DC Pandey Solutions: Refraction of Light- 1

# DC Pandey Solutions: Refraction of Light- 1 | DC Pandey Solutions for NEET Physics PDF Download

``` 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
```

122 docs

## FAQs on DC Pandey Solutions: Refraction of Light- 1 - DC Pandey Solutions for NEET Physics

 1. What is refraction of light?
Ans. Refraction of light is the bending of light as it passes from one medium to another, caused by a change in its speed. This bending of light occurs due to the change in the refractive index of the mediums.
 2. How does refraction of light occur?
Ans. When light passes from one medium to another with a different refractive index, its speed changes. Due to this change in speed, the direction of light also changes, resulting in the phenomenon of refraction.
 3. What are some examples of refraction of light in everyday life?
Ans. Some examples of refraction of light in everyday life include the bending of a pencil in a glass of water, the apparent bending of a straw in a glass, the formation of a rainbow, and the apparent bending of a stick when partially submerged in water.
 4. How does the refractive index affect the refraction of light?
Ans. The refractive index of a medium determines the degree of bending experienced by light when it passes from one medium to another. A higher refractive index means greater bending of light, while a lower refractive index means less bending of light.
 5. What are some applications of refraction of light?
Ans. Refraction of light finds applications in various fields. Some common applications include the formation of lenses for glasses, telescopes, and microscopes, the functioning of fiber optics in telecommunications, and the correction of vision problems through the use of contact lenses or refractive surgeries like LASIK.

## DC Pandey Solutions for NEET Physics

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