# DC Pandey Solutions: Refraction of Light- 1 Notes | Study DC Pandey Solutions for NEET Physics - NEET

## NEET: DC Pandey Solutions: Refraction of Light- 1 Notes | Study DC Pandey Solutions for NEET Physics - NEET

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``` Page 1

27 Refraction of Light
Introductory  Exercise 27.1
1. Let real depth of dust par ti cle is x  and
thick ness of slab is t
From Ist surface
m =
Real depth
App.depth
1.5
cm
=
x
6
Þ  x = 9 cm …(i)
From other face
m =
- t x
4
Þ t x - = ´ 4 1.5
Þ t x = + = + = 6 9 6 16 cm
2.
1 2
4
3
m = Þ
m
m
2
1
4
3
= …(i)
2 3
3
2
m = Þ
m
m
3
2
3
2
= …(ii)
From Eqs. (i) and (ii), we get
m
m
3
1
4
3
3
2
2 = ´ =
3. Frequency re main same.
Let v
1
is velocity in medium  (1) and v
2
in
Medium (2)
We have
m
1
1
=
c
v
and m
2
2
=
c
v
Þ
m
m
1
2
2
1
=
v
v
Þ v v
2
1
2
1
=
m
m
Similarly, wavelength l
m
m
l
2
1
2
1
=
4. From v n
a a
= l
Þ l
a
a
a
v
n
= =
´
´
= ´
+
-
3 10
6 10
5 10
8
14
7
m
= 50 nm
m
l
l
= = = =
a
m
500
300
5
3
1.67
Introductory Exercise 27.2
1. Since light rays are com ing from glass to air
ap pl y ing
m m m m
2 1 2 1
v u R
- =
-
Þ
1
10
1
v
- =
-
-
1.5 1.5
1.5
Þ
1 1
30 10
1 4
30 v
= - + =
- + 1.5 .5
Þ v = =
30
3.5
8.57 cm
2.
m m m m
2 1 2 1
v u R
- =
-
(a)
1.5 0.5
m
-
-
=
1
20 6 ( )
On solving v = 45 cm
(b)
1.5 0.5
v
-
-
=
1
10 6 ( )
On solving we get v = - 90 cm
t
x
2nd
face
1st
face
t – x
Page 2

27 Refraction of Light
Introductory  Exercise 27.1
1. Let real depth of dust par ti cle is x  and
thick ness of slab is t
From Ist surface
m =
Real depth
App.depth
1.5
cm
=
x
6
Þ  x = 9 cm …(i)
From other face
m =
- t x
4
Þ t x - = ´ 4 1.5
Þ t x = + = + = 6 9 6 16 cm
2.
1 2
4
3
m = Þ
m
m
2
1
4
3
= …(i)
2 3
3
2
m = Þ
m
m
3
2
3
2
= …(ii)
From Eqs. (i) and (ii), we get
m
m
3
1
4
3
3
2
2 = ´ =
3. Frequency re main same.
Let v
1
is velocity in medium  (1) and v
2
in
Medium (2)
We have
m
1
1
=
c
v
and m
2
2
=
c
v
Þ
m
m
1
2
2
1
=
v
v
Þ v v
2
1
2
1
=
m
m
Similarly, wavelength l
m
m
l
2
1
2
1
=
4. From v n
a a
= l
Þ l
a
a
a
v
n
= =
´
´
= ´
+
-
3 10
6 10
5 10
8
14
7
m
= 50 nm
m
l
l
= = = =
a
m
500
300
5
3
1.67
Introductory Exercise 27.2
1. Since light rays are com ing from glass to air
ap pl y ing
m m m m
2 1 2 1
v u R
- =
-
Þ
1
10
1
v
- =
-
-
1.5 1.5
1.5
Þ
1 1
30 10
1 4
30 v
= - + =
- + 1.5 .5
Þ v = =
30
3.5
8.57 cm
2.
m m m m
2 1 2 1
v u R
- =
-
(a)
1.5 0.5
m
-
-
=
1
20 6 ( )
On solving v = 45 cm
(b)
1.5 0.5
v
-
-
=
1
10 6 ( )
On solving we get v = - 90 cm
t
x
2nd
face
1st
face
t – x
(c)
1.5 0.5
v
-
-
=
1
3 6 ( )
On solving v = - 6.0 cm
3. Light rays are coming from glass to air

m m m m
2 1 2 1
v u R
- =
-
1 4
3 10
1 4 3
15 v
-
-
=
-
- ( )
/
( )
1 4
30
1
45 v
+ = on solving v = - 9 cm
4. Applying
u
v u R
2 1 2 1
- =
- m m m
Þ
1.44 0.44
1.25 v
-
¥
=
1
On solving v = 0.795 cm
5.
m m m m
2 1 2 1
v u R
- =
-
1.635 0.635
2.50) v
-
-
=
-
1
9 ( ) (
on solving v = 6.993 cm
Lateral magnification m
v
u
= -
Þ = - = -
6.993
9
0.777
Introductory Exercise 27.3
1. We have
1 1 1
1
1 1
1 2
v u f R R
- = = - -
é
ë
ê
ù
û
ú
( ) m
Þ - + = - - -
é
ë
ê
ù
û
ú
1
20
1
60
1
1 1
( ) 1.65
R R
Þ
- +
= ´
- 3 1
60
2
0.65
R
Þ R = ´ = 60 39 0.65 cm
2. Using
1 1 1
v u f
- = , we get
1
50
1 1
30 -
- =
x
Þ - = +
1 1
30
1
50 x
On solving x = - 18.75 cm
m
v
u
=
-
=
50
18.75
Height of filament image = ´ 2
50
18.75

= 5.3 cm
3.
1
1
1 1
f R R
= - +
é
ë
ê
ù
û
ú
( ) m
If lens faces becomes opposite three is no
change in radius of curvature hence focal
length does not change.
4. Using for mula
1 1 1
v u f
- = when u ® 0
1
v ® 0
when u f v ® ® ¥ , hence im age moves from
sur face to ¥ .
5.
1
1
1 1
1 2
f R R
= - -
é
ë
ê
ù
û
ú
( ) m
Þ
1
1
1 1 2
20 f R R
= -
-
-
é
ë
ê
ù
û
ú
= ´
-
( ) 1.3 0.3
Þ f = -
100
3
cm
(a) When immersed in a liquid of 1.8
refractive index
1
1
2 2
20
1
f R
= -
æ
è
ç
ö
ø
÷
-
é
ë
ê
ù
û
ú
=
-
´
- 1.3
1.8
0.5
1.8
f ¢ = 36 cm
(b) The minimum distance is equal to the
focal length = 36 cm
20
O
10 cm
Page 3

27 Refraction of Light
Introductory  Exercise 27.1
1. Let real depth of dust par ti cle is x  and
thick ness of slab is t
From Ist surface
m =
Real depth
App.depth
1.5
cm
=
x
6
Þ  x = 9 cm …(i)
From other face
m =
- t x
4
Þ t x - = ´ 4 1.5
Þ t x = + = + = 6 9 6 16 cm
2.
1 2
4
3
m = Þ
m
m
2
1
4
3
= …(i)
2 3
3
2
m = Þ
m
m
3
2
3
2
= …(ii)
From Eqs. (i) and (ii), we get
m
m
3
1
4
3
3
2
2 = ´ =
3. Frequency re main same.
Let v
1
is velocity in medium  (1) and v
2
in
Medium (2)
We have
m
1
1
=
c
v
and m
2
2
=
c
v
Þ
m
m
1
2
2
1
=
v
v
Þ v v
2
1
2
1
=
m
m
Similarly, wavelength l
m
m
l
2
1
2
1
=
4. From v n
a a
= l
Þ l
a
a
a
v
n
= =
´
´
= ´
+
-
3 10
6 10
5 10
8
14
7
m
= 50 nm
m
l
l
= = = =
a
m
500
300
5
3
1.67
Introductory Exercise 27.2
1. Since light rays are com ing from glass to air
ap pl y ing
m m m m
2 1 2 1
v u R
- =
-
Þ
1
10
1
v
- =
-
-
1.5 1.5
1.5
Þ
1 1
30 10
1 4
30 v
= - + =
- + 1.5 .5
Þ v = =
30
3.5
8.57 cm
2.
m m m m
2 1 2 1
v u R
- =
-
(a)
1.5 0.5
m
-
-
=
1
20 6 ( )
On solving v = 45 cm
(b)
1.5 0.5
v
-
-
=
1
10 6 ( )
On solving we get v = - 90 cm
t
x
2nd
face
1st
face
t – x
(c)
1.5 0.5
v
-
-
=
1
3 6 ( )
On solving v = - 6.0 cm
3. Light rays are coming from glass to air

m m m m
2 1 2 1
v u R
- =
-
1 4
3 10
1 4 3
15 v
-
-
=
-
- ( )
/
( )
1 4
30
1
45 v
+ = on solving v = - 9 cm
4. Applying
u
v u R
2 1 2 1
- =
- m m m
Þ
1.44 0.44
1.25 v
-
¥
=
1
On solving v = 0.795 cm
5.
m m m m
2 1 2 1
v u R
- =
-
1.635 0.635
2.50) v
-
-
=
-
1
9 ( ) (
on solving v = 6.993 cm
Lateral magnification m
v
u
= -
Þ = - = -
6.993
9
0.777
Introductory Exercise 27.3
1. We have
1 1 1
1
1 1
1 2
v u f R R
- = = - -
é
ë
ê
ù
û
ú
( ) m
Þ - + = - - -
é
ë
ê
ù
û
ú
1
20
1
60
1
1 1
( ) 1.65
R R
Þ
- +
= ´
- 3 1
60
2
0.65
R
Þ R = ´ = 60 39 0.65 cm
2. Using
1 1 1
v u f
- = , we get
1
50
1 1
30 -
- =
x
Þ - = +
1 1
30
1
50 x
On solving x = - 18.75 cm
m
v
u
=
-
=
50
18.75
Height of filament image = ´ 2
50
18.75

= 5.3 cm
3.
1
1
1 1
f R R
= - +
é
ë
ê
ù
û
ú
( ) m
If lens faces becomes opposite three is no
change in radius of curvature hence focal
length does not change.
4. Using for mula
1 1 1
v u f
- = when u ® 0
1
v ® 0
when u f v ® ® ¥ , hence im age moves from
sur face to ¥ .
5.
1
1
1 1
1 2
f R R
= - -
é
ë
ê
ù
û
ú
( ) m
Þ
1
1
1 1 2
20 f R R
= -
-
-
é
ë
ê
ù
û
ú
= ´
-
( ) 1.3 0.3
Þ f = -
100
3
cm
(a) When immersed in a liquid of 1.8
refractive index
1
1
2 2
20
1
f R
= -
æ
è
ç
ö
ø
÷
-
é
ë
ê
ù
û
ú
=
-
´
- 1.3
1.8
0.5
1.8
f ¢ = 36 cm
(b) The minimum distance is equal to the
focal length = 36 cm
20
O
10 cm
6. Using
1 1 1
v u f
- =
1 1
20
12
10 v
-
-
=
( )
On solving v = 20 cm
Magnification = - = -
v
u
1
Hence the image of same size and inverted.
Let the distance between second lens is x
Since magnification is unity image distance
also x using again
1 1 1
v u f
- =
we get
1 1 1 1
10 x x f
-
-
= =
( )
Þ x = 20 cm
Hence the distance between two lenses
= + = 20 20 40 cm cm cm
7.
1 1 1
v f u
= + …(i)
1 1 1
1
v f u du
= +
+
…(ii)
1 1
v v
u du u
u duu ¢
- =
+ -
+
( )
( )
on solving, we get
v v
v
du
u u du
- ¢
¢
=
+ ( )
thickness dv
v
u
du =
-
2
2
10. Size of im age = ´ = 6
2
3
2 cm.
11. Let im age distance is u
| | m = 3 Þ v u = 3
1
3
1 1
12 u u
+ = Þ v = 16 cm
12. Since im age is up right and diminished
hence lens is concave. Now
u v - = 20 …(i)
m
v
u
= =
1
2
Þ
1
2
20
=
- u
u
Þ u = 40 cm and v = 20 cm
Þ Using
1 1 1
v u f
- =
- + =
1
20
1
40
1
f
Þ f = - 40 cm
13. The im age co in cide it self if light falls
nor mally on plane mir ror hence ob ject must
be on focus i e . . + 10 cm.
8.
1 1 2 2 1
2 1
2
2 1
1
v u R R
+ = -
- ( ) ( ) m /m m /m
1 1 2 4 3 2 4 3 1
v
+ = -
-
0.2 0.4 0.4
( / ) ( / )
On solving v = 12 cm
9. Since shift in po si tion D t = 0.1 m
Hence real depth = + ( ) 0.1 0.2 m
= 0.3 m
and apparent depth= 0.2 m
m =
real depth
apparent depth
= =
0.3
0.2
1.5
21
Page 4

27 Refraction of Light
Introductory  Exercise 27.1
1. Let real depth of dust par ti cle is x  and
thick ness of slab is t
From Ist surface
m =
Real depth
App.depth
1.5
cm
=
x
6
Þ  x = 9 cm …(i)
From other face
m =
- t x
4
Þ t x - = ´ 4 1.5
Þ t x = + = + = 6 9 6 16 cm
2.
1 2
4
3
m = Þ
m
m
2
1
4
3
= …(i)
2 3
3
2
m = Þ
m
m
3
2
3
2
= …(ii)
From Eqs. (i) and (ii), we get
m
m
3
1
4
3
3
2
2 = ´ =
3. Frequency re main same.
Let v
1
is velocity in medium  (1) and v
2
in
Medium (2)
We have
m
1
1
=
c
v
and m
2
2
=
c
v
Þ
m
m
1
2
2
1
=
v
v
Þ v v
2
1
2
1
=
m
m
Similarly, wavelength l
m
m
l
2
1
2
1
=
4. From v n
a a
= l
Þ l
a
a
a
v
n
= =
´
´
= ´
+
-
3 10
6 10
5 10
8
14
7
m
= 50 nm
m
l
l
= = = =
a
m
500
300
5
3
1.67
Introductory Exercise 27.2
1. Since light rays are com ing from glass to air
ap pl y ing
m m m m
2 1 2 1
v u R
- =
-
Þ
1
10
1
v
- =
-
-
1.5 1.5
1.5
Þ
1 1
30 10
1 4
30 v
= - + =
- + 1.5 .5
Þ v = =
30
3.5
8.57 cm
2.
m m m m
2 1 2 1
v u R
- =
-
(a)
1.5 0.5
m
-
-
=
1
20 6 ( )
On solving v = 45 cm
(b)
1.5 0.5
v
-
-
=
1
10 6 ( )
On solving we get v = - 90 cm
t
x
2nd
face
1st
face
t – x
(c)
1.5 0.5
v
-
-
=
1
3 6 ( )
On solving v = - 6.0 cm
3. Light rays are coming from glass to air

m m m m
2 1 2 1
v u R
- =
-
1 4
3 10
1 4 3
15 v
-
-
=
-
- ( )
/
( )
1 4
30
1
45 v
+ = on solving v = - 9 cm
4. Applying
u
v u R
2 1 2 1
- =
- m m m
Þ
1.44 0.44
1.25 v
-
¥
=
1
On solving v = 0.795 cm
5.
m m m m
2 1 2 1
v u R
- =
-
1.635 0.635
2.50) v
-
-
=
-
1
9 ( ) (
on solving v = 6.993 cm
Lateral magnification m
v
u
= -
Þ = - = -
6.993
9
0.777
Introductory Exercise 27.3
1. We have
1 1 1
1
1 1
1 2
v u f R R
- = = - -
é
ë
ê
ù
û
ú
( ) m
Þ - + = - - -
é
ë
ê
ù
û
ú
1
20
1
60
1
1 1
( ) 1.65
R R
Þ
- +
= ´
- 3 1
60
2
0.65
R
Þ R = ´ = 60 39 0.65 cm
2. Using
1 1 1
v u f
- = , we get
1
50
1 1
30 -
- =
x
Þ - = +
1 1
30
1
50 x
On solving x = - 18.75 cm
m
v
u
=
-
=
50
18.75
Height of filament image = ´ 2
50
18.75

= 5.3 cm
3.
1
1
1 1
f R R
= - +
é
ë
ê
ù
û
ú
( ) m
If lens faces becomes opposite three is no
change in radius of curvature hence focal
length does not change.
4. Using for mula
1 1 1
v u f
- = when u ® 0
1
v ® 0
when u f v ® ® ¥ , hence im age moves from
sur face to ¥ .
5.
1
1
1 1
1 2
f R R
= - -
é
ë
ê
ù
û
ú
( ) m
Þ
1
1
1 1 2
20 f R R
= -
-
-
é
ë
ê
ù
û
ú
= ´
-
( ) 1.3 0.3
Þ f = -
100
3
cm
(a) When immersed in a liquid of 1.8
refractive index
1
1
2 2
20
1
f R
= -
æ
è
ç
ö
ø
÷
-
é
ë
ê
ù
û
ú
=
-
´
- 1.3
1.8
0.5
1.8
f ¢ = 36 cm
(b) The minimum distance is equal to the
focal length = 36 cm
20
O
10 cm
6. Using
1 1 1
v u f
- =
1 1
20
12
10 v
-
-
=
( )
On solving v = 20 cm
Magnification = - = -
v
u
1
Hence the image of same size and inverted.
Let the distance between second lens is x
Since magnification is unity image distance
also x using again
1 1 1
v u f
- =
we get
1 1 1 1
10 x x f
-
-
= =
( )
Þ x = 20 cm
Hence the distance between two lenses
= + = 20 20 40 cm cm cm
7.
1 1 1
v f u
= + …(i)
1 1 1
1
v f u du
= +
+
…(ii)
1 1
v v
u du u
u duu ¢
- =
+ -
+
( )
( )
on solving, we get
v v
v
du
u u du
- ¢
¢
=
+ ( )
thickness dv
v
u
du =
-
2
2
10. Size of im age = ´ = 6
2
3
2 cm.
11. Let im age distance is u
| | m = 3 Þ v u = 3
1
3
1 1
12 u u
+ = Þ v = 16 cm
12. Since im age is up right and diminished
hence lens is concave. Now
u v - = 20 …(i)
m
v
u
= =
1
2
Þ
1
2
20
=
- u
u
Þ u = 40 cm and v = 20 cm
Þ Using
1 1 1
v u f
- =
- + =
1
20
1
40
1
f
Þ f = - 40 cm
13. The im age co in cide it self if light falls
nor mally on plane mir ror hence ob ject must
be on focus i e . . + 10 cm.
8.
1 1 2 2 1
2 1
2
2 1
1
v u R R
+ = -
- ( ) ( ) m /m m /m
1 1 2 4 3 2 4 3 1
v
+ = -
-
0.2 0.4 0.4
( / ) ( / )
On solving v = 12 cm
9. Since shift in po si tion D t = 0.1 m
Hence real depth = + ( ) 0.1 0.2 m
= 0.3 m
and apparent depth= 0.2 m
m =
real depth
apparent depth
= =
0.3
0.2
1.5
21
AIEEE Corner
1.
We have r i + ° + = ° 90 180 Þ r i = - 90
From Snell’s law 1.5 = =
-
sin
sin
sin
sin ( )
i
r
i
i 90
Þ tan i = 1.5
Þ         i =
-
tan ( )
1
1.5
2. n
v
u
w
w
= = =
air
0.229
343
1498
Critical angle q = = °
-
sin ( . )
1
0229 13.2
3. Sped in glycrine v
c
n
g
g
= =
´ 3 10
8
1.47
t
v
g
1
8
8
20 20
3 10
10 = =
´
´
= ´
-
1.47
9.8 s
Speed in glycrine v
c
n
g
g
= =
´ 3 10
8
1.63
t
v
c
2
8
8
20 20
3 10
10 = =
´
´
- ´
-
1.63
10.8
~
t t
2 1
8
1086 10 - = - ´
-
( . ) 9.8
= ´
-
1.67 s 10
8
4. (a) t
v
1
6
1
6
8
1 10 1 10
3 10
=
´
=
´
´
- -
m
1.2
m
/
=
´
´
-
1.2 10
3 10
6
8
Þ t
1
14
10 = ´
-
0.4 s
t
2
6
8
14
1 0
3 1 0
1 0 =
´
´
= ´
-
-
1.5
0.5 s
t
3
6
8
6
8
14
1 10
3 10
10
3 10
10 =
´
´
=
´
´
= ´
- -
-
/1.8
1.8
0.6
Hence t
1
is least and t
1
14
10 = ´
-
0.4 s
(b) Total number of wavelengths
= + +
1 1
1 2 3
m
l
m
l
m
l
m 1.5 m m
/ / / n n n
=
´
+
´ 1000
600
100
600
1.2 nm
nm
1.5 nm
nm
+
´ ´ 1 1000
600
1.8 nm
nm
= =
4500
600
7.5
5. The given wave equation is
E y t
x
( , ) =
´
- ´ ´
é
ë
ê
ù
û
ú
-
E
y
t
ax
sin
2
5 10
3 10 2
7
14
p
l
Comparing with standard equation
E y t E ky t
x
( , ) sin [ ] = -
0
w
k =
´
= ´ ´
-
2
5 10
2 3 10
7
14
p
, w p
v
k
= =
´ ´
´
= ´
-
w p
p
2 3 10
2 5 10
10
14
7
8
/
1.5 m/s
Refractive index n
c
v
= =
´
´
=
3 10
10
2
8
8
1.5
Wavelength in this way l
p
n
k
=
2
Þ l
p
n
=
´
= ´
-
2
25 5 10
5 10
7
7
/
m
Þ l
n
= 500 nm
If vacuum, wavelength is l then
l
l
n
n
=
Þ l l = = ´ = n
n
2 500 1000nm
6. Refraction from plane and spher i cal
surfaces
22
Reflected ray
Reflected ray
Incident ray
90–i
90°
i i
r
Page 5

27 Refraction of Light
Introductory  Exercise 27.1
1. Let real depth of dust par ti cle is x  and
thick ness of slab is t
From Ist surface
m =
Real depth
App.depth
1.5
cm
=
x
6
Þ  x = 9 cm …(i)
From other face
m =
- t x
4
Þ t x - = ´ 4 1.5
Þ t x = + = + = 6 9 6 16 cm
2.
1 2
4
3
m = Þ
m
m
2
1
4
3
= …(i)
2 3
3
2
m = Þ
m
m
3
2
3
2
= …(ii)
From Eqs. (i) and (ii), we get
m
m
3
1
4
3
3
2
2 = ´ =
3. Frequency re main same.
Let v
1
is velocity in medium  (1) and v
2
in
Medium (2)
We have
m
1
1
=
c
v
and m
2
2
=
c
v
Þ
m
m
1
2
2
1
=
v
v
Þ v v
2
1
2
1
=
m
m
Similarly, wavelength l
m
m
l
2
1
2
1
=
4. From v n
a a
= l
Þ l
a
a
a
v
n
= =
´
´
= ´
+
-
3 10
6 10
5 10
8
14
7
m
= 50 nm
m
l
l
= = = =
a
m
500
300
5
3
1.67
Introductory Exercise 27.2
1. Since light rays are com ing from glass to air
ap pl y ing
m m m m
2 1 2 1
v u R
- =
-
Þ
1
10
1
v
- =
-
-
1.5 1.5
1.5
Þ
1 1
30 10
1 4
30 v
= - + =
- + 1.5 .5
Þ v = =
30
3.5
8.57 cm
2.
m m m m
2 1 2 1
v u R
- =
-
(a)
1.5 0.5
m
-
-
=
1
20 6 ( )
On solving v = 45 cm
(b)
1.5 0.5
v
-
-
=
1
10 6 ( )
On solving we get v = - 90 cm
t
x
2nd
face
1st
face
t – x
(c)
1.5 0.5
v
-
-
=
1
3 6 ( )
On solving v = - 6.0 cm
3. Light rays are coming from glass to air

m m m m
2 1 2 1
v u R
- =
-
1 4
3 10
1 4 3
15 v
-
-
=
-
- ( )
/
( )
1 4
30
1
45 v
+ = on solving v = - 9 cm
4. Applying
u
v u R
2 1 2 1
- =
- m m m
Þ
1.44 0.44
1.25 v
-
¥
=
1
On solving v = 0.795 cm
5.
m m m m
2 1 2 1
v u R
- =
-
1.635 0.635
2.50) v
-
-
=
-
1
9 ( ) (
on solving v = 6.993 cm
Lateral magnification m
v
u
= -
Þ = - = -
6.993
9
0.777
Introductory Exercise 27.3
1. We have
1 1 1
1
1 1
1 2
v u f R R
- = = - -
é
ë
ê
ù
û
ú
( ) m
Þ - + = - - -
é
ë
ê
ù
û
ú
1
20
1
60
1
1 1
( ) 1.65
R R
Þ
- +
= ´
- 3 1
60
2
0.65
R
Þ R = ´ = 60 39 0.65 cm
2. Using
1 1 1
v u f
- = , we get
1
50
1 1
30 -
- =
x
Þ - = +
1 1
30
1
50 x
On solving x = - 18.75 cm
m
v
u
=
-
=
50
18.75
Height of filament image = ´ 2
50
18.75

= 5.3 cm
3.
1
1
1 1
f R R
= - +
é
ë
ê
ù
û
ú
( ) m
If lens faces becomes opposite three is no
change in radius of curvature hence focal
length does not change.
4. Using for mula
1 1 1
v u f
- = when u ® 0
1
v ® 0
when u f v ® ® ¥ , hence im age moves from
sur face to ¥ .
5.
1
1
1 1
1 2
f R R
= - -
é
ë
ê
ù
û
ú
( ) m
Þ
1
1
1 1 2
20 f R R
= -
-
-
é
ë
ê
ù
û
ú
= ´
-
( ) 1.3 0.3
Þ f = -
100
3
cm
(a) When immersed in a liquid of 1.8
refractive index
1
1
2 2
20
1
f R
= -
æ
è
ç
ö
ø
÷
-
é
ë
ê
ù
û
ú
=
-
´
- 1.3
1.8
0.5
1.8
f ¢ = 36 cm
(b) The minimum distance is equal to the
focal length = 36 cm
20
O
10 cm
6. Using
1 1 1
v u f
- =
1 1
20
12
10 v
-
-
=
( )
On solving v = 20 cm
Magnification = - = -
v
u
1
Hence the image of same size and inverted.
Let the distance between second lens is x
Since magnification is unity image distance
also x using again
1 1 1
v u f
- =
we get
1 1 1 1
10 x x f
-
-
= =
( )
Þ x = 20 cm
Hence the distance between two lenses
= + = 20 20 40 cm cm cm
7.
1 1 1
v f u
= + …(i)
1 1 1
1
v f u du
= +
+
…(ii)
1 1
v v
u du u
u duu ¢
- =
+ -
+
( )
( )
on solving, we get
v v
v
du
u u du
- ¢
¢
=
+ ( )
thickness dv
v
u
du =
-
2
2
10. Size of im age = ´ = 6
2
3
2 cm.
11. Let im age distance is u
| | m = 3 Þ v u = 3
1
3
1 1
12 u u
+ = Þ v = 16 cm
12. Since im age is up right and diminished
hence lens is concave. Now
u v - = 20 …(i)
m
v
u
= =
1
2
Þ
1
2
20
=
- u
u
Þ u = 40 cm and v = 20 cm
Þ Using
1 1 1
v u f
- =
- + =
1
20
1
40
1
f
Þ f = - 40 cm
13. The im age co in cide it self if light falls
nor mally on plane mir ror hence ob ject must
be on focus i e . . + 10 cm.
8.
1 1 2 2 1
2 1
2
2 1
1
v u R R
+ = -
- ( ) ( ) m /m m /m
1 1 2 4 3 2 4 3 1
v
+ = -
-
0.2 0.4 0.4
( / ) ( / )
On solving v = 12 cm
9. Since shift in po si tion D t = 0.1 m
Hence real depth = + ( ) 0.1 0.2 m
= 0.3 m
and apparent depth= 0.2 m
m =
real depth
apparent depth
= =
0.3
0.2
1.5
21
AIEEE Corner
1.
We have r i + ° + = ° 90 180 Þ r i = - 90
From Snell’s law 1.5 = =
-
sin
sin
sin
sin ( )
i
r
i
i 90
Þ tan i = 1.5
Þ         i =
-
tan ( )
1
1.5
2. n
v
u
w
w
= = =
air
0.229
343
1498
Critical angle q = = °
-
sin ( . )
1
0229 13.2
3. Sped in glycrine v
c
n
g
g
= =
´ 3 10
8
1.47
t
v
g
1
8
8
20 20
3 10
10 = =
´
´
= ´
-
1.47
9.8 s
Speed in glycrine v
c
n
g
g
= =
´ 3 10
8
1.63
t
v
c
2
8
8
20 20
3 10
10 = =
´
´
- ´
-
1.63
10.8
~
t t
2 1
8
1086 10 - = - ´
-
( . ) 9.8
= ´
-
1.67 s 10
8
4. (a) t
v
1
6
1
6
8
1 10 1 10
3 10
=
´
=
´
´
- -
m
1.2
m
/
=
´
´
-
1.2 10
3 10
6
8
Þ t
1
14
10 = ´
-
0.4 s
t
2
6
8
14
1 0
3 1 0
1 0 =
´
´
= ´
-
-
1.5
0.5 s
t
3
6
8
6
8
14
1 10
3 10
10
3 10
10 =
´
´
=
´
´
= ´
- -
-
/1.8
1.8
0.6
Hence t
1
is least and t
1
14
10 = ´
-
0.4 s
(b) Total number of wavelengths
= + +
1 1
1 2 3
m
l
m
l
m
l
m 1.5 m m
/ / / n n n
=
´
+
´ 1000
600
100
600
1.2 nm
nm
1.5 nm
nm
+
´ ´ 1 1000
600
1.8 nm
nm
= =
4500
600
7.5
5. The given wave equation is
E y t
x
( , ) =
´
- ´ ´
é
ë
ê
ù
û
ú
-
E
y
t
ax
sin
2
5 10
3 10 2
7
14
p
l
Comparing with standard equation
E y t E ky t
x
( , ) sin [ ] = -
0
w
k =
´
= ´ ´
-
2
5 10
2 3 10
7
14
p
, w p
v
k
= =
´ ´
´
= ´
-
w p
p
2 3 10
2 5 10
10
14
7
8
/
1.5 m/s
Refractive index n
c
v
= =
´
´
=
3 10
10
2
8
8
1.5
Wavelength in this way l
p
n
k
=
2
Þ l
p
n
=
´
= ´
-
2
25 5 10
5 10
7
7
/
m
Þ l
n
= 500 nm
If vacuum, wavelength is l then
l
l
n
n
=
Þ l l = = ´ = n
n
2 500 1000nm
6. Refraction from plane and spher i cal
surfaces
22
Reflected ray
Reflected ray
Incident ray
90–i
90°
i i
r
We have
sin
sin
60°
=
r
1.8
Þ                 sin
sin
r =
° 60
1.8
Þ sin r =
´
=
3
2 1.8
0.48
Þ        r =
-
sin ( )
1
0.48
Þ        r
~
- ° 28.7
Now
MO
r
6
= tan
Þ MO r = 6 tan
Similarly ON = 6 tan r
Þ MN MO ON r = + = = ° 12 12 tan tan( ) 28.7
Þ MN = 6.6 cm
7. From Snell’s law
4
3
45
=
° sin
sin r
Solving we get r = ° 32
EF DE r = = ° tan tan 3 32
= 1.88 m
Total length of shadow = + 1 1.88
= 2.88 m
8. The sit u a tion is shown in figure
For first surface
m m m m
2 1 2 1
v u R
- =
-
Þ
1.5
2.5)
0.5
v
-
-
=
1
10 (
Þ
1.5
2.5 v
= - = -
1
20
1 7
20
Þ     v = -
30
7
cm
This image acts as a virtual object for 2nd
surface
u
2
20
30
7
170
7
= - +
æ
è
ç
ö
ø
÷
= - cm
and   R = - 10 cm

m m m m
2 1 2 1
v u r
- =
-
Þ
1
170 7 10 v
+ =
-
-
1.5 0.5
/
Þ
1 1
20 170 v
= -
10.5
Þ    v = - 85 cm
Hence final image will produced at -65cm
from Ist surface.
9. Here v = - 1 cm
R = - 2 cm
Applying
m m m m
2 1 2 1
v u R
- =
-
Þ
1
1
1
2 2
1
4 -
- =
-
-
=
-
-
=
1.5 1.5 0.5
x
Þ
-
=
1.5
x
5
4
Þ  x =
-
= -
6
5
1.2 cm
23
2 cm
0.25 cm 10 cm
2 1
45°
45°
D
B
1 m
A
3 m
C E F
r
N
N
2
N
1
60°
r
r r
6 cm
N
3
M
O
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