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9 | 3739-3742 | For example, in ideal GaAs
the ratio of Ga:As is 1:1 but in Ga-rich or As-rich GaAs it could
respectively be Ga1 1 As0 9 or Ga0 9 As1 |
9 | 3740-3743 | 1 As0 9 or Ga0 9 As1 1 |
9 | 3741-3744 | 9 or Ga0 9 As1 1 In general, the presence of
defects control the properties of semiconductors in many ways |
9 | 3742-3745 | 9 As1 1 In general, the presence of
defects control the properties of semiconductors in many ways EXERCISES
14 |
9 | 3743-3746 | 1 In general, the presence of
defects control the properties of semiconductors in many ways EXERCISES
14 1
In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the
dopants |
9 | 3744-3747 | In general, the presence of
defects control the properties of semiconductors in many ways EXERCISES
14 1
In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the
dopants (b) Electrons are minority carriers and pentavalent atoms are the
dopants |
9 | 3745-3748 | EXERCISES
14 1
In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the
dopants (b) Electrons are minority carriers and pentavalent atoms are the
dopants (c) Holes are minority carriers and pentavalent atoms are the
dopants |
9 | 3746-3749 | 1
In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the
dopants (b) Electrons are minority carriers and pentavalent atoms are the
dopants (c) Holes are minority carriers and pentavalent atoms are the
dopants (d) Holes are majority carriers and trivalent atoms are the dopants |
9 | 3747-3750 | (b) Electrons are minority carriers and pentavalent atoms are the
dopants (c) Holes are minority carriers and pentavalent atoms are the
dopants (d) Holes are majority carriers and trivalent atoms are the dopants 14 |
9 | 3748-3751 | (c) Holes are minority carriers and pentavalent atoms are the
dopants (d) Holes are majority carriers and trivalent atoms are the dopants 14 2
Which of the statements given in Exercise 14 |
9 | 3749-3752 | (d) Holes are majority carriers and trivalent atoms are the dopants 14 2
Which of the statements given in Exercise 14 1 is true for p-type
semiconductos |
9 | 3750-3753 | 14 2
Which of the statements given in Exercise 14 1 is true for p-type
semiconductos 14 |
9 | 3751-3754 | 2
Which of the statements given in Exercise 14 1 is true for p-type
semiconductos 14 3
Carbon, silicon and germanium have four valence electrons each |
9 | 3752-3755 | 1 is true for p-type
semiconductos 14 3
Carbon, silicon and germanium have four valence electrons each These are characterised by valence and conduction bands separated
Rationalised 2023-24
Physics
342
by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge |
9 | 3753-3756 | 14 3
Carbon, silicon and germanium have four valence electrons each These are characterised by valence and conduction bands separated
Rationalised 2023-24
Physics
342
by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge Which
of the following statements is true |
9 | 3754-3757 | 3
Carbon, silicon and germanium have four valence electrons each These are characterised by valence and conduction bands separated
Rationalised 2023-24
Physics
342
by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge Which
of the following statements is true (a) (Eg)Si < (Eg)Ge < (Eg)C
(b) (Eg)C < (Eg)Ge > (Eg)Si
(c) (Eg)C > (Eg)Si > (Eg)Ge
(d) (Eg)C = (Eg)Si = (Eg)Ge
14 |
9 | 3755-3758 | These are characterised by valence and conduction bands separated
Rationalised 2023-24
Physics
342
by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge Which
of the following statements is true (a) (Eg)Si < (Eg)Ge < (Eg)C
(b) (Eg)C < (Eg)Ge > (Eg)Si
(c) (Eg)C > (Eg)Si > (Eg)Ge
(d) (Eg)C = (Eg)Si = (Eg)Ge
14 4
In an unbiased p-n junction, holes diffuse from the p-region to
n-region because
(a) free electrons in the n-region attract them |
9 | 3756-3759 | Which
of the following statements is true (a) (Eg)Si < (Eg)Ge < (Eg)C
(b) (Eg)C < (Eg)Ge > (Eg)Si
(c) (Eg)C > (Eg)Si > (Eg)Ge
(d) (Eg)C = (Eg)Si = (Eg)Ge
14 4
In an unbiased p-n junction, holes diffuse from the p-region to
n-region because
(a) free electrons in the n-region attract them (b) they move across the junction by the potential difference |
9 | 3757-3760 | (a) (Eg)Si < (Eg)Ge < (Eg)C
(b) (Eg)C < (Eg)Ge > (Eg)Si
(c) (Eg)C > (Eg)Si > (Eg)Ge
(d) (Eg)C = (Eg)Si = (Eg)Ge
14 4
In an unbiased p-n junction, holes diffuse from the p-region to
n-region because
(a) free electrons in the n-region attract them (b) they move across the junction by the potential difference (c) hole concentration in p-region is more as compared to n-region |
9 | 3758-3761 | 4
In an unbiased p-n junction, holes diffuse from the p-region to
n-region because
(a) free electrons in the n-region attract them (b) they move across the junction by the potential difference (c) hole concentration in p-region is more as compared to n-region (d) All the above |
9 | 3759-3762 | (b) they move across the junction by the potential difference (c) hole concentration in p-region is more as compared to n-region (d) All the above 14 |
9 | 3760-3763 | (c) hole concentration in p-region is more as compared to n-region (d) All the above 14 5
When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier |
9 | 3761-3764 | (d) All the above 14 5
When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier (b) reduces the majority carrier current to zero |
9 | 3762-3765 | 14 5
When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier (b) reduces the majority carrier current to zero (c) lowers the potential barrier |
9 | 3763-3766 | 5
When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier (b) reduces the majority carrier current to zero (c) lowers the potential barrier (d) None of the above |
9 | 3764-3767 | (b) reduces the majority carrier current to zero (c) lowers the potential barrier (d) None of the above 14 |
9 | 3765-3768 | (c) lowers the potential barrier (d) None of the above 14 6
In half-wave rectification, what is the output frequency if the input
frequency is 50 Hz |
9 | 3766-3769 | (d) None of the above 14 6
In half-wave rectification, what is the output frequency if the input
frequency is 50 Hz What is the output frequency of a full-wave rectifier
for the same input frequency |
9 | 3767-3770 | 14 6
In half-wave rectification, what is the output frequency if the input
frequency is 50 Hz What is the output frequency of a full-wave rectifier
for the same input frequency Rationalised 2023-24
Notes
Rationalised 2023-24
344
Physics
APPENDICES
APPENDIX A 1
THE GREEK ALPHABET
APPENDIX A 2
COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES
Rationalised 2023-24
345
Answers
Appendices
APPENDIX A 3
SOME IMPORTANT CONSTANTS
OTHER USEFUL CONSTANTS
Rationalised 2023-24
346
Physics
ANSWERS
CHAPTER 9
9 |
9 | 3768-3771 | 6
In half-wave rectification, what is the output frequency if the input
frequency is 50 Hz What is the output frequency of a full-wave rectifier
for the same input frequency Rationalised 2023-24
Notes
Rationalised 2023-24
344
Physics
APPENDICES
APPENDIX A 1
THE GREEK ALPHABET
APPENDIX A 2
COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES
Rationalised 2023-24
345
Answers
Appendices
APPENDIX A 3
SOME IMPORTANT CONSTANTS
OTHER USEFUL CONSTANTS
Rationalised 2023-24
346
Physics
ANSWERS
CHAPTER 9
9 1
v = –54 cm |
9 | 3769-3772 | What is the output frequency of a full-wave rectifier
for the same input frequency Rationalised 2023-24
Notes
Rationalised 2023-24
344
Physics
APPENDICES
APPENDIX A 1
THE GREEK ALPHABET
APPENDIX A 2
COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES
Rationalised 2023-24
345
Answers
Appendices
APPENDIX A 3
SOME IMPORTANT CONSTANTS
OTHER USEFUL CONSTANTS
Rationalised 2023-24
346
Physics
ANSWERS
CHAPTER 9
9 1
v = –54 cm The image is real, inverted and magnified |
9 | 3770-3773 | Rationalised 2023-24
Notes
Rationalised 2023-24
344
Physics
APPENDICES
APPENDIX A 1
THE GREEK ALPHABET
APPENDIX A 2
COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES
Rationalised 2023-24
345
Answers
Appendices
APPENDIX A 3
SOME IMPORTANT CONSTANTS
OTHER USEFUL CONSTANTS
Rationalised 2023-24
346
Physics
ANSWERS
CHAPTER 9
9 1
v = –54 cm The image is real, inverted and magnified The size of the
image is 5 |
9 | 3771-3774 | 1
v = –54 cm The image is real, inverted and magnified The size of the
image is 5 0 cm |
9 | 3772-3775 | The image is real, inverted and magnified The size of the
image is 5 0 cm As u ® f, v ® ¥; for u < f, image is virtual |
9 | 3773-3776 | The size of the
image is 5 0 cm As u ® f, v ® ¥; for u < f, image is virtual 9 |
9 | 3774-3777 | 0 cm As u ® f, v ® ¥; for u < f, image is virtual 9 2
v = 6 |
9 | 3775-3778 | As u ® f, v ® ¥; for u < f, image is virtual 9 2
v = 6 7 cm |
9 | 3776-3779 | 9 2
v = 6 7 cm Magnification = 5/9, i |
9 | 3777-3780 | 2
v = 6 7 cm Magnification = 5/9, i e |
9 | 3778-3781 | 7 cm Magnification = 5/9, i e , the size of the image is 2 |
9 | 3779-3782 | Magnification = 5/9, i e , the size of the image is 2 5 cm |
9 | 3780-3783 | e , the size of the image is 2 5 cm As
u ® ¥; v ® f (but never beyond) while m ® 0 |
9 | 3781-3784 | , the size of the image is 2 5 cm As
u ® ¥; v ® f (but never beyond) while m ® 0 9 |
9 | 3782-3785 | 5 cm As
u ® ¥; v ® f (but never beyond) while m ® 0 9 3
1 |
9 | 3783-3786 | As
u ® ¥; v ® f (but never beyond) while m ® 0 9 3
1 33; 1 |
9 | 3784-3787 | 9 3
1 33; 1 7 cm
9 |
9 | 3785-3788 | 3
1 33; 1 7 cm
9 4
nga = 1 |
9 | 3786-3789 | 33; 1 7 cm
9 4
nga = 1 51; nwa = 1 |
9 | 3787-3790 | 7 cm
9 4
nga = 1 51; nwa = 1 32; ngw = 1 |
9 | 3788-3791 | 4
nga = 1 51; nwa = 1 32; ngw = 1 144; which gives sin r = 0 |
9 | 3789-3792 | 51; nwa = 1 32; ngw = 1 144; which gives sin r = 0 6181 i |
9 | 3790-3793 | 32; ngw = 1 144; which gives sin r = 0 6181 i e |
9 | 3791-3794 | 144; which gives sin r = 0 6181 i e ,
r ~ 38° |
9 | 3792-3795 | 6181 i e ,
r ~ 38° 9 |
9 | 3793-3796 | e ,
r ~ 38° 9 5
r = 0 |
9 | 3794-3797 | ,
r ~ 38° 9 5
r = 0 8 × tan ic and sin
1/1 |
9 | 3795-3798 | 9 5
r = 0 8 × tan ic and sin
1/1 33
0 |
9 | 3796-3799 | 5
r = 0 8 × tan ic and sin
1/1 33
0 75
ci
=
≅
, where r is the radius (in m)
of the largest circle from which light comes out and ic is the critical
angle for water-air interface, Area = 2 |
9 | 3797-3800 | 8 × tan ic and sin
1/1 33
0 75
ci
=
≅
, where r is the radius (in m)
of the largest circle from which light comes out and ic is the critical
angle for water-air interface, Area = 2 6 m2
9 |
9 | 3798-3801 | 33
0 75
ci
=
≅
, where r is the radius (in m)
of the largest circle from which light comes out and ic is the critical
angle for water-air interface, Area = 2 6 m2
9 6
n ≅ 1 |
9 | 3799-3802 | 75
ci
=
≅
, where r is the radius (in m)
of the largest circle from which light comes out and ic is the critical
angle for water-air interface, Area = 2 6 m2
9 6
n ≅ 1 53 and Dm for prism in water ≅ 10°
9 |
9 | 3800-3803 | 6 m2
9 6
n ≅ 1 53 and Dm for prism in water ≅ 10°
9 7
R = 22 cm
9 |
9 | 3801-3804 | 6
n ≅ 1 53 and Dm for prism in water ≅ 10°
9 7
R = 22 cm
9 8
Here the object is virtual and the image is real |
9 | 3802-3805 | 53 and Dm for prism in water ≅ 10°
9 7
R = 22 cm
9 8
Here the object is virtual and the image is real u = +12 cm (object on
right; virtual)
(a)
f = +20 cm |
9 | 3803-3806 | 7
R = 22 cm
9 8
Here the object is virtual and the image is real u = +12 cm (object on
right; virtual)
(a)
f = +20 cm Image is real and at 7 |
9 | 3804-3807 | 8
Here the object is virtual and the image is real u = +12 cm (object on
right; virtual)
(a)
f = +20 cm Image is real and at 7 5 cm from the lens on its right
side |
9 | 3805-3808 | u = +12 cm (object on
right; virtual)
(a)
f = +20 cm Image is real and at 7 5 cm from the lens on its right
side (b)
f = –16 cm |
9 | 3806-3809 | Image is real and at 7 5 cm from the lens on its right
side (b)
f = –16 cm Image is real and at 48 cm from the lens on its right side |
9 | 3807-3810 | 5 cm from the lens on its right
side (b)
f = –16 cm Image is real and at 48 cm from the lens on its right side 9 |
9 | 3808-3811 | (b)
f = –16 cm Image is real and at 48 cm from the lens on its right side 9 9
v = 8 |
9 | 3809-3812 | Image is real and at 48 cm from the lens on its right side 9 9
v = 8 4 cm, image is erect and virtual |
9 | 3810-3813 | 9 9
v = 8 4 cm, image is erect and virtual It is diminished to a size
1 |
9 | 3811-3814 | 9
v = 8 4 cm, image is erect and virtual It is diminished to a size
1 8 cm |
9 | 3812-3815 | 4 cm, image is erect and virtual It is diminished to a size
1 8 cm As u ® ¥, v ® f (but never beyond f while m ® 0) |
9 | 3813-3816 | It is diminished to a size
1 8 cm As u ® ¥, v ® f (but never beyond f while m ® 0) Note that when the object is placed at the focus of the concave lens
(21 cm), the image is located at 10 |
9 | 3814-3817 | 8 cm As u ® ¥, v ® f (but never beyond f while m ® 0) Note that when the object is placed at the focus of the concave lens
(21 cm), the image is located at 10 5 cm (not at infinity as one might
wrongly think) |
9 | 3815-3818 | As u ® ¥, v ® f (but never beyond f while m ® 0) Note that when the object is placed at the focus of the concave lens
(21 cm), the image is located at 10 5 cm (not at infinity as one might
wrongly think) 9 |
9 | 3816-3819 | Note that when the object is placed at the focus of the concave lens
(21 cm), the image is located at 10 5 cm (not at infinity as one might
wrongly think) 9 10
A diverging lens of focal length 60 cm
9 |
9 | 3817-3820 | 5 cm (not at infinity as one might
wrongly think) 9 10
A diverging lens of focal length 60 cm
9 11
(a)
ve = –25 cm and fe = 6 |
9 | 3818-3821 | 9 10
A diverging lens of focal length 60 cm
9 11
(a)
ve = –25 cm and fe = 6 25 cm give ue = –5 cm; vO = (15 – 5) cm =
10 cm,
fO = uO = – 2 |
9 | 3819-3822 | 10
A diverging lens of focal length 60 cm
9 11
(a)
ve = –25 cm and fe = 6 25 cm give ue = –5 cm; vO = (15 – 5) cm =
10 cm,
fO = uO = – 2 5 cm; Magnifying power = 20
(b)
uO = – 2 |
9 | 3820-3823 | 11
(a)
ve = –25 cm and fe = 6 25 cm give ue = –5 cm; vO = (15 – 5) cm =
10 cm,
fO = uO = – 2 5 cm; Magnifying power = 20
(b)
uO = – 2 59 cm |
9 | 3821-3824 | 25 cm give ue = –5 cm; vO = (15 – 5) cm =
10 cm,
fO = uO = – 2 5 cm; Magnifying power = 20
(b)
uO = – 2 59 cm Magnifying power = 13 |
9 | 3822-3825 | 5 cm; Magnifying power = 20
(b)
uO = – 2 59 cm Magnifying power = 13 5 |
9 | 3823-3826 | 59 cm Magnifying power = 13 5 9 |
9 | 3824-3827 | Magnifying power = 13 5 9 12
Angular magnification of the eye-piece for image at 25 cm
25
2 5
1
11 |
9 | 3825-3828 | 5 9 12
Angular magnification of the eye-piece for image at 25 cm
25
2 5
1
11 ; |
| |
9 | 3826-3829 | 9 12
Angular magnification of the eye-piece for image at 25 cm
25
2 5
1
11 ; |
| 25 cm
2 27cm
11
ue
=
=
; vO = 7 |
9 | 3827-3830 | 12
Angular magnification of the eye-piece for image at 25 cm
25
2 5
1
11 ; |
| 25 cm
2 27cm
11
ue
=
=
; vO = 7 2 cm
Separation = 9 |
9 | 3828-3831 | ; |
| 25 cm
2 27cm
11
ue
=
=
; vO = 7 2 cm
Separation = 9 47 cm; Magnifying power = 88
9 |
9 | 3829-3832 | 25 cm
2 27cm
11
ue
=
=
; vO = 7 2 cm
Separation = 9 47 cm; Magnifying power = 88
9 13
24; 150 cm
9 |
9 | 3830-3833 | 2 cm
Separation = 9 47 cm; Magnifying power = 88
9 13
24; 150 cm
9 14
(a)
Angular magnification = 1500
(b)
Diameter of the image = 13 |
9 | 3831-3834 | 47 cm; Magnifying power = 88
9 13
24; 150 cm
9 14
(a)
Angular magnification = 1500
(b)
Diameter of the image = 13 7 cm |
9 | 3832-3835 | 13
24; 150 cm
9 14
(a)
Angular magnification = 1500
(b)
Diameter of the image = 13 7 cm Rationalised 2023-24
347
Answers
9 |
9 | 3833-3836 | 14
(a)
Angular magnification = 1500
(b)
Diameter of the image = 13 7 cm Rationalised 2023-24
347
Answers
9 15
Apply mirror equation and the condition:
(a)
f < 0 (concave mirror); u < 0 (object on left)
(b)
f > 0; u < 0
(c)
f > 0 (convex mirror) and u < 0
(d)
f < 0 (concave mirror); f < u < 0
to deduce the desired result |
9 | 3834-3837 | 7 cm Rationalised 2023-24
347
Answers
9 15
Apply mirror equation and the condition:
(a)
f < 0 (concave mirror); u < 0 (object on left)
(b)
f > 0; u < 0
(c)
f > 0 (convex mirror) and u < 0
(d)
f < 0 (concave mirror); f < u < 0
to deduce the desired result 9 |
9 | 3835-3838 | Rationalised 2023-24
347
Answers
9 15
Apply mirror equation and the condition:
(a)
f < 0 (concave mirror); u < 0 (object on left)
(b)
f > 0; u < 0
(c)
f > 0 (convex mirror) and u < 0
(d)
f < 0 (concave mirror); f < u < 0
to deduce the desired result 9 16
The pin appears raised by 5 |
9 | 3836-3839 | 15
Apply mirror equation and the condition:
(a)
f < 0 (concave mirror); u < 0 (object on left)
(b)
f > 0; u < 0
(c)
f > 0 (convex mirror) and u < 0
(d)
f < 0 (concave mirror); f < u < 0
to deduce the desired result 9 16
The pin appears raised by 5 0 cm |
9 | 3837-3840 | 9 16
The pin appears raised by 5 0 cm It can be seen with an explicit ray
diagram that the answer is independent of the location of the slab
(for small angles of incidence) |
9 | 3838-3841 | 16
The pin appears raised by 5 0 cm It can be seen with an explicit ray
diagram that the answer is independent of the location of the slab
(for small angles of incidence) 9 |
Subsets and Splits