Reflection questions

CPP-1

Class - XI

Batches - PHONON

REFLECTION 1.

Sol.

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 Distance between object and cancave mirror First reflection from concave mirror : 22.5 cm R = 20 cm 1 1 1 f = 10 cm  = O v1 ( x) (10) x 1 1 1   v ( x) =  10 1 Plane Concave 1 1 1 10  x  v =  = x 10 10 x 1 10 x 10  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 :

v1 =

  10 x     10  x  – 22.5 = 22.5 – x   



x–

10 x = 22.5 + 22.5 = 45 (10  x )

10 x  x 2  10 x = 45 10  x  –x2 = 45 (10 – x) 2  x – 45x + 450 = 0



x=

45  2025  1800 45  452  4(450) = 2 2

45  15 60 30 45  225 = = or 2 2 2 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 = mT = 

v u

Page 1

Longitudinal magnification = mL =  here : du = 2mm; 

dv = mT2 du

u = 15 cm; f = 10 cm; dv = ?

1 1 1  = f v u

1 1 1  =– v 15 10  v = –30 cm



1 (30) =  = –2 v (15) mL = –mT2 = –(–2)2 = –4

mT = 

dv = –4 du  dv = 8 mm  Amplitude of image oscillation is 8 mm.



3.

Ans.

A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it. When its 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 1 1 1  = v u f

Sol.

Differentiate w.r.t. time t d 1 d 1 d 1      = dt  f  dt  v  dt  u   

 

1 dv 1 du  =0 v 2 dt u 2 dt



dv v 2 du = 2 dt u dt

here :

du = + 4 cm/sec; dt

1 1 Mirror formula :  v u



=

u = –20 cm;

dv =? dt

1 f

1 1 1  = v 20 12 v = –30 cm

(30)2 dv = (+4) = – 9 cm/sec dt (20)2 So image is moving at 9 cm/sec away from mirror

=

4.

Sol.

In the figure shown consider the first reflection at the plane mirror and second at the convex mirror. AB is object : velocity (A) The second image is real, inverted of 1/5th magnification. A B C (B*) The second image is virtual and erect with magnification 1/2. 10cm 10cm 120 cm 50 cm (C*) The second image moves towards the convex mirror. (D) The second image moves away from the convex mirror. First reflection at plane mirror : Velocity Image of A = A'  10 cm behind P. B O P A Image of B = B'  40 cm behind P. 30 cm Second reflection at convex mirror 10 cm 10 cm R = 120 A'B' becomes object for convex mirror 50 cm f = 60

Page 2

 for image of 

A' = A" u = –60, f = +60 1 1 1  = f v u

   for image of 

1 1 1  = v 60 60 v = +30 cm B' = B" u = –90, f = +60

1 1 1  = f v u 1 1 1  = v 90 60  v = +45 cm So second image  A"B"  Behind convex mirror  Virtual  Size = 15 cm  Magnification is 1/2



dv v 2 du = 2 dt u dt

More over :

Since : first image moves towards convex mirror so

du dv positive. So must be negative i.e. second image also dt dt

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 a a  u = b   ; f =  2 2  

Sol.

1 1 1  = v u f





2 1 1  = a a v  b   2   1 2 2 4b 2[2b  a  a] =  = = v a (2b  a) a(2b  a) a(2b  a)

 2ab  a 2   v =    4b  Distance of image from focus 

= v

a 2

2ab  a 2 a a2   = 4b 2 = 4b =+ 6.

a2 4b

Ans.

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 3

Sol.

For the spherical mirror hI v m= h =  u O h  v = u  I   hO 



hI O and I are on same side of principle axis means h is positive, so v and u have opposite sign. O hI O and I are on opposite side of principle axis means h is negative, so v and u have same sign. O Option (A) is correct. Option (B) is correct. Option (C) is incorrect

7.

Sol.

8.

Sol.

Two plane mirrors are placed as shown in the figure and a point object 'O' is (2, 4) placed at the origin. (2, 3) (a) How many images will be formed. (1, 1.25) (b) Find the position(s) of image(s). (2, 2) P (c) Will the incident ray passing through a point 'P' (1, 1.25) take part in image formation. (0, 0) O (2, 0) object Ans. [(a) 1; (b) (4, 0); (c) No] (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. 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] u = +40 cm f= 

80 = –40 cm 2

1 1 1  = v u f

1 1 1  = v (40) (40)

  9.

Sol.

v = –20 cm Ans.

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] u = –60 cm f = –40 cm (a)

1 1 1  = v u f



1+

u u = f v f v = u f u



differentiate mirror formula w.r.t. time :



1 dv 1 du  =0 v 2 dt u 2 dt Page 4

2



dv  v  du =   dt  v  dt



dv  f  du =   dt  u  f  dt

2

2

 40  =   (+10)  60  40  = – (4) (+10) dv = – 40 cm/sec dt

Ans.

(b) Magnification formula : hi v  f   m= h =  =  u o u f 



 f   ho hi =   u f 



 f  dh0 dhi  =  dt  u  f  dt

 40  =   (+10)  60  40  dhi = –20 cm/sec Ans. dt

30º M

30°

Sol.

Find the angle of deviation (both clockwise and anticlockwise) suffered by a ray incident on a plane mirror, at an angle of incidence 30º. Ans. [120º anticlockwise and 240º clockwise] ACW = 180° – 60° = 120° Anticlockwise CW = 180° + 60° = 240° Clockwise

30°

10.

Page 5

CPP-2

Batches - PHONON

Class - XI

REFLECTION 1.

Sol.

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. 1 1 1  = f v u 1 dv 1 du   2  2 =0 v dt u dt

 v 2  du dv =  2   u  dt dt   For spherical mirror : u is negative and particle is moving towards mirror i.e. u is increasing and

du is positive, dt

dv must be negative or v should decrease. dt Case-1  Image real  v negative  decreasing v means image moving away from mirror Case-2  Image virtual  v positive  decreasing v means image moving towards mirror So, image may move towards the mirror

therefore

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. [

245 cm = 61.25 cm] 4 1 1 1  = V U F

Sol.



1 1 1 = + (–25) V 35



35  25 V= cm 10

Distance of mirror

35

____ V–35 2 V

x

 V – 35  (x) =   + 35  2  = 61.25 cm

3.

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

Page 6

(90 – ) +  = 90º

Sol. or

= 

4.

Sol.

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 is 60º. Find the angle '' shown in figure. 60º

Ans. [60º] If light is incident on first mirro at angle  then in triangle ABC B 2.+ 2[90–(30+)] + = 180º 90–  2 + 120 – 2 + = 180º 60º  = 60º

A

30+ C

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. The height of his image is : (A) 0.25 m (B) 0.33 m (C*) 0.5 m (D) 0.67 m 1 1 1  = f V u

Sol.





6.

1 1 1  = V (– f ) f

V=

f 2

m=

1 1  height of image = m. 2 2

A concave mirror of radius of curvature 20 cm forms image of the sun. The diameter of the sun subtends an angle 1º on the earth. Then the diameter of the image is (in cm) : (A) 2 /9 (B) /9 (C) 20 (D*) /18 r 1 = tan   ( R / 2)  2

Sol.

or

r=

 R 1 × × 2 2 180



D=

2R    = cm 4 180 18

d

r

1/2 R/2

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 ray emerges 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 parallel for all values of .

Page 7

90–

90– 90– 90–

8.

A flat mirror M is arranged parallel to a wall W at a distance L from it. The light produced by a point source S kept 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 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 larger then v. (D*) The width of the light patch on the wall remains the same.

Light patch

Sol.

Light – patch will remain the same. 9.

Sol.

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

I2(–x,y)

x

O I3(–x,–y)

10.

I1(x,–y)

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

Sol.

Page 8

CPP-3

Batches - PHONON

Class - XI

REFLECTION 1.

Sol.

y axis

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.  (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 60º, – 1 sin 60º);   (b) Velocity of image = (1 cos 60º i , + 1 sin 60º j ) m/s]

M

30º object

(a) Co-ordinate of point from diagram (+1 cos 60º, – 1 sin 60º)

(0,0)

3 (1/2,– ) 2

(1,0) 1m 30º 60º 30º

1cos60º 1/2 1cos60º 1m 1/2

(b) Speed of object parallel to mirro

= 1 cos 30 =

Speed of object perpendicualr to mirror

= 1 sin 30 =

x axis

(1, 0)

x

P

3 2 1 2

O

velocity of image along x axis

1m

30º 3/2

1 3 cos30 – cos60º 2 2

=

=

1/2

30º

3 1 1 – = 4 4 2

velocity of image along y axis =

2.

Sol.

1 3 3 3 3 sin30 + sin60 = + = 2 2 4 4 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 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 m= –

hI V = h0 u

If object is placed at centre of convature |V | = | u | | hI | = + | h0 | d | hI | d | h0 | = + dt dt

u O

Page 9

d | hI | = velocity of image wrst mirror,  to primapal axis dt d | h0 | = velocity of object worst mirro,  to priampal axis dt

VI – u = u VI = 2u 3.

Sun ray are incident at an angle of 24º 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 78º or 12º to the horizontal]

Sol.

Incident ray normal to plane mirror 90–12=78 12º

Reflected ray

normal to plane mirror Incident ray 78º

24º

4.

Reflected ray 90–78=12º

A square ABCD of 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 1 1 1 + = f V u

Sol.

1 1 1 1 1 1 + (–15) = (–10)  =– + V V 10 15

V = – 30 cm | m|= long t a of image

V 30 = =2 u 15

C'D' = A'B' = 2 mm 1 1 1 + = f V u

–

1 dv 1 – = 0 V 2 du u 2

dv V = du u

2

=4 Page 10

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.

Sol.

In the figure shown a thin parallel beam of light is incident on a plane mirror m1 at small angle ''. m2 is a concave mirror of focal length 'f '. After three successive reflections 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  After Ist reflection (from plane mirror) Parallel beam of light strick on a concave mirror.

y

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

6.

Sol.

x

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 the line AB at distance 15 cm from M1. The distance between the mirrors 20 cm. Considering two successive reflections first on M1 and then on M2. The distance of final image from the line AB is : (A) 3 cm (B*) 1.5 cm (C) 4.5 cm (D) 1 cm for M1 1 1 1  = V (–15) –10 1 1 1 =– + V 10 15

f =10 cm 15cm

I1 3cm D

C

V = – 30 cm

M1

To M2

10cm

B

A

M2 20cm

1 1 1  = V 10 –10

V = – 5cm m= –

7.

Sol.

(–5cm) h2 V 1 = – (10cm) = =  h1 = 1.5 cm u 3 cm 2

A light ray I is incident on a plane mirror M. The mirror is rotated in the direction as shown in the figure by an arrow at frequency 9/ rps. The light reflected by the mirror is received on the wall W at a distance 10 m from the axis of rotation. When the angle of incidence becomes 37º 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 ' of reflected light is

9  ' = 2  2  = 36 rad/sec   v' = R'

53º v

R

R

53º 10m Page 11

Let velocity of spot on wall is v cos 53º = R' v=

8.

10 36 36  25 × = = 1000 rad/sec cos53º cos 53º 9

Two plane mirrors of length L are separated by distance L and a man M2 is standing at distance L from the connecting line of mirrors as shown in figure. A man 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 : 4L u 3L (B) u 6L (C*) u 9L (D) u

(A)

Sol.

9L 3L

M2

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

9.

Sol.

6L U

As shown in the figure, an object O is at the position (– 10, 2) with respect to the origin P. The concave mirror M1 has radius of curvature 30 cm. A plane mirror M2 is kept at a distance 40 cm infront of the concave mirror. Considering first reflection on the concave mirror M1 and second on the plane mirror M2. 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 1 1 1  = f v u

40 cm

1 1 = v 30

B

(30,6)

R=30cm

1  1  2 –1 + =  = v  –10  –30 15

1 1 1 3 3 = – = – v 10 15 30 30

y

(–10,2) O x

45º 90º

(x cordinate) is v = 30 cm Page 12

m=

v 30 = = +3 u –10

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.

Sol.

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] 10 cm 10 cm v A = uA vB = ? C F uB = –10 cm A B 1 1 1  = (10) vB (10)  vB =  20 cm So image of AB is infinitely large.

Page 13