Chapter 20 : Dispersion and Spectra - HC Verma Solution, Physics Class 11 Notes | EduRev

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JEE : Chapter 20 : Dispersion and Spectra - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


20.1
SOLUTIONS TO CONCEPTS
CHAPTER – 20
1. Given that,
Refractive index of flint glass = ?
f
= 1.620
Refractive index of crown glass = ?
c
= 1.518
Refracting angle of flint prism = A
f
= 6.0°
For zero net deviation of mean ray 
( ?
f 
– 1)A
f
= ( ?
c
– 1) A
c
? A
c
= 
f
c
f
A
1
1
? ?
? ?
= ?
?
?
) 0 . 6 (
1 518 . 1
1 620 . 1
= 7.2°
2. Given that
?
r
= 1.56, ?
y
= 1.60, and ?
v
= 1.68
(a) Dispersive power = ? = 
1
y
r v
? ?
? ? ?
= 
1 60 . 1
56 . 1 68 . 1
?
?
= 0.2
(b) Angular dispersion = ( ?
v
- ?
r
)A = 0.12 × 6° = 7.2° ?
3. The focal length of a lens is given by
f
1
= ( ? – 1) 
?
?
?
?
?
?
?
?
?
2 1
R
1
R
1
? ( ? – 1) = 
?
?
?
?
?
?
?
?
?
?
2 1
R
1
R
1
1
f
1
  = 
f
K
…(1)
So, ?
r
– 1 = 
100
K
…(2)
?
y
– 1 = 
98
K
…(3)
And ?
v
– 1 = 
96
K
(4)
So, Dispersive power = ??=
1
y
r v
? ?
? ? ?
=
) 1 (
) 1 ( ) 1 (
y
r v
? ?
? ? ? ? ?
= 
98
K
100
K
96
K
?
= 
9600
4 98 ?
?= 0.0408 ?
4. Given that, ?
v
– ?
r
= 0.014
Again, ?
y
= 
depth Apparent
depth al Re
= 
30 . 1
00 . 2
= 1.515
So, dispersive power = 
1
y
r v
? ?
? ? ?
= 
1 515 . 1
014 . 0
?
= 0.027
5. Given that, ?
r
= 1.61, ?
v
= 1.65, ? = 0.07 and ?
y
= 4°
Now, ? = 
1
y
r v
? ?
? ? ?
?
? 0.07 = 
1
61 . 1 65 . 1
y
? ?
?
?
? ?
y
– 1 = 
07 . 0
04 . 0
= 
7
4
Again, ? = ( ? – 1) A 
? A = 
1
y
y
? ?
?
= 
) 7 / 4 (
4
= 7° 
2cm
Image
Object
1.32cm
Page 2


20.1
SOLUTIONS TO CONCEPTS
CHAPTER – 20
1. Given that,
Refractive index of flint glass = ?
f
= 1.620
Refractive index of crown glass = ?
c
= 1.518
Refracting angle of flint prism = A
f
= 6.0°
For zero net deviation of mean ray 
( ?
f 
– 1)A
f
= ( ?
c
– 1) A
c
? A
c
= 
f
c
f
A
1
1
? ?
? ?
= ?
?
?
) 0 . 6 (
1 518 . 1
1 620 . 1
= 7.2°
2. Given that
?
r
= 1.56, ?
y
= 1.60, and ?
v
= 1.68
(a) Dispersive power = ? = 
1
y
r v
? ?
? ? ?
= 
1 60 . 1
56 . 1 68 . 1
?
?
= 0.2
(b) Angular dispersion = ( ?
v
- ?
r
)A = 0.12 × 6° = 7.2° ?
3. The focal length of a lens is given by
f
1
= ( ? – 1) 
?
?
?
?
?
?
?
?
?
2 1
R
1
R
1
? ( ? – 1) = 
?
?
?
?
?
?
?
?
?
?
2 1
R
1
R
1
1
f
1
  = 
f
K
…(1)
So, ?
r
– 1 = 
100
K
…(2)
?
y
– 1 = 
98
K
…(3)
And ?
v
– 1 = 
96
K
(4)
So, Dispersive power = ??=
1
y
r v
? ?
? ? ?
=
) 1 (
) 1 ( ) 1 (
y
r v
? ?
? ? ? ? ?
= 
98
K
100
K
96
K
?
= 
9600
4 98 ?
?= 0.0408 ?
4. Given that, ?
v
– ?
r
= 0.014
Again, ?
y
= 
depth Apparent
depth al Re
= 
30 . 1
00 . 2
= 1.515
So, dispersive power = 
1
y
r v
? ?
? ? ?
= 
1 515 . 1
014 . 0
?
= 0.027
5. Given that, ?
r
= 1.61, ?
v
= 1.65, ? = 0.07 and ?
y
= 4°
Now, ? = 
1
y
r v
? ?
? ? ?
?
? 0.07 = 
1
61 . 1 65 . 1
y
? ?
?
?
? ?
y
– 1 = 
07 . 0
04 . 0
= 
7
4
Again, ? = ( ? – 1) A 
? A = 
1
y
y
? ?
?
= 
) 7 / 4 (
4
= 7° 
2cm
Image
Object
1.32cm
Chapter 20
20.2
6. Given that, ?
r
= 38.4°, ?
y
= 38.7° and ?
v
= 39.2°
Dispersive power = 
1
y
r v
? ?
? ? ?
= 
) 1 (
) 1 ( ) 1 (
y
r v
? ?
? ? ? ? ?
= 
?
?
?
?
?
? ?
?
?
?
?
?
? ?
? ?
?
?
?
?
? ?
A
A A
v
r v
[ ? ? = ( ? – 1) A]
= 
y
r v
?
? ? ?
= 
7 . 38
4 . 38 2 . 39 ?
= 0.0204
7. Two prisms of identical geometrical shape are combined.
Let A = Angle of the prisms
??
v
= 1.52 and ?
v
= 1.62, ?
v
= 1°
?
v
= ( ?
v
– 1)A – ( ??
v
– 1) A [since A = A ?]
? ?
v
= ( ?
v
– ??
v
)A 
? A = 
v v
v
? ? ? ?
?
= 
52 . 1 62 . 1
1
?
= 10°
8. Total deviation for yellow ray produced by the prism combination is
?
y
= ?
cy
– ?
fy
+ ?
cy
= 2 ??
cy
– ?
fy
= 2( ?
cy
– 1)A – ( ?
cy
– 1)A ?
Similarly the angular dispersion produced by the combination is 
?
v
– ?
r
= [( ?
vc
– 1)A – ( ?
vf
– 1)A ? + ( ?
vc
– 1)A] – [( ?
rc
– 1) A – ( ?
rf
– 1)A ? + ( ?
r
– 1) A)]
= 2( ?
vc
– 1)A – ( ?
vf
– 1)A ?
(a) For net angular dispersion to be zero,
?
v
– ?
r
= 0
? 2( ?
vc
– 1)A = ( ?
vf
– 1)A ??
?
A
A ?
= 
) (
) ( 2
rf vf
rc cv
? ? ?
? ? ?
= 
) (
) ( 2
r v
r v
? ? ? ? ?
? ? ?
(b) For net deviation in the yellow ray to be zero,
?
y
= 0
? 2( ?
cy
– 1)A = ( ?
fy
– 1)A ?
?
A
A ?
= 
) 1 (
) 1 ( 2
fy
cy
? ?
? ?
=  
) 1 (
) 1 ( 2
y
y
? ? ?
? ?
?
9. Given that, ?
cr
= 1.515, ?
cv
= 1.525 and ?
fr
= 1.612, ?
fv
= 1.632  and A = 5°
Since, they are similarly directed, the total deviation produced is given by,
? = ?
c
+ ?
r
= ( ?
c
– 1)A + ( ?
r
– 1) A = ( ?
c
+ ?
r
– 2)A
So, angular dispersion of the combination is given by,
?
v
– ?
y
= ( ?
cv
+ ?
fv
– 2)A – ( ?
cr
+ ?
fr
– 2)A 
= ( ?
cv
+ ?
fv
– ?
cr
– ?
fr
)A = (1.525 + 1.632 – 1.515 – 1.612)  5 = 0.15° ?
10. Given that, A ? = 6°, ?? = 0.07, ??
y
= 1.50
A = ? ? = 0.08, ?
y
= 1.60
The combination produces no deviation in the mean ray.
(a) ?
y
= ( ?
y
– 1)A – ( ??
y
– 1)A ? = 0 [Prism must be oppositely directed] 
? (1.60 – 1)A = ((1.50 – 1)A ?
? A = 
60 . 0
6 50 . 0 ? ?
= 5° ?
(b) When a beam of white light passes through it,
Net angular dispersion = ( ?
y
– 1) ?A – ( ??
y
– 1) ??A ?
? (1.60 – 1)(0.08)(5°) –  (1.50 –  1)(0.07)(6°)
? 0.24° – 0.21° = 0.03°
(c) If the prisms are similarly directed, 
?
y
= ( ?
y
– 1)A + ( ??
y
– 1)A
= (1.60 – 1)5° + (1.50 – 1)6° = 3° + 3° = 6° ?
(d) Similarly, if the prisms are similarly directed, the net angular dispersion is given by,
?
v
– ?
r
= ( ?
y
– 1) ?A – ( ??
y
– 1) ??A ? = 0.24° + 0.21° = 0.45° ?
5°
6°
5° 6°
5° 5°
Prism2
Prism1
Crown Crown
A A
A ?
Flint
Page 3


20.1
SOLUTIONS TO CONCEPTS
CHAPTER – 20
1. Given that,
Refractive index of flint glass = ?
f
= 1.620
Refractive index of crown glass = ?
c
= 1.518
Refracting angle of flint prism = A
f
= 6.0°
For zero net deviation of mean ray 
( ?
f 
– 1)A
f
= ( ?
c
– 1) A
c
? A
c
= 
f
c
f
A
1
1
? ?
? ?
= ?
?
?
) 0 . 6 (
1 518 . 1
1 620 . 1
= 7.2°
2. Given that
?
r
= 1.56, ?
y
= 1.60, and ?
v
= 1.68
(a) Dispersive power = ? = 
1
y
r v
? ?
? ? ?
= 
1 60 . 1
56 . 1 68 . 1
?
?
= 0.2
(b) Angular dispersion = ( ?
v
- ?
r
)A = 0.12 × 6° = 7.2° ?
3. The focal length of a lens is given by
f
1
= ( ? – 1) 
?
?
?
?
?
?
?
?
?
2 1
R
1
R
1
? ( ? – 1) = 
?
?
?
?
?
?
?
?
?
?
2 1
R
1
R
1
1
f
1
  = 
f
K
…(1)
So, ?
r
– 1 = 
100
K
…(2)
?
y
– 1 = 
98
K
…(3)
And ?
v
– 1 = 
96
K
(4)
So, Dispersive power = ??=
1
y
r v
? ?
? ? ?
=
) 1 (
) 1 ( ) 1 (
y
r v
? ?
? ? ? ? ?
= 
98
K
100
K
96
K
?
= 
9600
4 98 ?
?= 0.0408 ?
4. Given that, ?
v
– ?
r
= 0.014
Again, ?
y
= 
depth Apparent
depth al Re
= 
30 . 1
00 . 2
= 1.515
So, dispersive power = 
1
y
r v
? ?
? ? ?
= 
1 515 . 1
014 . 0
?
= 0.027
5. Given that, ?
r
= 1.61, ?
v
= 1.65, ? = 0.07 and ?
y
= 4°
Now, ? = 
1
y
r v
? ?
? ? ?
?
? 0.07 = 
1
61 . 1 65 . 1
y
? ?
?
?
? ?
y
– 1 = 
07 . 0
04 . 0
= 
7
4
Again, ? = ( ? – 1) A 
? A = 
1
y
y
? ?
?
= 
) 7 / 4 (
4
= 7° 
2cm
Image
Object
1.32cm
Chapter 20
20.2
6. Given that, ?
r
= 38.4°, ?
y
= 38.7° and ?
v
= 39.2°
Dispersive power = 
1
y
r v
? ?
? ? ?
= 
) 1 (
) 1 ( ) 1 (
y
r v
? ?
? ? ? ? ?
= 
?
?
?
?
?
? ?
?
?
?
?
?
? ?
? ?
?
?
?
?
? ?
A
A A
v
r v
[ ? ? = ( ? – 1) A]
= 
y
r v
?
? ? ?
= 
7 . 38
4 . 38 2 . 39 ?
= 0.0204
7. Two prisms of identical geometrical shape are combined.
Let A = Angle of the prisms
??
v
= 1.52 and ?
v
= 1.62, ?
v
= 1°
?
v
= ( ?
v
– 1)A – ( ??
v
– 1) A [since A = A ?]
? ?
v
= ( ?
v
– ??
v
)A 
? A = 
v v
v
? ? ? ?
?
= 
52 . 1 62 . 1
1
?
= 10°
8. Total deviation for yellow ray produced by the prism combination is
?
y
= ?
cy
– ?
fy
+ ?
cy
= 2 ??
cy
– ?
fy
= 2( ?
cy
– 1)A – ( ?
cy
– 1)A ?
Similarly the angular dispersion produced by the combination is 
?
v
– ?
r
= [( ?
vc
– 1)A – ( ?
vf
– 1)A ? + ( ?
vc
– 1)A] – [( ?
rc
– 1) A – ( ?
rf
– 1)A ? + ( ?
r
– 1) A)]
= 2( ?
vc
– 1)A – ( ?
vf
– 1)A ?
(a) For net angular dispersion to be zero,
?
v
– ?
r
= 0
? 2( ?
vc
– 1)A = ( ?
vf
– 1)A ??
?
A
A ?
= 
) (
) ( 2
rf vf
rc cv
? ? ?
? ? ?
= 
) (
) ( 2
r v
r v
? ? ? ? ?
? ? ?
(b) For net deviation in the yellow ray to be zero,
?
y
= 0
? 2( ?
cy
– 1)A = ( ?
fy
– 1)A ?
?
A
A ?
= 
) 1 (
) 1 ( 2
fy
cy
? ?
? ?
=  
) 1 (
) 1 ( 2
y
y
? ? ?
? ?
?
9. Given that, ?
cr
= 1.515, ?
cv
= 1.525 and ?
fr
= 1.612, ?
fv
= 1.632  and A = 5°
Since, they are similarly directed, the total deviation produced is given by,
? = ?
c
+ ?
r
= ( ?
c
– 1)A + ( ?
r
– 1) A = ( ?
c
+ ?
r
– 2)A
So, angular dispersion of the combination is given by,
?
v
– ?
y
= ( ?
cv
+ ?
fv
– 2)A – ( ?
cr
+ ?
fr
– 2)A 
= ( ?
cv
+ ?
fv
– ?
cr
– ?
fr
)A = (1.525 + 1.632 – 1.515 – 1.612)  5 = 0.15° ?
10. Given that, A ? = 6°, ?? = 0.07, ??
y
= 1.50
A = ? ? = 0.08, ?
y
= 1.60
The combination produces no deviation in the mean ray.
(a) ?
y
= ( ?
y
– 1)A – ( ??
y
– 1)A ? = 0 [Prism must be oppositely directed] 
? (1.60 – 1)A = ((1.50 – 1)A ?
? A = 
60 . 0
6 50 . 0 ? ?
= 5° ?
(b) When a beam of white light passes through it,
Net angular dispersion = ( ?
y
– 1) ?A – ( ??
y
– 1) ??A ?
? (1.60 – 1)(0.08)(5°) –  (1.50 –  1)(0.07)(6°)
? 0.24° – 0.21° = 0.03°
(c) If the prisms are similarly directed, 
?
y
= ( ?
y
– 1)A + ( ??
y
– 1)A
= (1.60 – 1)5° + (1.50 – 1)6° = 3° + 3° = 6° ?
(d) Similarly, if the prisms are similarly directed, the net angular dispersion is given by,
?
v
– ?
r
= ( ?
y
– 1) ?A – ( ??
y
– 1) ??A ? = 0.24° + 0.21° = 0.45° ?
5°
6°
5° 6°
5° 5°
Prism2
Prism1
Crown Crown
A A
A ?
Flint
Chapter 20
20.3
11. Given that, ??
v
– ??
r
= 0.014 and ?
v
– ?
r
= 0.024
A ? = 5.3° and A = 3.7°
(a) When the prisms are oppositely directed, 
angular dispersion = ( ?
v
– ?
r
)A – ( ??
v
– ??
r
)A ?
= 0.024 × 3.7° – 0.014 × 5.3° = 0.0146°
(b) When they are similarly directed, 
angular dispersion = ( ?
v
– ?
r
)A + ( ??
v
– ??
r
)A ?
= 0.024 × 3.7° + 0.014 × 5.3° = 0.163°
? ? ? ? ?
3.7° 5.3°
5.3°
3.7°
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