Reflection questions

5/24/2018 Reflection questions

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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 cmSol. Distance between object and cancave mirror

O

x

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

1

v=

1 1

10x =

10

10

x

x

v1=10

10

x

xit is negativeImage 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

x10

(10 )

x

x= 22.5 + 22.5 = 45

210 10

10

x 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

=

60

2or

30

2x = 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 = mT=v

u

CPP-1 Class - XI Batches -PHONON

REFLECTION


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Longitudinal magnification = mL=dv

du = mT

2

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

1 1

v u =

1

f

1 1

15v = 1

10

v= 30 cm

mT=1

v =

( 30)

( 15)

= 2

mL= mT2= (2)2= 4

dv

du = 4

dv= 8 mmAmplitude 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 1

v u =

1

f

Differentiate w.r.t. time t

1 1d d

dt v dt u

=1d

dt f

2 2

1 1dv du

dt dt v u = 0

dv

dt=

2

2

v du

dtu

here : du

dt= + 4 cm/sec; u= 20 cm;

dv

dt= ?

Mirror formula :1 1

v u =

1

f

1 1

20v =

1

12

v = 30 cm

=dv

dt=

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 at

the convex mirror.ABis object :

(A) The second image is real, inverted of 1/5thmagnification.velocity

10cm 10cm

50 cm120 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 cm

10 cm

30 cm

10 cm

O

R= 120f= 60

A B

Velocity

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

Second reflection at convex mirror

A'B' becomes object for convex mirror


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for image of A' =A" u = 60, f= +60

1 1

v u =

1

f

1 1

60v =

1

60 v = +30 cmfor image of B' =B" u = 90, f= +60

1 1

v u =

1

f

1 1

90v =

1

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

VirtualSize = 15 cm

Magnification is 1/2

More over :dv

dt=

2

2

v du

dtu

Since : first image moves towards convex mirror sodu

dtpositive. So

dv

dtmust 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 the

image from the focus is :

(A) b2/ 4a (B) a/ b2 (C*) a2/ 4b (D) 4b/ a2

Sol. u =2

ab

; f=2

a

1 1

v u =

1

f

1 1

2

avb

=2

a

1

v=

2 2

(2 )

a b a=

2[2 ]

(2 )

b a a

a b a=

4

(2 )

b

a b a

v =22

4

ab a

b

Distance of image from focus

=2

a

v

=

22

4 2

ab a a

b=

2

4

a

b

= +2

4

a

bAns.

6. Iis the image of a point object Oformed by spherical mirror, then which of the following statement is incorrect :

(A) If OandIare on same side of the principal axis, then they have to be on opposite sides of the mirror.(B) If OandIare on opposite sides of the principal axis, then they have to be on same side of the mirror.

(C*) If OandIare on opposite side of the principal axis, then they have to be on opposite side of the mirror.

(D) If Ois on principal axis thenIhas to lie on principal axis only.


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Sol. For the spherical mirror

m =I

O

h

h=

v

u

v = IO

hu

h

OandIare on same side of principle axis means

I

O

h

h is positive, so vand uhave opposite sign.

OandIare on opposite side of principle axis meansI

O

h

his negative, so vand uhave 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' is

placed 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 image

formation.

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 OPdoes 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 extensions

of 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=80

2 = 40 cm

1 1

v u =

1

f

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 principal

axis. 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 cm

f= 40 cm

(a)1 1

v u =

1

f

1 +u

v=

u

f

v

u=

f

u f

differentiate mirror formula w.r.t. time :

2 2

1 1dv du

dt dt v u = 0


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dv

dt=

2v du

v dt

dv

dt=

2f du

u f dt

=240

60 40

(+10)

= (4) (+10)

dv

dt= 40 cm/sec Ans.

(b) Magnification formula :

m=i

o

h

h=

v

u =

f

u f

hi =

o

fh

u f

idh

dt=

0dhf

u f dt

=40

60 40

(+10)

idh

dt= 20 cm/sec Ans.

10. Find the angle of deviation (both clockwise and anticlockwise) suffered by a ray

incident on a plane mirror, at an angle of incidence 30.

30

MAns. [120 anticlockwise and 240 clockwise]Sol.

ACW = 180 60

30

30= 120 Anticlockwise

CW

= 180 + 60

= 240 Clockwise


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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 1

v u =

1

f

2 2

1 1dv du

dt dt v u = 0

dv

dt=

2

2

v du

dtu

For spherical mirror :uis negative and particle is moving towards mirror i.e. uis increasing anddu

dtis positive,

thereforedv

dtmust be negative or vshould decrease.

Case-1Image real vnegative decreasing vmeans image moving away from mirrorCase-2Image virtual vpositive decreasing vmeans 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 to

principal axis) be placed for the image it forms (due to rays falling on it after reflection from the concave mirror) to

coincide with the point source ?

Ans. [2454

cm = 61.25 cm]

Sol.1 1

V U =

1

F

35

Vx

V35____2

1

V=

1

(25) +1

35

V =35 25

10

cm

Distance of mirror (x) = 35

2

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) = 45(B*) = 30(C) = 60(D) All three


REFLECTION


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Sol. (90 ) + = 90or

=

4. A light ray is incident on a plane mirror, which after getting reflected strikes

another plane mirror, as shown in figure. The angle between the two mirrors is60. Find the angle '' shown in figure.

60Ans. [60]

Sol. If light is incident on first mirro at angle thenin triangleABC

2. + 2[90(30+)] + = 180

60

B

C

90

30+

A

2+ 120 2+ = 180 = 60

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 1

V u =

1

f

1 1

( )V f =

1

f

V =2

f

m=1

2height of image =

1

2m.

6. A concave mirror of radius of curvature 20 cm forms image of the sun. The diameter of the sun subtends an angle 1

on the earth. Then the diameter of the image is (in cm) :

(A) 2 /9 (B) /9 (C) 20 (D*) /18

Sol. ( / 2)

r

R= tan

1

2

rd

R/2

1/2

or r=2

R

1

2

180

D=2

4 180

R =

18

cm

7. Two plane mirrors are arranged at right angles to each other as shown in figure. A ray

of 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) 60 (B) 30

(C) 45 (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 .


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90

90

9090

8. A flat mirrorMis arranged parallel to a wall Wat a distanceLfrom it. The light

produced by a point sourceSkept on the wall is reflected by the mirror and produces

a light patch on the wall. The mirror moves with velocity v towards the wall :

(A) The patch of light will move with the speed von 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 larger

then 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 90 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,I2andI3) 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.ABis object.

Sol.


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1. A point object is placed at (0, 0) and a plane mirror 'M' is placed,

inclined 30 with thexaxis.

(a) Find the position of image.

y axis

M

x axisobject (1, 0)

30

(b) If the object starts moving with velocity 1 i

m/s and the

mirror is fixed find the velocity of image.

Ans. [(a) Position of image = (1 cos 60, 1 sin 60);

(b) Velocity of image = (1 cos 60i

, + 1 sin 60j

) m/s]

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

1m

1cos601/2

1cos601/2

P

(0,0)(1,0)

1m

603030 x

(+1 cos 60, 1 sin 60)

(1/2,3

2)

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

2

301mO

301/2

3 / 2

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

2

velocity of image alongxaxis

=3

2cos30

1

2cos60

=3

4

1

4=

1

2

velocity of image alongyaxis

=3

2sin30 +

1

2sin60 =

3

4+

3

4=

3

2

2. A point object 'O' is at the centre of curvature of a concave mirror. The mirror starts to move with speedu, in a direction

perpendicular 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=

V

u = 0

Ih

h

If object is placed at centre of convature u

O|V | = | u|

|hI | = + | h0|

| |I

d h

dt= +

0| |d h

dt


REFLECTION


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| |I

d h

dt= velocity of image wrst mirror,to primapal axis

0| |d h

dt= velocity of object worst mirro, to priampal axis

VI u=u

VI

= 2u

3. Sun ray are incident at an angle of 24 with the horizontal. How can they be directed parallel to the horizon using a

plane mirror ?

Ans. [Mirror should be placed on the path of the rays at an of 78 or 12 to the horizontal]

Sol.

9012=78

12

Incident ray

normal toplane mirror

Reflected ray

9078=12

7824

Incident ray

normal toplane mirror

Reflected ray

4. A squareABCDof side 1mm is kept at distance 15 cm infront of the concave mirror as

shown in the figure. The focal length of the mirror is 10 cm. The length of the perimeter of

its image will be :

(A) 8 mm (B) 2 mm

(C*) 12 mm (D) 6 mm

Sol.1

V+

1

u=

1

f

1

V+

1

(15) =1

(10) 1

V=

1

10+

1

15

V = 30 cm

| m | =V

u=

30

15= 2

long ta C'D' = A'B' = 2 mm

of image

1

V+

1

u=

1

f

21

V

dv

du 2

1

u= 0

dv

du=

2V

u= 4


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So long th of image

A'O' = B'C' = 4 mm

Than perimeter will be

A'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 m1at

small angle ''. m2is a concave mirror of focal length 'f '. After three successivereflections of this beam thexandycoordinates of the image is :

(A)x=f d,y=f (B)x= d+f,y=f(C)x=fd,y= f (D*)x=df, y = f

Sol. After Istreflection (from plane mirror)

Parallel beam of light strick on a concave mirror.

x

y

After IIndreflection (from concave mirror)

Image is formed at focus and shifted by a distancef (to principal axis)For IIIrdreflection

Object is placed at a distance |f. d | from a plane mirror.

6. M1&M2are two concave mirrors of the same focal length 10 cm.AB&

CDare their principal axes respectively. A point object Ois kept on the

lineABat distance 15 cm from M1. The distance between the mirrors

20 cm. Considering two successive reflections first on M1and then on

M2. The distance of final image from the lineABis :

(A) 3 cm (B*) 1.5 cm

(C) 4.5 cm (D) 1 cm

Sol. forM1

1 1

(15)V

=1

10

1

V=

1

10+

1

15

10cm15cm

B

C D

3cm

I1A

M1 M220cm

f =10 cm

V = 30 cm

ToM2

1 1

10V =

1

10

V = 5cm

m= V

u=

(5 )

(10 )

cm

cm=

1

2=

2

3

h

cmh

1= 1.5 cm

7. A light rayIis incident on a plane mirrorM. The mirror is rotated in the

direction as shown in the figure by an arrow at frequency 9/rps. Thelight reflected by the mirror is received on the wall Wat a distance 10 m

from the axis of rotation. When the angle of incidence becomes 37 the

speed 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 R v53

10m

53

R

v' =R'


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Let velocity of spot on wall is

vcos 53 =R'

v=10

cos53

36

cos53=

36 25

9

= 1000 rad/sec

8. Two plane mirrors of length Lare separated by distance Land a man M2 is

standing at distanceLfrom the connecting line of mirrors as shown in figure. AmanM1is walking in a straight line at distance 2Lparallel to mirrors at speed u,

then manM2at Owill be able to see image ofM1for time :

(A)4Lu

(B)3Lu

(C*)6Lu

(D)9Lu

Sol.

3L9L M2

Length of shoded region whereM2is visible is 9L 3L= 6L

If speed of person is Uthen tinle for which object is visible with G.

t =6L

U

9. As shown in the figure, an object Ois at the position ( 10, 2) with respect to the

origin P. The concave mirrorM1has radius of curvature 30 cm. A plane mirrorM2is kept at a distance 40 cm infront of the concave mirror. Considering first reflection

on the concave mirrorM1and second on the plane mirrorM2. Find the coordinates

of the second image w.r.t. The origin P:(A*) ( 46, 70) (B) 20, 70

(C) 46, 50 (D) 20, 50

Sol.1 1

v u =

1

f

(10,2)

O

x45

90

By

R=30cm

40 cm(30,6)

1

v+

1

10

=2

30=

1

15

1

v=

1

10

1

15=

3

30

3

30

1

v=

1

30

(xcordinate) is v = 30 cm


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m=v

u=

30

10= +3

Soycoordinate isy= +6 cm.

Writing equation of lineABmirror isy=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 end

farther from the pole is 20 cm away from it. Find the length of the image :

Ans. [Infinitely large]

Sol. vA = uA10 cm

10 cm

A B

F

20 cm

CvB = ?

uB = 10 cm

1 1

( 10)vB

=

1

( 10) vB =

So image ofABis infinitely large.

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