Chapter 26 - Reflection of Light (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Chapter 26 - Reflection of Light (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


8. Here it is given that height of the boy
HF = 1.5 m
Length of mirror = = AB 0.75 m
The ray diagram is shown in above figure.
H is the Head of the boy and F is the feet. It
also shows the paths of the rays that leaves
the head of the man enter his eyes (E). After
reflection from the mirror at point A, and
the rays that leave his feet and enter his
eyes after reflected at point B.
From figure CE HE = =
1
2
0.05 m
   CF HF HC HF CE = - = =
          = - = 1.50 0.05 1.45 m
The distance of the bottom edge of mirror
above the floor is
BP KF CF KC CD AB = = - = -
       = - = 1.45 0.75 0.7 m
But according to question BD = 0.8 m (given)
which is greater than 0.7 m, the height
required to see full image. Hence the boy
cannot see his feet.
Option (c) is correct
9. Since the image is magnified hence mirror is 
concave mirror.
Here m
v
u
= - = 3 Þ v u = - 3
Þ | | | | v u u = - = 3 3 
but       | | v u - = 80
Þ | | 3 80 u u - = Þ  u = 40 cm'
Using mirror formula, we get
     
1 1 1
v u f
+ =
- - =
1
3
1 1
u u f
Þ f
u
=
-3
4
Þ f =
- ´
= -
3 40
2
30  cm
Mirror is concave and focal length is 30 cm.
Correct option is (a).
10. Here m
n
v
u
= + = -
1
Þ v
u
n
= -
From mirror formula 
1 1 1
f v u
= + , 
we get,
1 1 1
f u n u
=
-
+
( / )
Þ u n f = - - ( ) 1
Hence the correct option is (d).
11. Differentiating mirror formula, we get
dv
dt
v
u
du
dt
=
2
2
      [ Q here 
du
dt
 is –ve]
Using mirror formula
1 1 1
v u f
+ = ,
we get        
1 1 1
v f u
= -
Here u = - 60 cm,  f = - 24 cm
Putting these we get,    v = 40 cm
Hence,      
dv
dt
= ´ =
40
60
9 4
2
2
 cm/s
Hence the speed of the image is 4 cm/s
away from the mirror. 
Hence correct option is (c).
12. The wrong statement is (d)
 9
B
Mirror
A
D F
K
E
H
C
1.5m
0.1 m
Page 2


8. Here it is given that height of the boy
HF = 1.5 m
Length of mirror = = AB 0.75 m
The ray diagram is shown in above figure.
H is the Head of the boy and F is the feet. It
also shows the paths of the rays that leaves
the head of the man enter his eyes (E). After
reflection from the mirror at point A, and
the rays that leave his feet and enter his
eyes after reflected at point B.
From figure CE HE = =
1
2
0.05 m
   CF HF HC HF CE = - = =
          = - = 1.50 0.05 1.45 m
The distance of the bottom edge of mirror
above the floor is
BP KF CF KC CD AB = = - = -
       = - = 1.45 0.75 0.7 m
But according to question BD = 0.8 m (given)
which is greater than 0.7 m, the height
required to see full image. Hence the boy
cannot see his feet.
Option (c) is correct
9. Since the image is magnified hence mirror is 
concave mirror.
Here m
v
u
= - = 3 Þ v u = - 3
Þ | | | | v u u = - = 3 3 
but       | | v u - = 80
Þ | | 3 80 u u - = Þ  u = 40 cm'
Using mirror formula, we get
     
1 1 1
v u f
+ =
- - =
1
3
1 1
u u f
Þ f
u
=
-3
4
Þ f =
- ´
= -
3 40
2
30  cm
Mirror is concave and focal length is 30 cm.
Correct option is (a).
10. Here m
n
v
u
= + = -
1
Þ v
u
n
= -
From mirror formula 
1 1 1
f v u
= + , 
we get,
1 1 1
f u n u
=
-
+
( / )
Þ u n f = - - ( ) 1
Hence the correct option is (d).
11. Differentiating mirror formula, we get
dv
dt
v
u
du
dt
=
2
2
      [ Q here 
du
dt
 is –ve]
Using mirror formula
1 1 1
v u f
+ = ,
we get        
1 1 1
v f u
= -
Here u = - 60 cm,  f = - 24 cm
Putting these we get,    v = 40 cm
Hence,      
dv
dt
= ´ =
40
60
9 4
2
2
 cm/s
Hence the speed of the image is 4 cm/s
away from the mirror. 
Hence correct option is (c).
12. The wrong statement is (d)
 9
B
Mirror
A
D F
K
E
H
C
1.5m
0.1 m
13. Let  v
m
is the speed of mirror, v
p
 is the speed
of particle and v
p
 is the speed of the
observer, then speed of the image measured
by observer is given by
v v v v
op m p o
= + - 2 [ ]
Þ v
op
= + - 2 10 4 2 [ ]
      = - = 28 2 26 cm/s
Hence correct option is (d).
¢ Assertion and Reason
1. Assertion is wrong since when a virtual
object is placed at a distance less than the
focal length its real image is formed.
Hence answer is (d).
2. Using mirror formula 
1 1 1
v u f
+ = we get 
1 1
20
1
20 v
- = Þ v=10 cm
ie image is virtual exect and since m
v
u
= =
1
2
.
Hence image is diminished, thus assertion
is true.
If u = + 20 cm for virtual object v = ¥ hence
reason is true but reason is not correct
explanation of assertion. Hence answer is (b).
3. Using mirror formula 
1 1 1
v u f
+ = we get 
1 1 1
v f u
= -
If u is front of mirror u is negative and f is
negative for concave mirror.
Þ 
1 1 1
v f u
= - +  Þ v
uf
f u
=
-
Þ u f ®  Þ v ® ¥
Hence assertion is true also in refractive
image and object moves in opposite
direction. Hence both assertion and reason
are true and reason correctly explain the
assertion Correct answer is (a).
4. Real view mirror of vehicles is convex
mirror, hence assertion is true.
It never makes real image of real object
reason is also true but convex mirror is used
because since its field of view is greatest.
Hence both assertion and reason are true
but reason is not correct explanation of
assertion. Correct answer is (b).
5. Since m = - 2 hence it is definitely a concave
mirror since only concave mirror form
magnified image. Since concave mirror form
only real image of real object hence reason is
also true. Hence it may true but when object
is placed between C F m and , < 1.
Hence correct answer be (a) or (b).
6.
Hence assertion is true.
For normal incidence i = 0 hence d = ° 180 .
hence assertion is true but reason is false.
hence correct option is (c).
7.
10
i
i
Reflected ray
d =  180°– 2i
Incident ray
i
i
r
M
2
M
1
r
Page 3


8. Here it is given that height of the boy
HF = 1.5 m
Length of mirror = = AB 0.75 m
The ray diagram is shown in above figure.
H is the Head of the boy and F is the feet. It
also shows the paths of the rays that leaves
the head of the man enter his eyes (E). After
reflection from the mirror at point A, and
the rays that leave his feet and enter his
eyes after reflected at point B.
From figure CE HE = =
1
2
0.05 m
   CF HF HC HF CE = - = =
          = - = 1.50 0.05 1.45 m
The distance of the bottom edge of mirror
above the floor is
BP KF CF KC CD AB = = - = -
       = - = 1.45 0.75 0.7 m
But according to question BD = 0.8 m (given)
which is greater than 0.7 m, the height
required to see full image. Hence the boy
cannot see his feet.
Option (c) is correct
9. Since the image is magnified hence mirror is 
concave mirror.
Here m
v
u
= - = 3 Þ v u = - 3
Þ | | | | v u u = - = 3 3 
but       | | v u - = 80
Þ | | 3 80 u u - = Þ  u = 40 cm'
Using mirror formula, we get
     
1 1 1
v u f
+ =
- - =
1
3
1 1
u u f
Þ f
u
=
-3
4
Þ f =
- ´
= -
3 40
2
30  cm
Mirror is concave and focal length is 30 cm.
Correct option is (a).
10. Here m
n
v
u
= + = -
1
Þ v
u
n
= -
From mirror formula 
1 1 1
f v u
= + , 
we get,
1 1 1
f u n u
=
-
+
( / )
Þ u n f = - - ( ) 1
Hence the correct option is (d).
11. Differentiating mirror formula, we get
dv
dt
v
u
du
dt
=
2
2
      [ Q here 
du
dt
 is –ve]
Using mirror formula
1 1 1
v u f
+ = ,
we get        
1 1 1
v f u
= -
Here u = - 60 cm,  f = - 24 cm
Putting these we get,    v = 40 cm
Hence,      
dv
dt
= ´ =
40
60
9 4
2
2
 cm/s
Hence the speed of the image is 4 cm/s
away from the mirror. 
Hence correct option is (c).
12. The wrong statement is (d)
 9
B
Mirror
A
D F
K
E
H
C
1.5m
0.1 m
13. Let  v
m
is the speed of mirror, v
p
 is the speed
of particle and v
p
 is the speed of the
observer, then speed of the image measured
by observer is given by
v v v v
op m p o
= + - 2 [ ]
Þ v
op
= + - 2 10 4 2 [ ]
      = - = 28 2 26 cm/s
Hence correct option is (d).
¢ Assertion and Reason
1. Assertion is wrong since when a virtual
object is placed at a distance less than the
focal length its real image is formed.
Hence answer is (d).
2. Using mirror formula 
1 1 1
v u f
+ = we get 
1 1
20
1
20 v
- = Þ v=10 cm
ie image is virtual exect and since m
v
u
= =
1
2
.
Hence image is diminished, thus assertion
is true.
If u = + 20 cm for virtual object v = ¥ hence
reason is true but reason is not correct
explanation of assertion. Hence answer is (b).
3. Using mirror formula 
1 1 1
v u f
+ = we get 
1 1 1
v f u
= -
If u is front of mirror u is negative and f is
negative for concave mirror.
Þ 
1 1 1
v f u
= - +  Þ v
uf
f u
=
-
Þ u f ®  Þ v ® ¥
Hence assertion is true also in refractive
image and object moves in opposite
direction. Hence both assertion and reason
are true and reason correctly explain the
assertion Correct answer is (a).
4. Real view mirror of vehicles is convex
mirror, hence assertion is true.
It never makes real image of real object
reason is also true but convex mirror is used
because since its field of view is greatest.
Hence both assertion and reason are true
but reason is not correct explanation of
assertion. Correct answer is (b).
5. Since m = - 2 hence it is definitely a concave
mirror since only concave mirror form
magnified image. Since concave mirror form
only real image of real object hence reason is
also true. Hence it may true but when object
is placed between C F m and , < 1.
Hence correct answer be (a) or (b).
6.
Hence assertion is true.
For normal incidence i = 0 hence d = ° 180 .
hence assertion is true but reason is false.
hence correct option is (c).
7.
10
i
i
Reflected ray
d =  180°– 2i
Incident ray
i
i
r
M
2
M
1
r
Deviation produced by M i
1
180 2 = ° -
Deviation produced by M r
2
180 2 = ° -
Total deviation produced = ° - + 360 2( ) i r
But from figure i r + = ° 90 , hence deviation 
= ° 180 for any value of i.
Hence assertion is true but reason is false.
Correct option is (c).
8. The correct option is (b).
9. The correct option is (a, b).
10. The correct option is (b).
¢ Objective Questions (Level 2)
1. v A
max
= w
Þ v
k
m
A
max
=
Q    w
2
=
k
m
 for SHM
Maximum speed of insect relative to its
image
= ^ = ° 2 2 60 v v
max max
sin
= A
k
m
3
Hence correct option is (c).
2. au g
n
= ¯
Height = x
Let after time t paperndicular distance
between mirror and source is x¢ we have
from figure
    AB AM MB SM SA MB = + = - +
but  SM MB =
   AB MB SA x SA = - = ¢ f - 2 2 tan
          = ¢ f - ¢ 2 2 x x tan tan q
Þ   AB x = ¢ f - 2 [tan tan ] q
         = ¢
¢
-
¢
é
ë
ê
ù
û
ú
2x
SM
x
SN
x
 = - 2 [ ] SM SN
Þ AB L = ´ 2 , 
where SM SN L - = = Length of mirror
Þ 
d
dt
AB
d
dt
L [ ] ( ) = = 2 0
Q Length of mirror is constant.
Hence the correct option is (d).
3. Here u = - 10 cm and v = - 20 cm
Using mirror formula
1 1 1
u v f
+ =  we get = - =
dv
v
du
u
2 2
0
Þ 
dv
du
v
u
= - = - = -
2
2
2
2
20
10
4
Þ dv du = - 4
Þ dv = - - - 4 ( ) 0.1 , here du = - 0.1
Þ dv = 0.4 cm, 
ie, 0.4 cm away from the mirror.
Hence the correct option is (a).
4. The first and second images are shown in
figure but according to question
 11
x
q q q q
S N A M B
L
q = 60°
M
(3–x)
L
1
L
2
3–x x
6 – x
I II
x
Page 4


8. Here it is given that height of the boy
HF = 1.5 m
Length of mirror = = AB 0.75 m
The ray diagram is shown in above figure.
H is the Head of the boy and F is the feet. It
also shows the paths of the rays that leaves
the head of the man enter his eyes (E). After
reflection from the mirror at point A, and
the rays that leave his feet and enter his
eyes after reflected at point B.
From figure CE HE = =
1
2
0.05 m
   CF HF HC HF CE = - = =
          = - = 1.50 0.05 1.45 m
The distance of the bottom edge of mirror
above the floor is
BP KF CF KC CD AB = = - = -
       = - = 1.45 0.75 0.7 m
But according to question BD = 0.8 m (given)
which is greater than 0.7 m, the height
required to see full image. Hence the boy
cannot see his feet.
Option (c) is correct
9. Since the image is magnified hence mirror is 
concave mirror.
Here m
v
u
= - = 3 Þ v u = - 3
Þ | | | | v u u = - = 3 3 
but       | | v u - = 80
Þ | | 3 80 u u - = Þ  u = 40 cm'
Using mirror formula, we get
     
1 1 1
v u f
+ =
- - =
1
3
1 1
u u f
Þ f
u
=
-3
4
Þ f =
- ´
= -
3 40
2
30  cm
Mirror is concave and focal length is 30 cm.
Correct option is (a).
10. Here m
n
v
u
= + = -
1
Þ v
u
n
= -
From mirror formula 
1 1 1
f v u
= + , 
we get,
1 1 1
f u n u
=
-
+
( / )
Þ u n f = - - ( ) 1
Hence the correct option is (d).
11. Differentiating mirror formula, we get
dv
dt
v
u
du
dt
=
2
2
      [ Q here 
du
dt
 is –ve]
Using mirror formula
1 1 1
v u f
+ = ,
we get        
1 1 1
v f u
= -
Here u = - 60 cm,  f = - 24 cm
Putting these we get,    v = 40 cm
Hence,      
dv
dt
= ´ =
40
60
9 4
2
2
 cm/s
Hence the speed of the image is 4 cm/s
away from the mirror. 
Hence correct option is (c).
12. The wrong statement is (d)
 9
B
Mirror
A
D F
K
E
H
C
1.5m
0.1 m
13. Let  v
m
is the speed of mirror, v
p
 is the speed
of particle and v
p
 is the speed of the
observer, then speed of the image measured
by observer is given by
v v v v
op m p o
= + - 2 [ ]
Þ v
op
= + - 2 10 4 2 [ ]
      = - = 28 2 26 cm/s
Hence correct option is (d).
¢ Assertion and Reason
1. Assertion is wrong since when a virtual
object is placed at a distance less than the
focal length its real image is formed.
Hence answer is (d).
2. Using mirror formula 
1 1 1
v u f
+ = we get 
1 1
20
1
20 v
- = Þ v=10 cm
ie image is virtual exect and since m
v
u
= =
1
2
.
Hence image is diminished, thus assertion
is true.
If u = + 20 cm for virtual object v = ¥ hence
reason is true but reason is not correct
explanation of assertion. Hence answer is (b).
3. Using mirror formula 
1 1 1
v u f
+ = we get 
1 1 1
v f u
= -
If u is front of mirror u is negative and f is
negative for concave mirror.
Þ 
1 1 1
v f u
= - +  Þ v
uf
f u
=
-
Þ u f ®  Þ v ® ¥
Hence assertion is true also in refractive
image and object moves in opposite
direction. Hence both assertion and reason
are true and reason correctly explain the
assertion Correct answer is (a).
4. Real view mirror of vehicles is convex
mirror, hence assertion is true.
It never makes real image of real object
reason is also true but convex mirror is used
because since its field of view is greatest.
Hence both assertion and reason are true
but reason is not correct explanation of
assertion. Correct answer is (b).
5. Since m = - 2 hence it is definitely a concave
mirror since only concave mirror form
magnified image. Since concave mirror form
only real image of real object hence reason is
also true. Hence it may true but when object
is placed between C F m and , < 1.
Hence correct answer be (a) or (b).
6.
Hence assertion is true.
For normal incidence i = 0 hence d = ° 180 .
hence assertion is true but reason is false.
hence correct option is (c).
7.
10
i
i
Reflected ray
d =  180°– 2i
Incident ray
i
i
r
M
2
M
1
r
Deviation produced by M i
1
180 2 = ° -
Deviation produced by M r
2
180 2 = ° -
Total deviation produced = ° - + 360 2( ) i r
But from figure i r + = ° 90 , hence deviation 
= ° 180 for any value of i.
Hence assertion is true but reason is false.
Correct option is (c).
8. The correct option is (b).
9. The correct option is (a, b).
10. The correct option is (b).
¢ Objective Questions (Level 2)
1. v A
max
= w
Þ v
k
m
A
max
=
Q    w
2
=
k
m
 for SHM
Maximum speed of insect relative to its
image
= ^ = ° 2 2 60 v v
max max
sin
= A
k
m
3
Hence correct option is (c).
2. au g
n
= ¯
Height = x
Let after time t paperndicular distance
between mirror and source is x¢ we have
from figure
    AB AM MB SM SA MB = + = - +
but  SM MB =
   AB MB SA x SA = - = ¢ f - 2 2 tan
          = ¢ f - ¢ 2 2 x x tan tan q
Þ   AB x = ¢ f - 2 [tan tan ] q
         = ¢
¢
-
¢
é
ë
ê
ù
û
ú
2x
SM
x
SN
x
 = - 2 [ ] SM SN
Þ AB L = ´ 2 , 
where SM SN L - = = Length of mirror
Þ 
d
dt
AB
d
dt
L [ ] ( ) = = 2 0
Q Length of mirror is constant.
Hence the correct option is (d).
3. Here u = - 10 cm and v = - 20 cm
Using mirror formula
1 1 1
u v f
+ =  we get = - =
dv
v
du
u
2 2
0
Þ 
dv
du
v
u
= - = - = -
2
2
2
2
20
10
4
Þ dv du = - 4
Þ dv = - - - 4 ( ) 0.1 , here du = - 0.1
Þ dv = 0.4 cm, 
ie, 0.4 cm away from the mirror.
Hence the correct option is (a).
4. The first and second images are shown in
figure but according to question
 11
x
q q q q
S N A M B
L
q = 60°
M
(3–x)
L
1
L
2
3–x x
6 – x
I II
x
( ) 6 4 - - = x x
Þ  2 2 = x   Þ   x = 1 m
Hence the correct option is (c).
5. For vertical part - + = -
1
20
1 1
5 v
Þ v = -
20
3
 
| |
/
m
v
u
v
=
½
½
½
½
=
½
½
½
½
½
½
20 3
20
 =
1
3
 
Þ L
v
=
10
3
 cm
For horizontal part first end is at C hence
its image is also at C ie at v = - 10 cm, for
other end
-
-
+ = -
1
20
1 1
5 v
 Þ v = -
20
3
Þ | | v =
20
3
L v u
H
= - = -
½
½
½
½
= | |
20
3
10
10
3
Þ L
H
=
10
3
The ratio L L
V H
: : = 1 1.
Hence correct option is (c)  1 : 1.
6. Here u = - 15 cm, f = - 10 cm
Using 
1 1 1
v u f
+ =
We get, v = - 30 cm
We have m
v
m
v v
u u
= =
-
-
( )
2 1
2 1
     = =
dv
du
2
Þ dv du = 2 ,
AB du = = 4 mm
Þ dv = ´ 2 4  mm  Þ  dv = 8 mm
Hence the correct option is (c).
7. If the mirror is rotated by an angle q in
anticlock, wise direction about an axis ¥ ^ to 
the plane mirror, the new angle of incidence
becomes i - q and angle of reflection also 
i-2q.
According to problem
i i + - = ° 2 45 q
2 45 2 i = ° + q = ° + ´ ° = ° 45 2 20 85
But angle of incidence = angle of reflection.
Hence the angle between origial incident
and reflected ray was 85°. Similarly is the
mirror is rotated clockwise the angle became 
5°.
Hence correct option is (c) 85° or 5°.
8. The person see his hair if the incident ray
statics from point A after reflected by mirror 
reach his eyes. Let O is point at minimum at
a distance x below the point A.
We have 2 60 x = cm Þ x = 3 cm
The distance of O from P is 
= - = 170 3 167 cm
Hence correct option is (a).
12
A B
2 mm 2 mm
F C
10 cm
P
20 cm
20 cm
q
q
x
x
Person 170 cm
B
A
O
P
164
E
Page 5


8. Here it is given that height of the boy
HF = 1.5 m
Length of mirror = = AB 0.75 m
The ray diagram is shown in above figure.
H is the Head of the boy and F is the feet. It
also shows the paths of the rays that leaves
the head of the man enter his eyes (E). After
reflection from the mirror at point A, and
the rays that leave his feet and enter his
eyes after reflected at point B.
From figure CE HE = =
1
2
0.05 m
   CF HF HC HF CE = - = =
          = - = 1.50 0.05 1.45 m
The distance of the bottom edge of mirror
above the floor is
BP KF CF KC CD AB = = - = -
       = - = 1.45 0.75 0.7 m
But according to question BD = 0.8 m (given)
which is greater than 0.7 m, the height
required to see full image. Hence the boy
cannot see his feet.
Option (c) is correct
9. Since the image is magnified hence mirror is 
concave mirror.
Here m
v
u
= - = 3 Þ v u = - 3
Þ | | | | v u u = - = 3 3 
but       | | v u - = 80
Þ | | 3 80 u u - = Þ  u = 40 cm'
Using mirror formula, we get
     
1 1 1
v u f
+ =
- - =
1
3
1 1
u u f
Þ f
u
=
-3
4
Þ f =
- ´
= -
3 40
2
30  cm
Mirror is concave and focal length is 30 cm.
Correct option is (a).
10. Here m
n
v
u
= + = -
1
Þ v
u
n
= -
From mirror formula 
1 1 1
f v u
= + , 
we get,
1 1 1
f u n u
=
-
+
( / )
Þ u n f = - - ( ) 1
Hence the correct option is (d).
11. Differentiating mirror formula, we get
dv
dt
v
u
du
dt
=
2
2
      [ Q here 
du
dt
 is –ve]
Using mirror formula
1 1 1
v u f
+ = ,
we get        
1 1 1
v f u
= -
Here u = - 60 cm,  f = - 24 cm
Putting these we get,    v = 40 cm
Hence,      
dv
dt
= ´ =
40
60
9 4
2
2
 cm/s
Hence the speed of the image is 4 cm/s
away from the mirror. 
Hence correct option is (c).
12. The wrong statement is (d)
 9
B
Mirror
A
D F
K
E
H
C
1.5m
0.1 m
13. Let  v
m
is the speed of mirror, v
p
 is the speed
of particle and v
p
 is the speed of the
observer, then speed of the image measured
by observer is given by
v v v v
op m p o
= + - 2 [ ]
Þ v
op
= + - 2 10 4 2 [ ]
      = - = 28 2 26 cm/s
Hence correct option is (d).
¢ Assertion and Reason
1. Assertion is wrong since when a virtual
object is placed at a distance less than the
focal length its real image is formed.
Hence answer is (d).
2. Using mirror formula 
1 1 1
v u f
+ = we get 
1 1
20
1
20 v
- = Þ v=10 cm
ie image is virtual exect and since m
v
u
= =
1
2
.
Hence image is diminished, thus assertion
is true.
If u = + 20 cm for virtual object v = ¥ hence
reason is true but reason is not correct
explanation of assertion. Hence answer is (b).
3. Using mirror formula 
1 1 1
v u f
+ = we get 
1 1 1
v f u
= -
If u is front of mirror u is negative and f is
negative for concave mirror.
Þ 
1 1 1
v f u
= - +  Þ v
uf
f u
=
-
Þ u f ®  Þ v ® ¥
Hence assertion is true also in refractive
image and object moves in opposite
direction. Hence both assertion and reason
are true and reason correctly explain the
assertion Correct answer is (a).
4. Real view mirror of vehicles is convex
mirror, hence assertion is true.
It never makes real image of real object
reason is also true but convex mirror is used
because since its field of view is greatest.
Hence both assertion and reason are true
but reason is not correct explanation of
assertion. Correct answer is (b).
5. Since m = - 2 hence it is definitely a concave
mirror since only concave mirror form
magnified image. Since concave mirror form
only real image of real object hence reason is
also true. Hence it may true but when object
is placed between C F m and , < 1.
Hence correct answer be (a) or (b).
6.
Hence assertion is true.
For normal incidence i = 0 hence d = ° 180 .
hence assertion is true but reason is false.
hence correct option is (c).
7.
10
i
i
Reflected ray
d =  180°– 2i
Incident ray
i
i
r
M
2
M
1
r
Deviation produced by M i
1
180 2 = ° -
Deviation produced by M r
2
180 2 = ° -
Total deviation produced = ° - + 360 2( ) i r
But from figure i r + = ° 90 , hence deviation 
= ° 180 for any value of i.
Hence assertion is true but reason is false.
Correct option is (c).
8. The correct option is (b).
9. The correct option is (a, b).
10. The correct option is (b).
¢ Objective Questions (Level 2)
1. v A
max
= w
Þ v
k
m
A
max
=
Q    w
2
=
k
m
 for SHM
Maximum speed of insect relative to its
image
= ^ = ° 2 2 60 v v
max max
sin
= A
k
m
3
Hence correct option is (c).
2. au g
n
= ¯
Height = x
Let after time t paperndicular distance
between mirror and source is x¢ we have
from figure
    AB AM MB SM SA MB = + = - +
but  SM MB =
   AB MB SA x SA = - = ¢ f - 2 2 tan
          = ¢ f - ¢ 2 2 x x tan tan q
Þ   AB x = ¢ f - 2 [tan tan ] q
         = ¢
¢
-
¢
é
ë
ê
ù
û
ú
2x
SM
x
SN
x
 = - 2 [ ] SM SN
Þ AB L = ´ 2 , 
where SM SN L - = = Length of mirror
Þ 
d
dt
AB
d
dt
L [ ] ( ) = = 2 0
Q Length of mirror is constant.
Hence the correct option is (d).
3. Here u = - 10 cm and v = - 20 cm
Using mirror formula
1 1 1
u v f
+ =  we get = - =
dv
v
du
u
2 2
0
Þ 
dv
du
v
u
= - = - = -
2
2
2
2
20
10
4
Þ dv du = - 4
Þ dv = - - - 4 ( ) 0.1 , here du = - 0.1
Þ dv = 0.4 cm, 
ie, 0.4 cm away from the mirror.
Hence the correct option is (a).
4. The first and second images are shown in
figure but according to question
 11
x
q q q q
S N A M B
L
q = 60°
M
(3–x)
L
1
L
2
3–x x
6 – x
I II
x
( ) 6 4 - - = x x
Þ  2 2 = x   Þ   x = 1 m
Hence the correct option is (c).
5. For vertical part - + = -
1
20
1 1
5 v
Þ v = -
20
3
 
| |
/
m
v
u
v
=
½
½
½
½
=
½
½
½
½
½
½
20 3
20
 =
1
3
 
Þ L
v
=
10
3
 cm
For horizontal part first end is at C hence
its image is also at C ie at v = - 10 cm, for
other end
-
-
+ = -
1
20
1 1
5 v
 Þ v = -
20
3
Þ | | v =
20
3
L v u
H
= - = -
½
½
½
½
= | |
20
3
10
10
3
Þ L
H
=
10
3
The ratio L L
V H
: : = 1 1.
Hence correct option is (c)  1 : 1.
6. Here u = - 15 cm, f = - 10 cm
Using 
1 1 1
v u f
+ =
We get, v = - 30 cm
We have m
v
m
v v
u u
= =
-
-
( )
2 1
2 1
     = =
dv
du
2
Þ dv du = 2 ,
AB du = = 4 mm
Þ dv = ´ 2 4  mm  Þ  dv = 8 mm
Hence the correct option is (c).
7. If the mirror is rotated by an angle q in
anticlock, wise direction about an axis ¥ ^ to 
the plane mirror, the new angle of incidence
becomes i - q and angle of reflection also 
i-2q.
According to problem
i i + - = ° 2 45 q
2 45 2 i = ° + q = ° + ´ ° = ° 45 2 20 85
But angle of incidence = angle of reflection.
Hence the angle between origial incident
and reflected ray was 85°. Similarly is the
mirror is rotated clockwise the angle became 
5°.
Hence correct option is (c) 85° or 5°.
8. The person see his hair if the incident ray
statics from point A after reflected by mirror 
reach his eyes. Let O is point at minimum at
a distance x below the point A.
We have 2 60 x = cm Þ x = 3 cm
The distance of O from P is 
= - = 170 3 167 cm
Hence correct option is (a).
12
A B
2 mm 2 mm
F C
10 cm
P
20 cm
20 cm
q
q
x
x
Person 170 cm
B
A
O
P
164
E
9. Acceleration of block 
a
mg
m m
AB
=
+
3
3
 =
3
4
g
Acceleration of block CD :
a
mg
m m
g
CD
=
+
=
2
2
2
3
Since the accelerations are in opposite
directions relative acceleration of one image
with respect to other is given by
a a
g g g
AB CD
+ = + =
3
4
2
13
17
12
Hence the correct option is (c).
10. Here 
BD
0.2
= ° tan 30
Þ      BD = ´ 0.2
1
3
No. of reflections = =
2 3
3
30
0.2 /
Hence, the correct option is (b).
11. Resolving velocity along parallel to mirror
and perpendicular to mirror, we get
v v
| |
sin = ° 37 and v v
^
= ° cos 37
From figure, we get
v v v
x
= ° ° + ° ° cos sin sin cos 37 37 37 37
= ° ° 2 37 37 vcos sin
v
x
= ´ ´ ´ = = 2 5
4
5
3
5
24
5
4.8
v v
y
= ° ´ ° cos cos 37 37
            - ° ´ ° v sin sin 37 37
v v
y
= ° - ° [cos sin ]
2 2
37 37
     = ´ +
æ
è
ç
ö
ø
÷
-
æ
è
ç
ö
ø
÷
5
4
5
3
5
4
5
3
5
     = ´ ´ = = 5
7
5
1
5
7
5
1.4
Hence velocity of image is given by
v i j
®
= + v v
x y
^ ^
Þ v i j
®
= + 4.8 1.4
^ ^
Hence the correct option is (c).
12. Since elevator start falling freely, the
relative acceleration of the particle in
elevator frame = - = g g 0
Hence, in elevator frame path of the
particle is a straight line.
The vertical component of velocity is 
u sin 45 2
1
2
1 ° = ´ = m/s
The separation between mirror and
particle in 0.5 s is
y v t
y
= = ´ = 1 0.5 0.5 m
 13
37°
37°
37°
v cos 37°
v sin 37°
y
x
0.2 m
B
D
30°
30°
2 3
B D
A
O
3 m
2 m
y
x
C
M N
q = 45°
u = 2 m/s
Read More
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