DC Pandey Solutions: Refraction of Light- 3 Notes | Study DC Pandey Solutions for NEET Physics - NEET

NEET: DC Pandey Solutions: Refraction of Light- 3 Notes | Study DC Pandey Solutions for NEET Physics - NEET

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 Page 1


8. F o cal length of com bi na tion
1 1 1 1 1
1 2
F F F f f
= + = -
convex convex
if f f
1 2
>
Þ
1 1
1 2
f f
<
Þ F = - negative
Hence assertion is true. Since power is a
measure of converging or divergence of a
lens. Hence reason is not true. Correct
option is (c).
9. Since glass slab pro duced a net shift. Hence
v is in creased. Thus mag ni fied im age is
ob tained but im age may be real or vir tual
de pend ing on the po si tion of slab.
Correct option is (b)
10. In this case im age dis tance of O O
1 2
and are
same from the lens. 
Q  
1 1 1
v u f
- = and reason is trure.
Hence correct option is (d).
11. As ser tion is false since only ray energe if
refractve in dex of the col our less than the
prism and an gle of in ci dence is less than
crit i cal an gle but rea son is true. Cor rect
op tion is (a).
12. If two ob ject is placed be tween pole and fo cus 
im age is real hence as ser tion is true. Also
rea son is correct.
Hence correct option is (b)
13. Since both as ser tion and rea son are true and 
rea son e x pla n a tion is cor rect.
Hence correct option is (a).
Objective Questions (Level 2)
¢ Single option correct
1. We have m =
Real depth
App.depth
Þ           
4
3
1
=
App.depth
Þ App. depth =
3
4
Hence the distance between bird and mirror
= + = 2
3
4
11
4
m
Since plane mirror form image behind the
mirror (for real object) at same distance as
object hence the distance between bird and
its image 
= + =
11
4
11
4
11
2
 m
Correct option is (d).
2.
m m m m
2 1 2 1
v u R
- =
-
Here m = - ¥  u
1
1 =
and m
2
= 1.5
Þ 
1.5 0.5
v R
+
¥
=
1
 Þ v R = 3
Hence correct option is (b).
3. From figure, r = ° 30
Q r = + - = ° - ° q a 90 120 90
 37
60°
r
r
60°
60°
Normal
O
Page 2


8. F o cal length of com bi na tion
1 1 1 1 1
1 2
F F F f f
= + = -
convex convex
if f f
1 2
>
Þ
1 1
1 2
f f
<
Þ F = - negative
Hence assertion is true. Since power is a
measure of converging or divergence of a
lens. Hence reason is not true. Correct
option is (c).
9. Since glass slab pro duced a net shift. Hence
v is in creased. Thus mag ni fied im age is
ob tained but im age may be real or vir tual
de pend ing on the po si tion of slab.
Correct option is (b)
10. In this case im age dis tance of O O
1 2
and are
same from the lens. 
Q  
1 1 1
v u f
- = and reason is trure.
Hence correct option is (d).
11. As ser tion is false since only ray energe if
refractve in dex of the col our less than the
prism and an gle of in ci dence is less than
crit i cal an gle but rea son is true. Cor rect
op tion is (a).
12. If two ob ject is placed be tween pole and fo cus 
im age is real hence as ser tion is true. Also
rea son is correct.
Hence correct option is (b)
13. Since both as ser tion and rea son are true and 
rea son e x pla n a tion is cor rect.
Hence correct option is (a).
Objective Questions (Level 2)
¢ Single option correct
1. We have m =
Real depth
App.depth
Þ           
4
3
1
=
App.depth
Þ App. depth =
3
4
Hence the distance between bird and mirror
= + = 2
3
4
11
4
m
Since plane mirror form image behind the
mirror (for real object) at same distance as
object hence the distance between bird and
its image 
= + =
11
4
11
4
11
2
 m
Correct option is (d).
2.
m m m m
2 1 2 1
v u R
- =
-
Here m = - ¥  u
1
1 =
and m
2
= 1.5
Þ 
1.5 0.5
v R
+
¥
=
1
 Þ v R = 3
Hence correct option is (b).
3. From figure, r = ° 30
Q r = + - = ° - ° q a 90 120 90
 37
60°
r
r
60°
60°
Normal
O
Hence m =
°
°
=
sin
sin
60
30
3
The correct option is (d).
4. The lens be come di verg ing if
m m m m
1 2 2 3
- > -
Þ m m m
1 3 2
2 + >
or 2
2 1 3
m m m < + 
Hence correct option is (b).
5.
1 1
16
1
1
v f
+ = Þ v
f
f
1
16
16
=
-
m
v f
f
f
f
1
1
16
16
16 16 16
=
-
=
-
-
=
-
- ( ) ( )
 …(i)
and 
1 1
6
1
2
v f
+ = Þ v
f
f
2
6
6
=
-
m
2
2
6
=
v
      ( Q image is virtual)
Þ m
f
f
f
f
2
6
6 6 6
=
-
=
- ( )
 …(ii)
But m m
1 2
= Þ 
-
-
=
-
f
f
f
f 16 6
Þ - + = - 6 16 f f Þ 2 22 f =
Þ         f = 11 cm
Hence correct option is (d).
6. Let real depth at any in stant t of the wa ter is 
h then volume of wa ter V R h = p
2
Þ 
dV
dt
R
dh
dt
= p
2
  …(i)
Let apparent depth at this instant is h¢
Q m =
Real depth
Apparent depth
Þ 
n
n
h
h
2
1 1
= Þ h
n
n
h
1
1
2
=
Now 
dh
dt
x
n
n
dh
dt
¢
= =
1
2
Þ 
dh
dt
xn
n
=
2
1
 …(ii)
From Eq. (i) and (ii), we get
dV
dt
R xn
n
=
p
2
2
1
Hence correct option is (b).
7.
Q q = ° 37 Þ i = - - = = ° [ ( )] 90 90 37 q q
Applying Snell’s law on face BC.
sin
sin
37 3
5
2
1
°
= =
r
m
m
Þ sin sin r = ° = ´ =
4
3
37
5
3
3
5
1
Þ       r = ° 90
Hence deviation d = ° + ° = ° 90 37 127
Correct option is (b).
8. Let refractive in dex of liq uid is m
For position of fish w.r.t. bird is
m =
Real depth
App.depth
 Þ m =
x
h
1
 …(i)
For position of bird w.r.t. fish is
1
2
m
=
y
h
 …(2)
From Eq. (i) and (ii) we get u
h
h
=
2
1
Hence correct option is (a).
9. 
Þ ( ) R t R - + =
2 2 2
3
Þ 2 9 Rt
~
-   Q R T >>
38
90–q
i
r
90
m = 5/3
A
90°
air
C
B
air
(R – t )
R
t
3
Page 3


8. F o cal length of com bi na tion
1 1 1 1 1
1 2
F F F f f
= + = -
convex convex
if f f
1 2
>
Þ
1 1
1 2
f f
<
Þ F = - negative
Hence assertion is true. Since power is a
measure of converging or divergence of a
lens. Hence reason is not true. Correct
option is (c).
9. Since glass slab pro duced a net shift. Hence
v is in creased. Thus mag ni fied im age is
ob tained but im age may be real or vir tual
de pend ing on the po si tion of slab.
Correct option is (b)
10. In this case im age dis tance of O O
1 2
and are
same from the lens. 
Q  
1 1 1
v u f
- = and reason is trure.
Hence correct option is (d).
11. As ser tion is false since only ray energe if
refractve in dex of the col our less than the
prism and an gle of in ci dence is less than
crit i cal an gle but rea son is true. Cor rect
op tion is (a).
12. If two ob ject is placed be tween pole and fo cus 
im age is real hence as ser tion is true. Also
rea son is correct.
Hence correct option is (b)
13. Since both as ser tion and rea son are true and 
rea son e x pla n a tion is cor rect.
Hence correct option is (a).
Objective Questions (Level 2)
¢ Single option correct
1. We have m =
Real depth
App.depth
Þ           
4
3
1
=
App.depth
Þ App. depth =
3
4
Hence the distance between bird and mirror
= + = 2
3
4
11
4
m
Since plane mirror form image behind the
mirror (for real object) at same distance as
object hence the distance between bird and
its image 
= + =
11
4
11
4
11
2
 m
Correct option is (d).
2.
m m m m
2 1 2 1
v u R
- =
-
Here m = - ¥  u
1
1 =
and m
2
= 1.5
Þ 
1.5 0.5
v R
+
¥
=
1
 Þ v R = 3
Hence correct option is (b).
3. From figure, r = ° 30
Q r = + - = ° - ° q a 90 120 90
 37
60°
r
r
60°
60°
Normal
O
Hence m =
°
°
=
sin
sin
60
30
3
The correct option is (d).
4. The lens be come di verg ing if
m m m m
1 2 2 3
- > -
Þ m m m
1 3 2
2 + >
or 2
2 1 3
m m m < + 
Hence correct option is (b).
5.
1 1
16
1
1
v f
+ = Þ v
f
f
1
16
16
=
-
m
v f
f
f
f
1
1
16
16
16 16 16
=
-
=
-
-
=
-
- ( ) ( )
 …(i)
and 
1 1
6
1
2
v f
+ = Þ v
f
f
2
6
6
=
-
m
2
2
6
=
v
      ( Q image is virtual)
Þ m
f
f
f
f
2
6
6 6 6
=
-
=
- ( )
 …(ii)
But m m
1 2
= Þ 
-
-
=
-
f
f
f
f 16 6
Þ - + = - 6 16 f f Þ 2 22 f =
Þ         f = 11 cm
Hence correct option is (d).
6. Let real depth at any in stant t of the wa ter is 
h then volume of wa ter V R h = p
2
Þ 
dV
dt
R
dh
dt
= p
2
  …(i)
Let apparent depth at this instant is h¢
Q m =
Real depth
Apparent depth
Þ 
n
n
h
h
2
1 1
= Þ h
n
n
h
1
1
2
=
Now 
dh
dt
x
n
n
dh
dt
¢
= =
1
2
Þ 
dh
dt
xn
n
=
2
1
 …(ii)
From Eq. (i) and (ii), we get
dV
dt
R xn
n
=
p
2
2
1
Hence correct option is (b).
7.
Q q = ° 37 Þ i = - - = = ° [ ( )] 90 90 37 q q
Applying Snell’s law on face BC.
sin
sin
37 3
5
2
1
°
= =
r
m
m
Þ sin sin r = ° = ´ =
4
3
37
5
3
3
5
1
Þ       r = ° 90
Hence deviation d = ° + ° = ° 90 37 127
Correct option is (b).
8. Let refractive in dex of liq uid is m
For position of fish w.r.t. bird is
m =
Real depth
App.depth
 Þ m =
x
h
1
 …(i)
For position of bird w.r.t. fish is
1
2
m
=
y
h
 …(2)
From Eq. (i) and (ii) we get u
h
h
=
2
1
Hence correct option is (a).
9. 
Þ ( ) R t R - + =
2 2 2
3
Þ 2 9 Rt
~
-   Q R T >>
38
90–q
i
r
90
m = 5/3
A
90°
air
C
B
air
(R – t )
R
t
3
Þ 2
900
3
300 R = =
mm
mm
mm
Þ R = = 150 15 mm cm
Hence correct option is (a).
10.
1 1 1
1 1
v v f
+ = Þ v
fv
v f
1
1
1
=
-
m
v
v
f
v f
1
1
1 1
=
-
=
-
- ( )
 …(i)
1 1 1
2 2
v v f
+ = Þ v
fv
v f
2
2
2
=
-
m
v
v
f
f
2
2
2 2
= =
- m
 …(ii)
Q m m
1 2
=
Þ 
-
-
=
f
v f
f
v f
1 2
 Þ f
v v
=
+
1 2
2
Hence the correct option is (d).
11.
1 1 1
1 2 1 2
F f f
d
f f
= + -
Þ F
f f
f f d
=
+ -
1 2
1 2
as d increases f f d
1 2
+ - decreases hence F
increases. Hence image move to right.
Correct option is (b).
12. In this case, minimum de vi a tion of ray 1 is
same as ray 2.
Hence correct option is (c).
13. For critical an gle at glass air sur face
sin q
m
c
g
= =
1 2
3
 …(i)
Now for glass water surface.
m
m
q
w
g
c
r
=
sin
sin
 Þ 
4 3
3 2
2
3
/
/ sin
=
´ r
Þ sin r =
4
3
Now for water air surface
m
m
w
a
r
r
=
¢
sin
sin
 Þ 
4
3
4
3
1
= ´
¢ sin r
Þ sin r¢ = 1 Þ r¢ = ° 90
Hence correct option is (d).
14. F or lim it ing an g le of in ci dent emer gent ra y
be come par al lel to the 2nd face
Q 
sin
sin
r¢
°
=
90
3
7
 Þ r¢ =
-
sin
1
3
7
Now  r r + ° + ¢ = ° 30 90
Þ    r r = ° - ¢ 60
Þ    r = ° -
-
( sin / ) 60 3 7
1
Now    m =
sin
sin
i
r
Þ 
7
3 60 3 7
1
=
-
-
sin
sin [ sin / ]
i
Þ i = ´ -
- -
sin { / sin ( sin / }
1 1
7 3 60 3 7
= ´ ° - °
ì
í
î
ü
ý
þ
-
sin sin ( )
1
7
3
60 21
= ´ °
-
sin { / sin }
1
7 3 19
= - °
-
sin {
~
1
30 0.49}
Hence correct option is (a).
15. The im age form the ob ject it self if the rays
in ci dent par al lel to op ti cal axis on the mir ror 
i.e., image of refraction is formed at ¥. It is
possible when O is placed at focus i e . ., 
d=10cm 
Hence correct option is (c).
16. The dot will ap pear at c for all val ues of m.
Since po si tion does not in same me dium.
Hence correct option is (b).
 39
i
30+r'
60°
90°–r'
r'
r
Page 4


8. F o cal length of com bi na tion
1 1 1 1 1
1 2
F F F f f
= + = -
convex convex
if f f
1 2
>
Þ
1 1
1 2
f f
<
Þ F = - negative
Hence assertion is true. Since power is a
measure of converging or divergence of a
lens. Hence reason is not true. Correct
option is (c).
9. Since glass slab pro duced a net shift. Hence
v is in creased. Thus mag ni fied im age is
ob tained but im age may be real or vir tual
de pend ing on the po si tion of slab.
Correct option is (b)
10. In this case im age dis tance of O O
1 2
and are
same from the lens. 
Q  
1 1 1
v u f
- = and reason is trure.
Hence correct option is (d).
11. As ser tion is false since only ray energe if
refractve in dex of the col our less than the
prism and an gle of in ci dence is less than
crit i cal an gle but rea son is true. Cor rect
op tion is (a).
12. If two ob ject is placed be tween pole and fo cus 
im age is real hence as ser tion is true. Also
rea son is correct.
Hence correct option is (b)
13. Since both as ser tion and rea son are true and 
rea son e x pla n a tion is cor rect.
Hence correct option is (a).
Objective Questions (Level 2)
¢ Single option correct
1. We have m =
Real depth
App.depth
Þ           
4
3
1
=
App.depth
Þ App. depth =
3
4
Hence the distance between bird and mirror
= + = 2
3
4
11
4
m
Since plane mirror form image behind the
mirror (for real object) at same distance as
object hence the distance between bird and
its image 
= + =
11
4
11
4
11
2
 m
Correct option is (d).
2.
m m m m
2 1 2 1
v u R
- =
-
Here m = - ¥  u
1
1 =
and m
2
= 1.5
Þ 
1.5 0.5
v R
+
¥
=
1
 Þ v R = 3
Hence correct option is (b).
3. From figure, r = ° 30
Q r = + - = ° - ° q a 90 120 90
 37
60°
r
r
60°
60°
Normal
O
Hence m =
°
°
=
sin
sin
60
30
3
The correct option is (d).
4. The lens be come di verg ing if
m m m m
1 2 2 3
- > -
Þ m m m
1 3 2
2 + >
or 2
2 1 3
m m m < + 
Hence correct option is (b).
5.
1 1
16
1
1
v f
+ = Þ v
f
f
1
16
16
=
-
m
v f
f
f
f
1
1
16
16
16 16 16
=
-
=
-
-
=
-
- ( ) ( )
 …(i)
and 
1 1
6
1
2
v f
+ = Þ v
f
f
2
6
6
=
-
m
2
2
6
=
v
      ( Q image is virtual)
Þ m
f
f
f
f
2
6
6 6 6
=
-
=
- ( )
 …(ii)
But m m
1 2
= Þ 
-
-
=
-
f
f
f
f 16 6
Þ - + = - 6 16 f f Þ 2 22 f =
Þ         f = 11 cm
Hence correct option is (d).
6. Let real depth at any in stant t of the wa ter is 
h then volume of wa ter V R h = p
2
Þ 
dV
dt
R
dh
dt
= p
2
  …(i)
Let apparent depth at this instant is h¢
Q m =
Real depth
Apparent depth
Þ 
n
n
h
h
2
1 1
= Þ h
n
n
h
1
1
2
=
Now 
dh
dt
x
n
n
dh
dt
¢
= =
1
2
Þ 
dh
dt
xn
n
=
2
1
 …(ii)
From Eq. (i) and (ii), we get
dV
dt
R xn
n
=
p
2
2
1
Hence correct option is (b).
7.
Q q = ° 37 Þ i = - - = = ° [ ( )] 90 90 37 q q
Applying Snell’s law on face BC.
sin
sin
37 3
5
2
1
°
= =
r
m
m
Þ sin sin r = ° = ´ =
4
3
37
5
3
3
5
1
Þ       r = ° 90
Hence deviation d = ° + ° = ° 90 37 127
Correct option is (b).
8. Let refractive in dex of liq uid is m
For position of fish w.r.t. bird is
m =
Real depth
App.depth
 Þ m =
x
h
1
 …(i)
For position of bird w.r.t. fish is
1
2
m
=
y
h
 …(2)
From Eq. (i) and (ii) we get u
h
h
=
2
1
Hence correct option is (a).
9. 
Þ ( ) R t R - + =
2 2 2
3
Þ 2 9 Rt
~
-   Q R T >>
38
90–q
i
r
90
m = 5/3
A
90°
air
C
B
air
(R – t )
R
t
3
Þ 2
900
3
300 R = =
mm
mm
mm
Þ R = = 150 15 mm cm
Hence correct option is (a).
10.
1 1 1
1 1
v v f
+ = Þ v
fv
v f
1
1
1
=
-
m
v
v
f
v f
1
1
1 1
=
-
=
-
- ( )
 …(i)
1 1 1
2 2
v v f
+ = Þ v
fv
v f
2
2
2
=
-
m
v
v
f
f
2
2
2 2
= =
- m
 …(ii)
Q m m
1 2
=
Þ 
-
-
=
f
v f
f
v f
1 2
 Þ f
v v
=
+
1 2
2
Hence the correct option is (d).
11.
1 1 1
1 2 1 2
F f f
d
f f
= + -
Þ F
f f
f f d
=
+ -
1 2
1 2
as d increases f f d
1 2
+ - decreases hence F
increases. Hence image move to right.
Correct option is (b).
12. In this case, minimum de vi a tion of ray 1 is
same as ray 2.
Hence correct option is (c).
13. For critical an gle at glass air sur face
sin q
m
c
g
= =
1 2
3
 …(i)
Now for glass water surface.
m
m
q
w
g
c
r
=
sin
sin
 Þ 
4 3
3 2
2
3
/
/ sin
=
´ r
Þ sin r =
4
3
Now for water air surface
m
m
w
a
r
r
=
¢
sin
sin
 Þ 
4
3
4
3
1
= ´
¢ sin r
Þ sin r¢ = 1 Þ r¢ = ° 90
Hence correct option is (d).
14. F or lim it ing an g le of in ci dent emer gent ra y
be come par al lel to the 2nd face
Q 
sin
sin
r¢
°
=
90
3
7
 Þ r¢ =
-
sin
1
3
7
Now  r r + ° + ¢ = ° 30 90
Þ    r r = ° - ¢ 60
Þ    r = ° -
-
( sin / ) 60 3 7
1
Now    m =
sin
sin
i
r
Þ 
7
3 60 3 7
1
=
-
-
sin
sin [ sin / ]
i
Þ i = ´ -
- -
sin { / sin ( sin / }
1 1
7 3 60 3 7
= ´ ° - °
ì
í
î
ü
ý
þ
-
sin sin ( )
1
7
3
60 21
= ´ °
-
sin { / sin }
1
7 3 19
= - °
-
sin {
~
1
30 0.49}
Hence correct option is (a).
15. The im age form the ob ject it self if the rays
in ci dent par al lel to op ti cal axis on the mir ror 
i.e., image of refraction is formed at ¥. It is
possible when O is placed at focus i e . ., 
d=10cm 
Hence correct option is (c).
16. The dot will ap pear at c for all val ues of m.
Since po si tion does not in same me dium.
Hence correct option is (b).
 39
i
30+r'
60°
90°–r'
r'
r
17. We have for to tal internal reflection
sin
sin
/
/
i
90
6 5
3 2
=
Þ sin i =
4
5
 Þ i =
æ
è
ç
ö
ø
÷
= °
-
sin
1
4
5
53
Hence the ray will not cross BC if i > ° 53
Þ 90 180 + + = ° q i
 q = - 90 i Q i > ° 53
Þ q < ° 37
Hence correct option is (a).
18. For reflection at curved sur face
1 1 3
2
1
1
10 v x
-
-
= -
æ
è
ç
ö
ø
÷
´
( )
Þ v
x
x
=
-
20
20
This image act as virtual object for
plane-glass-water surface
Þ 
m m m m
g w g w
x
x ¥
-
-
=
-
¥
( ) 20
20
Þ x = 20 cm
Hence answer is (c).
19. The ra tio of fo cal length in the sit u a tion II
and III is 1 : 1.
Hence correct option is (c).
20. We have 
1 1 1
OB OA f
-
-
=
( )
Þ 
1 1 1
OB OA f
+ =
Þ f
OB OA
OA OB
=
+
.
   Q  OB OA AB + = 
Þ f
OB OA
AB
=
.
 …(i)
Now AB AC BC
2 2 2
= +
( ) OA OB OC OA OB OC + = + + +
2 2 2 2 2
Þ 
OA OB OAOB C OA OB
2 2 2 2 2
2 20 + + = + +
Þ OC OA OB
2
= …(ii)
Putting this value in Eq. (i), we get
Þ f
OC
AB
=
2
Hence correct option is (c).
21. The shift pro duce
        Dt t
w g
= -
é
ë
ê
ê
ù
û
ú
ú
1
2
m
= -
é
ë
ê
ù
û
ú
36 1
1
9 8 /
 Q 
w g
m = =
3 2
4 3
9
8
/
/
=
´
=
36 1
9
4 cm
Hence correct option is (b).
22. m =
Real depth
App. depth
Þ 
4
3
=
real depth
10.5 cm
Þ Real depth = ´
4
3
10.5 cm = 14 cm
Hence correct option is (d).
23. Q y
0
1 = + cm and y
i
= - 2 cm
m
v
u
= - Þ 2 =
v
u
Now let x be the position of lens then 
v x = - 50 and v x = + ( ) 40 . 
Þ 2
50
40
=
-
+
x
x
 
Þ 80 2 50 + = - x x
Þ - = 3 30 x Þ x = - 10 cm
Hence correct option is (c).
40
3/2
B E
C D
q
q i
90°
6/5
A
Page 5


8. F o cal length of com bi na tion
1 1 1 1 1
1 2
F F F f f
= + = -
convex convex
if f f
1 2
>
Þ
1 1
1 2
f f
<
Þ F = - negative
Hence assertion is true. Since power is a
measure of converging or divergence of a
lens. Hence reason is not true. Correct
option is (c).
9. Since glass slab pro duced a net shift. Hence
v is in creased. Thus mag ni fied im age is
ob tained but im age may be real or vir tual
de pend ing on the po si tion of slab.
Correct option is (b)
10. In this case im age dis tance of O O
1 2
and are
same from the lens. 
Q  
1 1 1
v u f
- = and reason is trure.
Hence correct option is (d).
11. As ser tion is false since only ray energe if
refractve in dex of the col our less than the
prism and an gle of in ci dence is less than
crit i cal an gle but rea son is true. Cor rect
op tion is (a).
12. If two ob ject is placed be tween pole and fo cus 
im age is real hence as ser tion is true. Also
rea son is correct.
Hence correct option is (b)
13. Since both as ser tion and rea son are true and 
rea son e x pla n a tion is cor rect.
Hence correct option is (a).
Objective Questions (Level 2)
¢ Single option correct
1. We have m =
Real depth
App.depth
Þ           
4
3
1
=
App.depth
Þ App. depth =
3
4
Hence the distance between bird and mirror
= + = 2
3
4
11
4
m
Since plane mirror form image behind the
mirror (for real object) at same distance as
object hence the distance between bird and
its image 
= + =
11
4
11
4
11
2
 m
Correct option is (d).
2.
m m m m
2 1 2 1
v u R
- =
-
Here m = - ¥  u
1
1 =
and m
2
= 1.5
Þ 
1.5 0.5
v R
+
¥
=
1
 Þ v R = 3
Hence correct option is (b).
3. From figure, r = ° 30
Q r = + - = ° - ° q a 90 120 90
 37
60°
r
r
60°
60°
Normal
O
Hence m =
°
°
=
sin
sin
60
30
3
The correct option is (d).
4. The lens be come di verg ing if
m m m m
1 2 2 3
- > -
Þ m m m
1 3 2
2 + >
or 2
2 1 3
m m m < + 
Hence correct option is (b).
5.
1 1
16
1
1
v f
+ = Þ v
f
f
1
16
16
=
-
m
v f
f
f
f
1
1
16
16
16 16 16
=
-
=
-
-
=
-
- ( ) ( )
 …(i)
and 
1 1
6
1
2
v f
+ = Þ v
f
f
2
6
6
=
-
m
2
2
6
=
v
      ( Q image is virtual)
Þ m
f
f
f
f
2
6
6 6 6
=
-
=
- ( )
 …(ii)
But m m
1 2
= Þ 
-
-
=
-
f
f
f
f 16 6
Þ - + = - 6 16 f f Þ 2 22 f =
Þ         f = 11 cm
Hence correct option is (d).
6. Let real depth at any in stant t of the wa ter is 
h then volume of wa ter V R h = p
2
Þ 
dV
dt
R
dh
dt
= p
2
  …(i)
Let apparent depth at this instant is h¢
Q m =
Real depth
Apparent depth
Þ 
n
n
h
h
2
1 1
= Þ h
n
n
h
1
1
2
=
Now 
dh
dt
x
n
n
dh
dt
¢
= =
1
2
Þ 
dh
dt
xn
n
=
2
1
 …(ii)
From Eq. (i) and (ii), we get
dV
dt
R xn
n
=
p
2
2
1
Hence correct option is (b).
7.
Q q = ° 37 Þ i = - - = = ° [ ( )] 90 90 37 q q
Applying Snell’s law on face BC.
sin
sin
37 3
5
2
1
°
= =
r
m
m
Þ sin sin r = ° = ´ =
4
3
37
5
3
3
5
1
Þ       r = ° 90
Hence deviation d = ° + ° = ° 90 37 127
Correct option is (b).
8. Let refractive in dex of liq uid is m
For position of fish w.r.t. bird is
m =
Real depth
App.depth
 Þ m =
x
h
1
 …(i)
For position of bird w.r.t. fish is
1
2
m
=
y
h
 …(2)
From Eq. (i) and (ii) we get u
h
h
=
2
1
Hence correct option is (a).
9. 
Þ ( ) R t R - + =
2 2 2
3
Þ 2 9 Rt
~
-   Q R T >>
38
90–q
i
r
90
m = 5/3
A
90°
air
C
B
air
(R – t )
R
t
3
Þ 2
900
3
300 R = =
mm
mm
mm
Þ R = = 150 15 mm cm
Hence correct option is (a).
10.
1 1 1
1 1
v v f
+ = Þ v
fv
v f
1
1
1
=
-
m
v
v
f
v f
1
1
1 1
=
-
=
-
- ( )
 …(i)
1 1 1
2 2
v v f
+ = Þ v
fv
v f
2
2
2
=
-
m
v
v
f
f
2
2
2 2
= =
- m
 …(ii)
Q m m
1 2
=
Þ 
-
-
=
f
v f
f
v f
1 2
 Þ f
v v
=
+
1 2
2
Hence the correct option is (d).
11.
1 1 1
1 2 1 2
F f f
d
f f
= + -
Þ F
f f
f f d
=
+ -
1 2
1 2
as d increases f f d
1 2
+ - decreases hence F
increases. Hence image move to right.
Correct option is (b).
12. In this case, minimum de vi a tion of ray 1 is
same as ray 2.
Hence correct option is (c).
13. For critical an gle at glass air sur face
sin q
m
c
g
= =
1 2
3
 …(i)
Now for glass water surface.
m
m
q
w
g
c
r
=
sin
sin
 Þ 
4 3
3 2
2
3
/
/ sin
=
´ r
Þ sin r =
4
3
Now for water air surface
m
m
w
a
r
r
=
¢
sin
sin
 Þ 
4
3
4
3
1
= ´
¢ sin r
Þ sin r¢ = 1 Þ r¢ = ° 90
Hence correct option is (d).
14. F or lim it ing an g le of in ci dent emer gent ra y
be come par al lel to the 2nd face
Q 
sin
sin
r¢
°
=
90
3
7
 Þ r¢ =
-
sin
1
3
7
Now  r r + ° + ¢ = ° 30 90
Þ    r r = ° - ¢ 60
Þ    r = ° -
-
( sin / ) 60 3 7
1
Now    m =
sin
sin
i
r
Þ 
7
3 60 3 7
1
=
-
-
sin
sin [ sin / ]
i
Þ i = ´ -
- -
sin { / sin ( sin / }
1 1
7 3 60 3 7
= ´ ° - °
ì
í
î
ü
ý
þ
-
sin sin ( )
1
7
3
60 21
= ´ °
-
sin { / sin }
1
7 3 19
= - °
-
sin {
~
1
30 0.49}
Hence correct option is (a).
15. The im age form the ob ject it self if the rays
in ci dent par al lel to op ti cal axis on the mir ror 
i.e., image of refraction is formed at ¥. It is
possible when O is placed at focus i e . ., 
d=10cm 
Hence correct option is (c).
16. The dot will ap pear at c for all val ues of m.
Since po si tion does not in same me dium.
Hence correct option is (b).
 39
i
30+r'
60°
90°–r'
r'
r
17. We have for to tal internal reflection
sin
sin
/
/
i
90
6 5
3 2
=
Þ sin i =
4
5
 Þ i =
æ
è
ç
ö
ø
÷
= °
-
sin
1
4
5
53
Hence the ray will not cross BC if i > ° 53
Þ 90 180 + + = ° q i
 q = - 90 i Q i > ° 53
Þ q < ° 37
Hence correct option is (a).
18. For reflection at curved sur face
1 1 3
2
1
1
10 v x
-
-
= -
æ
è
ç
ö
ø
÷
´
( )
Þ v
x
x
=
-
20
20
This image act as virtual object for
plane-glass-water surface
Þ 
m m m m
g w g w
x
x ¥
-
-
=
-
¥
( ) 20
20
Þ x = 20 cm
Hence answer is (c).
19. The ra tio of fo cal length in the sit u a tion II
and III is 1 : 1.
Hence correct option is (c).
20. We have 
1 1 1
OB OA f
-
-
=
( )
Þ 
1 1 1
OB OA f
+ =
Þ f
OB OA
OA OB
=
+
.
   Q  OB OA AB + = 
Þ f
OB OA
AB
=
.
 …(i)
Now AB AC BC
2 2 2
= +
( ) OA OB OC OA OB OC + = + + +
2 2 2 2 2
Þ 
OA OB OAOB C OA OB
2 2 2 2 2
2 20 + + = + +
Þ OC OA OB
2
= …(ii)
Putting this value in Eq. (i), we get
Þ f
OC
AB
=
2
Hence correct option is (c).
21. The shift pro duce
        Dt t
w g
= -
é
ë
ê
ê
ù
û
ú
ú
1
2
m
= -
é
ë
ê
ù
û
ú
36 1
1
9 8 /
 Q 
w g
m = =
3 2
4 3
9
8
/
/
=
´
=
36 1
9
4 cm
Hence correct option is (b).
22. m =
Real depth
App. depth
Þ 
4
3
=
real depth
10.5 cm
Þ Real depth = ´
4
3
10.5 cm = 14 cm
Hence correct option is (d).
23. Q y
0
1 = + cm and y
i
= - 2 cm
m
v
u
= - Þ 2 =
v
u
Now let x be the position of lens then 
v x = - 50 and v x = + ( ) 40 . 
Þ 2
50
40
=
-
+
x
x
 
Þ 80 2 50 + = - x x
Þ - = 3 30 x Þ x = - 10 cm
Hence correct option is (c).
40
3/2
B E
C D
q
q i
90°
6/5
A
24. If the plane surface of plano-con v e x lens is
sil vered it be have the con cave mir ror of fo cal
length f
m
/2
Q f
m
= 10 cm
Þ f
e
= 5 cm hence R = 10 cm
Correct option is (c).
25. Since lens made real and mag ni fied im age,
hence it is a con vex lens when lens dipped in
wa ter its fo cal length.
1
1
1 1
1 2
f R R
w
g
= -
æ
è
ç
ç
ö
ø
÷
÷
+
æ
è
ç
ç
ö
ø
÷
÷
m
m
= -
æ
è
ç
ç
ö
ø
÷
÷
+
æ
è
ç
ç
ö
ø
÷
÷
=
-
+
æ
è
ç
ç
ö
ø
÷
4 3
3 2
1
1 1 1
9
1 1
1 2 1 2
/
/ R R R R
÷
Q f is - ve lens behave as concave, hence
the  image is virtual and magnified.
Correct option is (c).
26. The prism trans mit the light for which an gle 
of in ci dence( c), 2 9 0 c £ ° Þ c £ ° 4 5
Hence m = =
°
= =
1 1
4 5
2
s i n s i n C
1 . 4 1 4
Correct option is (b).
27.
1 1
16
1
v f
-
-
=
( )
 Q | | m
v
u
= = 3
For convex lens v u = 3
1
48
1
16
1
+ =
f
 Þ f = 12 cm (for real image)
Similarly when distance is 6 cm, 3 times
virtual image is formed hence mirror is
convex with focal length 12 cm.
Correct option is (c).
28. v n
g g
= l Þ 
c
n
g
g
m
l = …(i)
and 
c
n
w
w
m
l = …(ii)
Dividing (i) and (ii)
m
m
l
l
w
g
g
w
= Þ 
4
3
4
5 mg
=
Þ m
g
=
5
3
Hence correct option is (a).
29. x
f f d f d
f f d
=
+ -
+ -
1 2 1
1 2
( )
 and y
f d
f f d
=
-
+ -
( )
1
1 2
D
Here f f d
1 2
20 30 = = = cm cm ,
and D = = 55m 0.5 cm
Putting these values we get
x = 25 cm  and y = 0.25 cm
Correct option is (b).
30. Since for each q an gle of in ci dence at
glass-air boundry re mains 0° hence there
will never be to tal in ter nal reflection.
Correct option is (d).
31. Diameter = ´ m Orig i nal di am e ter
= ´ = -
4
3
1
4
3
cm cm
Hence correct option is (a).
32.
1
1
1
20
1
20
1
f
= - +
é
ë
ê
ù
û
ú
( ) 1.5
Þ f
1
20 = cm
Here u
1
30 = - cm
1 1 1
1 1
v u f
- = Þ 
1 1
20
1
30
1
v
= - Þ v
1
60 = cm
Magnification | | m
v
u
1
60
30
2 = - = = +
(Inverted image)
 41
30 cm
f = – 20 f = 20 cm
5 mm
90°
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