Ray Optics- 1 Practice Questions - DPP for NEET

 Page 1


(1) (a) Here, n = 
360 360
4
90
==
q
\   n is an even number.
Thus, number of images formed = n – 1 = 3. All these
three images lie on a circle with centre at C  (The point
of intersection of mirrors M
1
 and M
2 
) and whose radius
is equal to the distance between C and object.
(2) (c) Here, [n] = 5 Þ n – 1 £ 5 £ n
\ 
360
q
 – 1 £ 5 £ 
360
q
or,  q ³ 
360
6
 or  
360
5
q£
\ 60º
 
£ q £ 72º
(3) (d) For the given q = 50º,
n = 
360
q
 = 
360
50
 = 7.2
The integer value of (7.2) is 7.
Thus number of images formed is 7.
(4) (a) The situation is illustrated in figure.
90º-i
N
1
N
2
M
2
B
O
C
a
a
q
b
q
X
A M
1
i
i
XA is the incident ray . BC is the final reflected ray . It
is given that BC is parallel to mirror M
1
. Look at the
assignment of the angles carefully . Now N
2
 is normal
to mirror M
2
. Therefore b = q
Then from D OAB
q + b  + 90º – i = 180º
or q + q + 90º – i = 180º or  i = 2q – 90º
Thus if the angle of incidence is i = 2q - 90º, then the
final reflected ray will be parallel to the first mirror.
(5) (c) Small and erect image is formed only by convex mirror.
Plane mirror from images equal to object and concave
mirror form images bigger than object.
(6) (b) The image will be formed by the plane mirror at a 30
cm behind it, while the image by convex mirror will be
formed at 10 cm behind the convex mirror.
Since for convex mirror u = – 50 cm
v = 10 cm
1 1 1 154
f 50 10 50 50
-+
= +==
-
f = 
50
4
 = 12.5 cm
Therefore the radius of curvature  of convex mirror is
r = 2 f = 25 cm
(7) (a) The image of object O from mirrror M
1
 is I
1
 and the
image of I
1
 (the vitual object) from mirror M
2
 is I
3
.
The image of object O from mirror M
2
 is I
2
 and the
image of I
2
 (the virtual object) from mirror M
1
 is I
4
.
Notice that this interpretation, according to ray diagram
rules, is valid only for Fig. (A). All others are
inconsistent.
(8) (a) Angle betwen incident ray and mirror = 90º - 30º = 60º
30º
60º
60º
30°
30°
By law of reflection Ð i = Ð r
So angle of reflection Ð r = 30º.
Hence angle between mirror and reflected ray = 60º
(9) (b) As shown in figure, ray AB goes to mirror M
1
, gets
reflected and travels along BC and then gets reflected
by M
2
 and goes in CD direction. If the angle between
M
1
 and M
2
 be a, then
M
1
B
O
C
M
2
A
E
D
aa a
a
a
In  D OBC , ÐOBC and ÐOCB are equal to a
\  3a = 180º
  a = 60º
c
• O
M
2
M
1
I
1
I
3
I
2
X
Page 2


(1) (a) Here, n = 
360 360
4
90
==
q
\   n is an even number.
Thus, number of images formed = n – 1 = 3. All these
three images lie on a circle with centre at C  (The point
of intersection of mirrors M
1
 and M
2 
) and whose radius
is equal to the distance between C and object.
(2) (c) Here, [n] = 5 Þ n – 1 £ 5 £ n
\ 
360
q
 – 1 £ 5 £ 
360
q
or,  q ³ 
360
6
 or  
360
5
q£
\ 60º
 
£ q £ 72º
(3) (d) For the given q = 50º,
n = 
360
q
 = 
360
50
 = 7.2
The integer value of (7.2) is 7.
Thus number of images formed is 7.
(4) (a) The situation is illustrated in figure.
90º-i
N
1
N
2
M
2
B
O
C
a
a
q
b
q
X
A M
1
i
i
XA is the incident ray . BC is the final reflected ray . It
is given that BC is parallel to mirror M
1
. Look at the
assignment of the angles carefully . Now N
2
 is normal
to mirror M
2
. Therefore b = q
Then from D OAB
q + b  + 90º – i = 180º
or q + q + 90º – i = 180º or  i = 2q – 90º
Thus if the angle of incidence is i = 2q - 90º, then the
final reflected ray will be parallel to the first mirror.
(5) (c) Small and erect image is formed only by convex mirror.
Plane mirror from images equal to object and concave
mirror form images bigger than object.
(6) (b) The image will be formed by the plane mirror at a 30
cm behind it, while the image by convex mirror will be
formed at 10 cm behind the convex mirror.
Since for convex mirror u = – 50 cm
v = 10 cm
1 1 1 154
f 50 10 50 50
-+
= +==
-
f = 
50
4
 = 12.5 cm
Therefore the radius of curvature  of convex mirror is
r = 2 f = 25 cm
(7) (a) The image of object O from mirrror M
1
 is I
1
 and the
image of I
1
 (the vitual object) from mirror M
2
 is I
3
.
The image of object O from mirror M
2
 is I
2
 and the
image of I
2
 (the virtual object) from mirror M
1
 is I
4
.
Notice that this interpretation, according to ray diagram
rules, is valid only for Fig. (A). All others are
inconsistent.
(8) (a) Angle betwen incident ray and mirror = 90º - 30º = 60º
30º
60º
60º
30°
30°
By law of reflection Ð i = Ð r
So angle of reflection Ð r = 30º.
Hence angle between mirror and reflected ray = 60º
(9) (b) As shown in figure, ray AB goes to mirror M
1
, gets
reflected and travels along BC and then gets reflected
by M
2
 and goes in CD direction. If the angle between
M
1
 and M
2
 be a, then
M
1
B
O
C
M
2
A
E
D
aa a
a
a
In  D OBC , ÐOBC and ÐOCB are equal to a
\  3a = 180º
  a = 60º
c
• O
M
2
M
1
I
1
I
3
I
2
X
DPP/ P 49
135
(10) (a) Here, Object distance , u = – 15cm
focal length , f = – 5 cm
Object height, h
0
 = 0.2 cm
I
O
W e know , mirror formula,
1 11
v uf
+=
Þ V  =  
uf
uf -
          = 
( 15)( 5)
155
--
-+
 =  – 7.5 cm
Again, magnification,
v 7.51
m
u 152
-
=- =- =-
-
Now,  | m | = 
i
o
h
h
Þ h
i
 = | m | h
o
  = 
0.2
2
 cm = 0.1 cm
Thus, the image is formed at 7.5 cm from the pole of
the mirror and its size is 0.1 cm.
(11) (a)
Here, Object distance, u = – 30 cm
Focal length, f = + 20 cm
We know, mirror formula, 
1 11
v uf
+=
Þ v = 
uf 30 20
u f 30 20
-´
=
- --
 = + 12 cm
Again, magnification ,
m = 
v
u
-
  = 
cm 30
cm 12 -
 = 
2
5
Now , Image height  = m × object height
   = 
2
0.5
5
´ cm =  0.2 cm
Thus the image is formed behind the mirror at a distance
of 12 cm from the pole. Image height is 0.2 cm.
(12) (d) u = –50 cm , f = 25cm
1 11
25 50 v
=-+ ;
1 1 1 2 13
v 25 50 50 50
+
= + ==
v = 
50
3
 = 16.6 cm.
(13) (a) We know, 
1 12
v uR
= +
Þ v = 
uR
2uR -
2
dv (2u R).R uR.2 du
.
dt dt
(2u R)
--
=
-
      =  – 
2
R du
.
2u R dt
æö
ç÷
èø -
\  Speed of image 
dv
dt
 = 
2
R
2uR
æö
ç÷
èø -
.
du
dt
=
2
0
R
V
2u – R
æö
ç÷
èø
(14) (a) For shrot linear object du and dv represent size of
object and image respectively.
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
b
u
u+ du
2
1
We know, 
1 11
v uf
+=
Þ dv = – 
2
2
v
.du
u
Þ |db| = 
2
2
v
|du|
u
= 
2
f
.b
f – u
æö
ç÷
èø
15. (b)
1 11
.Also1
v uu
m
u f v u fv
\ =- = + Þ =+
1
u u vf
v f u fu
-
Þ- =- Þ=
-
so
f
m
fu
=
-
O P
F C I
Page 3


(1) (a) Here, n = 
360 360
4
90
==
q
\   n is an even number.
Thus, number of images formed = n – 1 = 3. All these
three images lie on a circle with centre at C  (The point
of intersection of mirrors M
1
 and M
2 
) and whose radius
is equal to the distance between C and object.
(2) (c) Here, [n] = 5 Þ n – 1 £ 5 £ n
\ 
360
q
 – 1 £ 5 £ 
360
q
or,  q ³ 
360
6
 or  
360
5
q£
\ 60º
 
£ q £ 72º
(3) (d) For the given q = 50º,
n = 
360
q
 = 
360
50
 = 7.2
The integer value of (7.2) is 7.
Thus number of images formed is 7.
(4) (a) The situation is illustrated in figure.
90º-i
N
1
N
2
M
2
B
O
C
a
a
q
b
q
X
A M
1
i
i
XA is the incident ray . BC is the final reflected ray . It
is given that BC is parallel to mirror M
1
. Look at the
assignment of the angles carefully . Now N
2
 is normal
to mirror M
2
. Therefore b = q
Then from D OAB
q + b  + 90º – i = 180º
or q + q + 90º – i = 180º or  i = 2q – 90º
Thus if the angle of incidence is i = 2q - 90º, then the
final reflected ray will be parallel to the first mirror.
(5) (c) Small and erect image is formed only by convex mirror.
Plane mirror from images equal to object and concave
mirror form images bigger than object.
(6) (b) The image will be formed by the plane mirror at a 30
cm behind it, while the image by convex mirror will be
formed at 10 cm behind the convex mirror.
Since for convex mirror u = – 50 cm
v = 10 cm
1 1 1 154
f 50 10 50 50
-+
= +==
-
f = 
50
4
 = 12.5 cm
Therefore the radius of curvature  of convex mirror is
r = 2 f = 25 cm
(7) (a) The image of object O from mirrror M
1
 is I
1
 and the
image of I
1
 (the vitual object) from mirror M
2
 is I
3
.
The image of object O from mirror M
2
 is I
2
 and the
image of I
2
 (the virtual object) from mirror M
1
 is I
4
.
Notice that this interpretation, according to ray diagram
rules, is valid only for Fig. (A). All others are
inconsistent.
(8) (a) Angle betwen incident ray and mirror = 90º - 30º = 60º
30º
60º
60º
30°
30°
By law of reflection Ð i = Ð r
So angle of reflection Ð r = 30º.
Hence angle between mirror and reflected ray = 60º
(9) (b) As shown in figure, ray AB goes to mirror M
1
, gets
reflected and travels along BC and then gets reflected
by M
2
 and goes in CD direction. If the angle between
M
1
 and M
2
 be a, then
M
1
B
O
C
M
2
A
E
D
aa a
a
a
In  D OBC , ÐOBC and ÐOCB are equal to a
\  3a = 180º
  a = 60º
c
• O
M
2
M
1
I
1
I
3
I
2
X
DPP/ P 49
135
(10) (a) Here, Object distance , u = – 15cm
focal length , f = – 5 cm
Object height, h
0
 = 0.2 cm
I
O
W e know , mirror formula,
1 11
v uf
+=
Þ V  =  
uf
uf -
          = 
( 15)( 5)
155
--
-+
 =  – 7.5 cm
Again, magnification,
v 7.51
m
u 152
-
=- =- =-
-
Now,  | m | = 
i
o
h
h
Þ h
i
 = | m | h
o
  = 
0.2
2
 cm = 0.1 cm
Thus, the image is formed at 7.5 cm from the pole of
the mirror and its size is 0.1 cm.
(11) (a)
Here, Object distance, u = – 30 cm
Focal length, f = + 20 cm
We know, mirror formula, 
1 11
v uf
+=
Þ v = 
uf 30 20
u f 30 20
-´
=
- --
 = + 12 cm
Again, magnification ,
m = 
v
u
-
  = 
cm 30
cm 12 -
 = 
2
5
Now , Image height  = m × object height
   = 
2
0.5
5
´ cm =  0.2 cm
Thus the image is formed behind the mirror at a distance
of 12 cm from the pole. Image height is 0.2 cm.
(12) (d) u = –50 cm , f = 25cm
1 11
25 50 v
=-+ ;
1 1 1 2 13
v 25 50 50 50
+
= + ==
v = 
50
3
 = 16.6 cm.
(13) (a) We know, 
1 12
v uR
= +
Þ v = 
uR
2uR -
2
dv (2u R).R uR.2 du
.
dt dt
(2u R)
--
=
-
      =  – 
2
R du
.
2u R dt
æö
ç÷
èø -
\  Speed of image 
dv
dt
 = 
2
R
2uR
æö
ç÷
èø -
.
du
dt
=
2
0
R
V
2u – R
æö
ç÷
èø
(14) (a) For shrot linear object du and dv represent size of
object and image respectively.
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
b
u
u+ du
2
1
We know, 
1 11
v uf
+=
Þ dv = – 
2
2
v
.du
u
Þ |db| = 
2
2
v
|du|
u
= 
2
f
.b
f – u
æö
ç÷
èø
15. (b)
1 11
.Also1
v uu
m
u f v u fv
\ =- = + Þ =+
1
u u vf
v f u fu
-
Þ- =- Þ=
-
so
f
m
fu
=
-
O P
F C I
DPP/ P 49
136
16. (b) Given u = – 15 cm,  f = – 10 cm,  O = 1 cm
1 11
v uf
+= , 
1 11
v fu
=- = 
11
10 15
-
--
\ v = – 30 cm
I v 30
O u 15
-
=- =-
-
 = – 2
I = – 2 × 1 = – 2 cm
Image is inverted and on the same side (real) of size 2
cm.
17. (b) As shown in the figure, when the object (O) is placed
between F and C, the image (I) is formed beyond C. It
is in this condition that when the student shifts his eyes
towards left, the image appears to the right of the object
pin.
I C O
F
Movement towards left
18. (a)
A
B
O
30º
30º
30º
60º
N
P
I
Q
Ð i = Ð r = 30º
\Ð OIQ = 60º
\ IOQ = 90º – 60º = 30º
19. (d) The image formed by a convex mirror is always virtual.
20. (d) From the ray diagram.
X
d/2
X
d/2
2L
L
A
B D
M
N
In DANM and DADB
ÐADB = ÐANM = 90°
ÐMAN = ÐBAN (laws of reflection)
Also ÐBAN = ÐABD
ÞÐMAN = ÐABD
\DANM is similar to  DADB
\
/2
2
xd
LL
= or  x = d
So, required distance  = d + d + d = 3d.
21. (a) m = –n  ;   m = 
f
fu -
–n = 
f
fu
-
--
   Þ nf + nu = –f
nu = –f –nf Þ u = 
(n 1)
n
-+
 f
22. (d) Here image can be real or virtual. If the image is real
f = –30, u = ?, m = –3
m = 
f
fu -
   Þ  –3 = 
30
30u
-
--
  ;  u = – 40 cm.
If the image is virtual
m = 
f
fu -
   Þ   3 = 
30
30u
-
--
Þ u = –20 cm.
23. (b) By keeping the incident ray is fixed, if plane mirror
rotates through an angle q reflected ray rotates through
an angle  2q
q
q
q
24. (d)
Real image
Virtual object
I
O
25. (b,c) Convex mirror and concave lens form, virtual image
for all positions of object.
 (26) (a);  (27) (d)
70 cm
700 cm 700 cm 100 cm 
O
I
2 1 21
–
–
µ µ µµ
v uR
=
, when x = 70 cm
1.5 1.2 1.5 – 1.2
–
–70 20 v
=
20 70 1.5
–1.2 20 0.3 70
v
´´
Þ=
´ +´
Page 4


(1) (a) Here, n = 
360 360
4
90
==
q
\   n is an even number.
Thus, number of images formed = n – 1 = 3. All these
three images lie on a circle with centre at C  (The point
of intersection of mirrors M
1
 and M
2 
) and whose radius
is equal to the distance between C and object.
(2) (c) Here, [n] = 5 Þ n – 1 £ 5 £ n
\ 
360
q
 – 1 £ 5 £ 
360
q
or,  q ³ 
360
6
 or  
360
5
q£
\ 60º
 
£ q £ 72º
(3) (d) For the given q = 50º,
n = 
360
q
 = 
360
50
 = 7.2
The integer value of (7.2) is 7.
Thus number of images formed is 7.
(4) (a) The situation is illustrated in figure.
90º-i
N
1
N
2
M
2
B
O
C
a
a
q
b
q
X
A M
1
i
i
XA is the incident ray . BC is the final reflected ray . It
is given that BC is parallel to mirror M
1
. Look at the
assignment of the angles carefully . Now N
2
 is normal
to mirror M
2
. Therefore b = q
Then from D OAB
q + b  + 90º – i = 180º
or q + q + 90º – i = 180º or  i = 2q – 90º
Thus if the angle of incidence is i = 2q - 90º, then the
final reflected ray will be parallel to the first mirror.
(5) (c) Small and erect image is formed only by convex mirror.
Plane mirror from images equal to object and concave
mirror form images bigger than object.
(6) (b) The image will be formed by the plane mirror at a 30
cm behind it, while the image by convex mirror will be
formed at 10 cm behind the convex mirror.
Since for convex mirror u = – 50 cm
v = 10 cm
1 1 1 154
f 50 10 50 50
-+
= +==
-
f = 
50
4
 = 12.5 cm
Therefore the radius of curvature  of convex mirror is
r = 2 f = 25 cm
(7) (a) The image of object O from mirrror M
1
 is I
1
 and the
image of I
1
 (the vitual object) from mirror M
2
 is I
3
.
The image of object O from mirror M
2
 is I
2
 and the
image of I
2
 (the virtual object) from mirror M
1
 is I
4
.
Notice that this interpretation, according to ray diagram
rules, is valid only for Fig. (A). All others are
inconsistent.
(8) (a) Angle betwen incident ray and mirror = 90º - 30º = 60º
30º
60º
60º
30°
30°
By law of reflection Ð i = Ð r
So angle of reflection Ð r = 30º.
Hence angle between mirror and reflected ray = 60º
(9) (b) As shown in figure, ray AB goes to mirror M
1
, gets
reflected and travels along BC and then gets reflected
by M
2
 and goes in CD direction. If the angle between
M
1
 and M
2
 be a, then
M
1
B
O
C
M
2
A
E
D
aa a
a
a
In  D OBC , ÐOBC and ÐOCB are equal to a
\  3a = 180º
  a = 60º
c
• O
M
2
M
1
I
1
I
3
I
2
X
DPP/ P 49
135
(10) (a) Here, Object distance , u = – 15cm
focal length , f = – 5 cm
Object height, h
0
 = 0.2 cm
I
O
W e know , mirror formula,
1 11
v uf
+=
Þ V  =  
uf
uf -
          = 
( 15)( 5)
155
--
-+
 =  – 7.5 cm
Again, magnification,
v 7.51
m
u 152
-
=- =- =-
-
Now,  | m | = 
i
o
h
h
Þ h
i
 = | m | h
o
  = 
0.2
2
 cm = 0.1 cm
Thus, the image is formed at 7.5 cm from the pole of
the mirror and its size is 0.1 cm.
(11) (a)
Here, Object distance, u = – 30 cm
Focal length, f = + 20 cm
We know, mirror formula, 
1 11
v uf
+=
Þ v = 
uf 30 20
u f 30 20
-´
=
- --
 = + 12 cm
Again, magnification ,
m = 
v
u
-
  = 
cm 30
cm 12 -
 = 
2
5
Now , Image height  = m × object height
   = 
2
0.5
5
´ cm =  0.2 cm
Thus the image is formed behind the mirror at a distance
of 12 cm from the pole. Image height is 0.2 cm.
(12) (d) u = –50 cm , f = 25cm
1 11
25 50 v
=-+ ;
1 1 1 2 13
v 25 50 50 50
+
= + ==
v = 
50
3
 = 16.6 cm.
(13) (a) We know, 
1 12
v uR
= +
Þ v = 
uR
2uR -
2
dv (2u R).R uR.2 du
.
dt dt
(2u R)
--
=
-
      =  – 
2
R du
.
2u R dt
æö
ç÷
èø -
\  Speed of image 
dv
dt
 = 
2
R
2uR
æö
ç÷
èø -
.
du
dt
=
2
0
R
V
2u – R
æö
ç÷
èø
(14) (a) For shrot linear object du and dv represent size of
object and image respectively.
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
b
u
u+ du
2
1
We know, 
1 11
v uf
+=
Þ dv = – 
2
2
v
.du
u
Þ |db| = 
2
2
v
|du|
u
= 
2
f
.b
f – u
æö
ç÷
èø
15. (b)
1 11
.Also1
v uu
m
u f v u fv
\ =- = + Þ =+
1
u u vf
v f u fu
-
Þ- =- Þ=
-
so
f
m
fu
=
-
O P
F C I
DPP/ P 49
136
16. (b) Given u = – 15 cm,  f = – 10 cm,  O = 1 cm
1 11
v uf
+= , 
1 11
v fu
=- = 
11
10 15
-
--
\ v = – 30 cm
I v 30
O u 15
-
=- =-
-
 = – 2
I = – 2 × 1 = – 2 cm
Image is inverted and on the same side (real) of size 2
cm.
17. (b) As shown in the figure, when the object (O) is placed
between F and C, the image (I) is formed beyond C. It
is in this condition that when the student shifts his eyes
towards left, the image appears to the right of the object
pin.
I C O
F
Movement towards left
18. (a)
A
B
O
30º
30º
30º
60º
N
P
I
Q
Ð i = Ð r = 30º
\Ð OIQ = 60º
\ IOQ = 90º – 60º = 30º
19. (d) The image formed by a convex mirror is always virtual.
20. (d) From the ray diagram.
X
d/2
X
d/2
2L
L
A
B D
M
N
In DANM and DADB
ÐADB = ÐANM = 90°
ÐMAN = ÐBAN (laws of reflection)
Also ÐBAN = ÐABD
ÞÐMAN = ÐABD
\DANM is similar to  DADB
\
/2
2
xd
LL
= or  x = d
So, required distance  = d + d + d = 3d.
21. (a) m = –n  ;   m = 
f
fu -
–n = 
f
fu
-
--
   Þ nf + nu = –f
nu = –f –nf Þ u = 
(n 1)
n
-+
 f
22. (d) Here image can be real or virtual. If the image is real
f = –30, u = ?, m = –3
m = 
f
fu -
   Þ  –3 = 
30
30u
-
--
  ;  u = – 40 cm.
If the image is virtual
m = 
f
fu -
   Þ   3 = 
30
30u
-
--
Þ u = –20 cm.
23. (b) By keeping the incident ray is fixed, if plane mirror
rotates through an angle q reflected ray rotates through
an angle  2q
q
q
q
24. (d)
Real image
Virtual object
I
O
25. (b,c) Convex mirror and concave lens form, virtual image
for all positions of object.
 (26) (a);  (27) (d)
70 cm
700 cm 700 cm 100 cm 
O
I
2 1 21
–
–
µ µ µµ
v uR
=
, when x = 70 cm
1.5 1.2 1.5 – 1.2
–
–70 20 v
=
20 70 1.5
–1.2 20 0.3 70
v
´´
Þ=
´ +´
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137
Þ v = – 700 cm
2 1 21
–
–
µ µ µµ
v uR
=
1.2 1.5 1.2 –1.5
–
900 –20 v
=
900 200 1.2
1.5 200 – 900 3
v
´´
Þ=
´´
Þ v = – 90 cm
Similarly , for x = 80 cm
v = 80 cm
and for x = 90 cm
v = 70 cm
28. (d)
29. (d) When an object is placed between two plane parallel
mirrors, then infinite number of images are formed.
Images are formed due to multiple reflections. At each
reflection, a part of light energy is absorbed. Therefore,
distant images get fainter.
30. (d) The size of the mirror does not affect the nature of the
image except that a bigger mirror forms a brighter
image.