance from the mirror is equal to its focal length. Theheight of his image is :(A) 0.25 m (B) 0.33 m (C*) 0.5 m (D) 0.67 m

Sol.1 1V u =

1f

1 1

(– )V f =

1f

V =2f

m =12

height of image = 12

m.

6. A concave mirror of radius of curvature 20 cm forms image of the sun. The diameter of the sun subtends an angle 1ºon the earth. Then the diameter of the image is (in cm) :(A) 2 /9 (B) /9 (C) 20 (D*) /18

Sol. ( / 2)r

R = tan12

rdR/2

1/2

or r =2R

× 12

× 180

D =24 180R = 18

cm

7. Two plane mirrors are arranged at right angles to each other as shown in figure. A rayof light is incident on the horizontal mirror at an angle . For what value of the rayemerges parallel to the incoming ray after reflection from the vertical mirror :(A) 60º (B) 30º(C) 45º (D*) All of the above

Sol. As shown in figure, both incident ray and reflected ray make same angle (90 – ) with horizontal and one anti parallelfor all values of .

Page 8: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 8

90–

90–90–90–

8. A flat mirror M is arranged parallel to a wall W at a distance L from it. The lightproduced by a point source S kept on the wall is reflected by the mirror and producesa light patch on the wall. The mirror moves with velocity v towards the wall :(A) The patch of light will move with the speed v on the wall.(B*) The patch of light will not move on the wall.(C) As the mirror comes closer the patch of light will become larger and shift away from the wall with speed largerthen v.(D*) The width of the light patch on the wall remains the same.

Sol. Light patch

Light – patch will remain the same.

9. A point object (placed between two plane mirrors whose reflecting surfaces make an angle of 90º with one another)and all its images lie on a :(A) Straight line (B) Parabola (C*) Circle (D) Ellipse

Sol. Object O' and image (I1, I2 and I3) lies on a circle where centre is a junction point of mirro.

O'

O

( , )x y

x

y

I x y2(– , )

I x y3(– ,– ) I x y1( ,– )

10. In the figure shown draw the field view of the image. AB is object.

Sol.

Page 9: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 9

1. A point object is placed at (0, 0) and a plane mirror 'M' is placed,inclined 30º with the x axis.(a) Find the position of image.

y axisM

x axisobject (1, 0)

30º

(b) If the object starts moving with velocity 1 i m/s and themirror is fixed find the velocity of image.Ans. [(a) Position of image = (1 cos 60º, – 1 sin 60º);(b) Velocity of image = (1 cos 60º i , + 1 sin 60º j

) m/s]

Sol. (a) Co-ordinate of point from diagram

1m

1cos60º1/2

1cos60º1/2

P

(0,0)(1,0)

1m60º30º

30º x(+1 cos 60º, – 1 sin 60º)

(1/2,–3

2)

(b) Speed of object parallel to mirro = 1 cos 30 = 3

2

30º1mO

30º1/23 / 2

Speed of object perpendicualr to mirror = 1 sin 30 = 12

velocity of image along x axis

=3

2cos30 –

12

cos60º

=34

– 14

= 12

velocity of image along y axis

=3

2 sin30 +

12

sin60 = 3

4 +

34

= 3

2

2. A point object 'O' is at the centre of curvature of a concave mirror. The mirror starts to move with speed u, in a directionperpendicular to the principal axis. Then the initial velocity of the image is :(A) 2u, in the direction opposite to that of mirror's velocity(B*) 2u, in the direction same as that of mirror's velocity(C) Zero(D) u, in the direction same as that of mirror's velocity

Sol. m = –Vu =

0

Ihh

If object is placed at centre of convature uO|V | = | u |

| hI | = + | h0 |

| |Id hdt = + 0| |d h

dt

CPP-3 Class - XI Batches - PHONONREFLECTION

Page 10: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 10

| |Id hdt = velocity of image wrst mirror, to primapal axis

0| |d hdt = velocity of object worst mirro, to priampal axis

VI – u = uVI = 2u

3. Sun ray are incident at an angle of 24º with the horizontal. How can they be directed parallel to the horizon using aplane mirror ?Ans. [Mirror should be placed on the path of the rays at an of 78º or 12º to the horizontal]

Sol.

90–12=7812º

Incident ray

normal toplane mirror

Reflected ray

90–78=12º78º24º

Incident ray

normal toplane mirror

Reflected ray

4. A square ABCD of side 1mm is kept at distance 15 cm infront of the concave mirror asshown in the figure. The focal length of the mirror is 10 cm. The length of the perimeter ofits image will be :(A) 8 mm (B) 2 mm(C*) 12 mm (D) 6 mm

Sol.1V +

1u =

1f

1V +

1(–15) =

1(–10)

1V = –

110 +

115

V = – 30 cm

| m | =Vu =

3015 = 2

long t a C'D' = A'B' = 2 mmof image

1V +

1u =

1f

– 2

1V

dvdu – 2

1u = 0

dvdu =

2Vu = 4

Page 11: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 11

So long th of imageA'O' = B'C' = 4 mm

Than perimeter will beA'B' + B'C' + C'O' + D'A' = 12 mm

5. In the figure shown a thin parallel beam of light is incident on a plane mirror m1 atsmall angle ''. m2 is a concave mirror of focal length 'f '. After three successivereflections of this beam the x and y coordinates of the image is :(A) x = f – d, y = f (B) x = d + f, y = f(C) x = f – d, y = – f (D*) x = d – f, y = – f

Sol. After Ist reflection (from plane mirror)Parallel beam of light strick on a concave mirror.

x

y

After IInd reflection (from concave mirror)Image is formed at focus and shifted by a distance f ( to principal axis)For IIIrd reflectionObject is placed at a distance | f. d | from a plane mirror.

6. M1 & M2 are two concave mirrors of the same focal length 10 cm. AB &CD are their principal axes respectively. A point object O is kept on theline AB at distance 15 cm from M1. The distance between the mirrors20 cm. Considering two successive reflections first on M1 and then onM2. The distance of final image from the line AB is :(A) 3 cm (B*) 1.5 cm(C) 4.5 cm (D) 1 cm

Sol. for M1

1 1(–15)V

=1

–10

1V = –

110 +

115

10cm15cmB

C D3cm

I1A

M1 M220cm

f =10 cm

V = – 30 cmTo M2

1 110V

=1

–10V = – 5cm

m = – Vu = –

(–5 )(10 )

cmcm =

12

= 2

3hcm h1 = 1.5 cm

7. A light ray I is incident on a plane mirror M. The mirror is rotated in thedirection as shown in the figure by an arrow at frequency 9/ rps. Thelight reflected by the mirror is received on the wall W at a distance 10 mfrom the axis of rotation. When the angle of incidence becomes 37º thespeed of the spot (a point) on the wall is :(A) 10 m/s (B*) 1000 m/s(C) 500 m/s (D) None of these

Sol. ' of reflected light is

' = 29 2

= 36 rad/sec Rv53º

10m

53º

R

v' = R'

Page 12: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 12

Let velocity of spot on wall isv cos 53º = R'

v =10

cos53º × 36

cos53º = 36 25

9

= 1000 rad/sec

8. Two plane mirrors of length L are separated by distance L and a man M2 isstanding at distance L from the connecting line of mirrors as shown in figure. Aman M1 is walking in a straight line at distance 2L parallel to mirrors at speed u,then man M2 at O will be able to see image of M1 for time :

(A) 4Lu

(B) 3Lu

(C*) 6Lu

(D) 9Lu

Sol.

3L9L M2

Length of shoded region where M2 is visible is 9L – 3L = 6LIf speed of person is U then tinle for which object is visible with G.

t =6LU

9. As shown in the figure, an object O is at the position (– 10, 2) with respect to theorigin P. The concave mirror M1 has radius of curvature 30 cm. A plane mirror M2is kept at a distance 40 cm infront of the concave mirror. Considering first reflectionon the concave mirror M1 and second on the plane mirror M2. Find the coordinatesof the second image w.r.t. The origin P :(A*) (– 46, – 70) (B) – 20, – 70(C) – 46, – 50 (D) – 20, – 50

Sol.1 1v u =

1f

(–10,2)O

x45º

90º

By

R=30cm

40 cm (30,6)1v +

1–10

=2

–30 = –115

1v =

110 –

115 =

330 –

330

1v =

130

(x cordinate) is v = 30 cm

Page 13: REFLECTION - FIITJEE Jaipurjaipur.fiitjee.com/phononcpp/Reflection-A.pdf · Second reflection from plane mirror image is formed at same distance from plane as is the distance of object

Page 13

m =vu =

30–10 = +3

So y coordinate is y = +6 cm.Writing equation of line AB mirror is y = x + c ...(1)Equation of line of mirror = y = – x – 40 ...(2)Solving (1) and (2) we set cordinates of intersection on mirror (– 8, –32)Now let image be at (a, b). Then mid point of (a, 6) and (30, 6) is point (– 8, – 32) on mirror i.e

10. A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that the endfarther from the pole is 20 cm away from it. Find the length of the image :Ans. [Infinitely large]

Sol. vA = uA

10 cm 10 cm

A BF

20 cm

CvB = ?uB = –10 cm

1 1( 10)vB

= 1

( 10) vB = So image of AB is infinitely large.