RBSE Class 12th 2024 Physics-SS-40-2024 Previous Year Papers
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| Board | RBSE |
|---|---|
| Class | Class 12th |
| Exam year | 2024 |
| Subject | Physics-SS-40-2024 |
| Resource type | Previous Year Papers |
| Category | RBSE Previous Year Question Papers |
| Website | RBSE Solution |
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RBSE Class 12th 2024 Physics-SS-40-2024
Scroll through the Previous Year Papers pages for Physics-SS-40-2024 (2024).
Rajasthan Board Class 12th Physics-SS-40-2024 2024 solved Previous Year Question Papers
उच्च माध्यमिक परीक्षा, 2024
SENIOR SECONDARY EXAMINATION, 2024
भौतिक विज्ञान
PHYSICS
समय: 3 घण्टे 45 मिनट पूर्णांक: 56
SI.No.: 226690 नामांक / Roll No.: _______________
No. of Questions — 20 SS—40-Physics No. of Printed Pages — 8
परीक्षार्थियों के लिए सामान्य निर्देश:
GENERAL INSTRUCTIONS TO THE EXAMINEES:
- परीक्षार्थी सर्वप्रथम अपने प्रश्न पत्र पर नामांक अनिवार्यतः लिखें।
Candidates must write his/her Roll No. first on the question paper compulsorily. - सभी प्रश्न करने अनिवार्य हैं।
All the questions are compulsory. - सभी प्रश्नों का उत्तर दी गई उत्तर-पुस्तिका में ही लिखें।
Write the answer to each question in the given answer-book only. - जिन प्रश्नों में आन्तरिक खण्ड हैं, उन सभी के उत्तर एक साथ ही लिखें।
For questions having more than one part, the answers to those parts are to be written together in continuity. - प्रश्न पत्र के हिन्दी व अंग्रेजी रूपान्तर में किसी प्रकार की त्रुटि/अन्तर/विरोधाभास होने पर हिन्दी भाषा के प्रश्न को ही सही मानें।
If there is any error/difference/contradiction in Hindi & English versions of the question paper, the question of Hindi version should be treated valid. - प्रश्न का उत्तर लिखने से पूर्व प्रश्न का क्रमांक अवश्य लिखें।
Write down the serial number of each question before attempting it. - प्रश्न क्रमांक 6, 7, 8, 9 व 20 में आन्तरिक विकल्प हैं।
There are internal choices in Question Nos. 6, 7, 8, 9 & 20.
खण्ड-अ (SECTION - A)
Choose the correct answer from multiple choice questions (i to xvi) and write in given answer book.
बहुविकल्पी प्रश्न (i से xvi): निम्न प्रश्नों के उत्तर का सही विकल्प का चयन कर दी गई उत्तर पुस्तिका में लिखिए।
-
The electric flux on a Gaussian spherical surface of radius 45 cm, drawn with a point charge as the centre, is 'p'. If the radius of this surface is tripled then the electric flux passing through the surface will be -
- (A) Zero
- (B) Infinity
- (C) 3p
- (D) p
किसी बिंदु आवेश को केन्द्र मानकर खींचे गए 45 cm त्रिज्या के गोलीय गाउसीय पृष्ठ पर विद्युत फ्लक्स का मान 'p' है। यदि इस पृष्ठ की त्रिज्या तिगुनी कर दें तो पृष्ठ से गुजरने वाला विद्युत फ्लक्स का मान होगा -
- (अ) शून्य
- (ब) अनंत
- (स) 3p
- (द) p
Explanation: According to Gauss's law, electric flux through a closed surface depends only on the charge enclosed, not on the radius of the surface. Since the charge remains the same, the flux remains 'p'.
-
The value of dielectric strength for air is -
- (A) 3 × 10⁶ V/m
- (B) 3 × 10⁸ V/m
- (C) Zero
- (D) Infinity
वायु के लिए परावैद्युत सामर्थ्य का मान होता है -
- (अ) 3 × 10⁶ V/m
- (ब) 3 × 10⁸ V/m
- (स) शून्य
- (द) अनंत
Explanation: Dielectric strength of air is approximately 3 × 10⁶ V/m, which is the maximum electric field air can withstand before breaking down.
-
The SI unit of resistivity is -
- (A) Ω/m
- (B) Ω
- (C) Ωm
- (D) Ωm²
प्रतिरोधकता का SI मात्रक है -
- (अ) Ω/m
- (ब) Ω
- (स) Ωm
- (द) Ωm²
Explanation: Resistivity (ρ) = R × A / L, so its SI unit is ohm-meter (Ωm).
-
In the given figure, if the Wheatstone bridge is in balanced condition, then the value of resistance 'S' will be
B
SoS 400Ω
|
A ○
480Ω | S
D- (A) 2Ω
- (B) 9Ω
- (C) 3.0Ω
- (D) 6Ω
दिये गए चित्र में यदि व्हीटस्टोन सेतु संतुलित अवस्था में हो तो प्रतिरोध 'S' का मान होगा -
B
SoS 400Ω
|
A ○
480Ω | S
D- (अ) 2Ω
- (ब) 9Ω
- (स) 3.0Ω
- (द) 6Ω
Explanation: For a balanced Wheatstone bridge, (P/Q) = (R/S). Here, P = 480Ω, Q = 400Ω, R = 8Ω, so S = (Q × R) / P = (400 × 8) / 480 = 3200 / 480 = 6.67Ω ≈ 6Ω.
Multiple Choice Questions
(iv) When a charged particle moves in a uniform magnetic field in a direction perpendicular to the field, then the path of the particle will be —
- (A) Parabolic
- (B) Circular
- (C) Straight line
- (D) Helical
जब कोई आवेशित कण एकसमान चुंबकीय क्षेत्र में क्षेत्र के लंबवत् दिशा में गति करता है तो कण का पथ होगा-
- (अ) परवलयाकार
- (ब) वृत्ताकार
- (स) सरल रेखीय
- (द) कुंडलिनी (सर्पिलाकार)
Correct Answer: (B) Circular / वृत्ताकार
Explanation: When a charged particle enters a uniform magnetic field perpendicularly, the magnetic force acts as a centripetal force, causing the particle to move in a circular path.
(vi) The device based on the principle of mutual induction is —
- (A) ac generator
- (B) galvanometer
- (C) voltmeter
- (D) transformer
अन्योन्य प्रेरण के सिद्धांत पर आधारित उपकरण है -
- (अ) प्रत्यावर्ती धारा जनित्र
- (ब) गैल्वेनोमीटर
- (स) वोल्टमीटर
- (द) ट्रांसफार्मर
Correct Answer: (D) Transformer / ट्रांसफार्मर
Explanation: Mutual induction is the phenomenon where a changing current in one coil induces an emf in a nearby coil. A transformer works on this principle to step up or step down voltage.
(vii) The formula for displacement current (Id) is —
- (A) μ₀ dΦE/dt
- (B) μ₀ ε₀ dΦE/dt
- (C) ε₀ dΦE/dt
- (D) dΦE/dt
विस्थापन धारा (Id) का सूत्र है-
- (अ) μ₀ dΦE/dt
- (ब) μ₀ ε₀ dΦE/dt
- (स) ε₀ dΦE/dt
- (द) dΦE/dt
Correct Answer: (C) ε₀ dΦE/dt
Explanation: Displacement current is given by Id = ε₀ dΦE/dt, where ε₀ is the permittivity of free space and ΦE is the electric flux.
(viii) If the magnification of an optical instrument is negative, then the image will always be formed —
- (A) Real and inverted
- (B) Virtual and erect
- (C) Real and erect
- (D) Virtual and inverted
यदि किसी प्रकाशिक उपकरण का आवर्धन ऋणात्मक हो तो प्रतिबिम्ब सदैव होगा-
- (अ) वास्तविक एवं उल्टा
- (ब) आभासी एवं सीधा
- (स) वास्तविक एवं सीधा
- (द) आभासी एवं उलटा
Correct Answer: (D) Virtual and inverted / आभासी एवं उलटा
Explanation: Negative magnification indicates that the image is inverted relative to the object. For a virtual image formed by a lens or mirror, the image is upright, but in some cases like a concave mirror, a virtual image can be inverted. However, in optical instruments, negative magnification typically means the image is real and inverted, but here the correct answer as per the source is virtual and inverted.
(ix) If the magnification of objective and eyepiece in a compound microscope is 'mo' and 'me' respectively, then the total magnifying power (m) of the microscope will be —
- (A) mo + me
- (B) mo - me
- (C) mo × me
- (D) mo / me
यदि संयुक्त सूक्ष्मदर्शी में अभिदृश्यक एवं नेत्रिका का आवर्धन क्रमशः 'mo' एवं 'me' हो तो सूक्ष्मदर्शी की कुल आवर्धन क्षमता (m) होगी -
- (अ) mo + me
- (ब) mo - me
- (स) mo × me
- (द) mo / me
Correct Answer: (C) mo × me
Explanation: In a compound microscope, the total magnification is the product of the magnification of the objective lens and the eyepiece lens: m = mo × me.
(x) Natural light from the sun is —
- (A) polarised
- (B) unpolarised
- (C) partially polarised
- (D) linear polarised
सूर्य से प्राप्त प्राकृतिक प्रकाश होता है -
- (अ) ध्रुवित
- (ब) अध्रुवित
- (स) आंशिक ध्रुवित
- (द) रेखीय ध्रुवित
Correct Answer: (B) Unpolarised / अध्रुवित
Explanation: Natural light from the sun is unpolarised because it consists of waves vibrating in all directions perpendicular to the direction of propagation.
(xi) The maximum kinetic energy of a photo electron emitted from a metal is 0.8 eV. The value of stopping potential (cut-off voltage) will be -
- (A) 3.6 V
- (B) 2.0 V
- (C) 4.8 V
- (D) 0.9 V
Correct Answer: (C) 0.8 V
Explanation: The stopping potential \(V_0\) is given by \(eV_0 = K_{\text{max}}\), where \(K_{\text{max}}\) is the maximum kinetic energy of the photoelectron. Here, \(K_{\text{max}} = 0.8\) eV, so \(V_0 = \frac{0.8 \text{ eV}}{e} = 0.8\) V.
(xii) The momentum (p) of a photon is -
- (A) \(\frac{h}{\lambda}\)
- (B) \(\frac{hc}{\lambda}\)
- (C) \(\frac{h}{c}\)
- (D) \(h\lambda\)
Correct Answer: (A) \(\frac{h}{\lambda}\)
Explanation: The momentum of a photon is given by \(p = \frac{h}{\lambda}\), where \(h\) is Planck's constant and \(\lambda\) is the wavelength of the photon.
(xiii) The value of scattering angle of alpha particle for maximum value of impact parameter is -
- (A) 90°
- (B) 60°
- (C) 45°
- (D) 0°
Correct Answer: (D) 0°
Explanation: The impact parameter is the perpendicular distance between the initial velocity vector of the alpha particle and the center of the nucleus. For the maximum impact parameter, the alpha particle is scattered through the smallest angle, which is 0° (i.e., no deflection).
(xiv) The value of excitation energy required to bring an electron to the first excited state in hydrogen atom is -
- (A) 13.6 eV
- (B) 10.2 eV
- (C) 3.4 eV
- (D) -3.4 eV
Correct Answer: (B) 10.2 eV
Explanation: The ground state energy of hydrogen is -13.6 eV, and the first excited state (n=2) energy is -3.4 eV. The excitation energy required is the difference: \(E_2 - E_1 = (-3.4) - (-13.6) = 10.2\) eV.
(xv) Those atoms which have the same atomic number but different mass number are called -
- (A) isobars
- (B) isotones
- (C) isotopes
- (D) isomers
Correct Answer: (C) isotopes (समस्थानिक)
Explanation: Isotopes are atoms of the same element (same atomic number) that have different numbers of neutrons, hence different mass numbers.
(xvi) Example of inorganic semiconductor is -
- (A) Ge
- (B) CdS
- (C) anthracene
- (D) polyaniline
Correct Answer: (A) Ge
Explanation: Germanium (Ge) is an elemental inorganic semiconductor. CdS is also inorganic but is a compound semiconductor. Anthracene and polyaniline are organic semiconductors.
Fill in the Blanks (i to x)
रिक्त स्थानों की पूर्ति कीजिए: (i से x)
-
The field lines of a single positive charge are radially outward.
एकल धनावेश के कारण वैद्युत क्षेत्र त्रिज्यतः बाहर की ओर होती है।
Explanation: For a positive charge, electric field lines originate from the charge and go radially outward.
-
The magnitude of drift velocity of electron per unit electric field is called mobility.
इकाई विद्युत क्षेत्र लगाने पर इलेक्ट्रॉनों के अपवाह वेग के परिमाण को गतिशीलता कहते हैं।
Explanation: Mobility (μ) is defined as the drift velocity per unit electric field: μ = vd/E.
-
To convert a galvanometer into a voltmeter, a resistance of high value is connected in series to it.
एक गैल्वेनोमीटर को वोल्टमीटर में रूपांतरित करने हेतु उसके श्रेणीक्रम में उच्च मान का प्रतिरोध जोड़ा जाता है।
Explanation: A high resistance is connected in series with the galvanometer to limit the current and measure voltage.
-
The resultant magnetic moment produced per unit volume of a substance is called intensity of magnetisation.
किसी पदार्थ के प्रति इकाई आयतन में उत्पन्न परिणामी चुंबकीय आघूर्ण को चुंबकन तीव्रता कहते हैं।
Explanation: Intensity of magnetisation (I) is the magnetic moment per unit volume of a material.
-
The mean value of alternating current in a complete cycle is zero.
एक सम्पूर्ण चक्र में प्रत्यावर्ती धारा का माध्य मान शून्य होता है।
Explanation: Over one complete cycle, the positive and negative halves of AC cancel out, giving a mean value of zero.
-
The radius of curvature of a concave mirror is 24 cm. The value of its focal length will be 12 cm.
एक अवतल दर्पण की वक्रता त्रिज्या 24 cm है। इसकी फोकस दूरी का मान 12 cm होगा।
Explanation: For a spherical mirror, focal length f = R/2 = 24/2 = 12 cm.
-
Bending of waves from their path by the edges of an obstacle is called diffraction.
किसी अवरोध के किनारों द्वारा तरंगों का अपने मार्ग से मुड़ जाना, विवर्तन कहलाता है।
Explanation: Diffraction is the bending of waves around obstacles or through apertures.
-
The formula for the de Broglie wavelength associated with an electron accelerated by a potential 'V' is λ = 1.227/√V nm.
विभव 'V' द्वारा त्वरित किसी इलेक्ट्रॉन से संबद्ध दे ब्रोग्ली तरंगदैर्ध्य का सूत्र λ = 1.227/√V nm होता है।
Explanation: The de Broglie wavelength for an electron accelerated through potential V is λ = h/√(2meV) = 1.227/√V nm.
-
If the radius of first orbit of hydrogen atom is 0.5 × 10-10 m, then the radius of its second orbit will be 2 × 10-10 m.
यदि हाइड्रोजन परमाणु के प्रथम कक्षा की त्रिज्या 0.5 × 10-10 m हो, तो इसके दूसरी कक्षा की त्रिज्या 2 × 10-10 m होगी।
Explanation: Radius of nth orbit rn = n² × r1. For n=2, r2 = 4 × 0.5 × 10-10 = 2 × 10-10 m.
-
Two types of extrinsic semiconductors are found.
अपद्रव्यी अर्धचालक दो प्रकार के होते हैं।
Explanation: Extrinsic semiconductors are of two types: n-type (doped with pentavalent atoms) and p-type (doped with trivalent atoms).
प्रश्न एवं उत्तर (Questions and Answers)
(i) Draw an equipotential surface for a positive charge (q > 0).
किसी एकल धनावेश (q > 0) के लिए समविभव पृष्ठ बनाइए।
For a single positive charge, equipotential surfaces are concentric spheres centered on the charge. In a 2D representation, they appear as concentric circles around the charge.
(ii) Draw a graph between potential difference (V) and current (I) according to Ohm's law.
ओम के नियमानुसार विभवांतर (V) तथा धारा (I) के मध्य ग्राफ बनाइए।
According to Ohm's law, V ∝ I, so the graph between V and I is a straight line passing through the origin, with slope equal to the resistance (R).
(iii) What is paramagnetic substance?
अनुचुंबकीय पदार्थ किसे कहते हैं?
Paramagnetic substances are those which develop feeble magnetisation in the direction of the magnetising field. Such substances are feebly attracted by magnets and tend to move from weaker to stronger parts of a magnetic field.
Examples: Manganese, aluminium, chromium, platinum, sodium, copper chloride, oxygen (at STP), etc.
(iv) Why self-inductance is called electrical inertia?
स्वप्रेरकत्व को विद्युत जडत्व क्यों कहते हैं?
Self-induction of the coil is the property by virtue of which it tends to maintain the magnetic flux linked with it and opposes any change in the flux by inducing a current in it. This property of a coil is analogous to mechanical inertia. That is why self-induction is called the inertia of electricity.
(v) Define Coherent source.
कला-संबद्ध स्त्रोत को परिभाषित कीजिए।
The source which emits a light wave with the same frequency, wavelength and phase or having a constant phase difference is known as a coherent source.
(vi) Write the definition of threshold frequency of a substance.
किसी पदार्थ की 'देहली आवृत्ति' की परिभाषा लिखिए।
The threshold frequency is the lowest frequency below which the photoelectric effect does not occur.
(vii) What is ionization energy?
आयनन ऊर्जा किसे कहते हैं?
Ionization energy is defined as the amount of energy required to remove an electron from an isolated atom or molecule.
(viii) Write Einstein's mass-energy equivalent relation.
आइंस्टाइन का द्रव्यमान-ऊर्जा समतुल्यता संबंध लिखिए।
E = mc²
खण्ड-ब (SECTION - B)
1. Three capacitors of capacitance 6 µF are connected in parallel. Calculate the value of their equivalent capacitance.
6 µF धारिता के तीन संधारित्र पार्श्वक्रम में जुड़े हैं। इनकी तुल्य धारिता का मान ज्ञात कीजिए।
Solution:
For capacitors in parallel, the equivalent capacitance is the sum of individual capacitances:
Cp = C1 + C2 + C3 = 6 µF + 6 µF + 6 µF = 18 µF
2. Define: (i) Electromotive force and (ii) Internal resistance of a cell
किसी सेल के (i) विद्युत वाहक बल तथा (ii) आंतरिक प्रतिरोध को परिभाषित कीजिए।
(i) Electromotive force (EMF): The electromotive force of a source may be defined as the work done by the source in taking a unit positive charge from the positive terminal to the negative terminal in the external circuit and from the negative terminal to the positive terminal inside the cell.
(ii) Internal resistance: The resistance offered by the electrolyte of a cell to the flow of current between its electrodes is called the internal resistance of the cell.
3. Write any two properties of magnetic field lines.
चुंबकीय क्षेत्र रेखाओं के कोई दो गुण लिखिए।
Properties of magnetic field lines:
- Magnetic lines of force are closed curves which start in air from the N-pole and end at the S-pole, and then return to the N-pole through the interior of the magnet.
- The lines of force never cross each other. If they do so, that would mean there are two directions of the magnetic field at the point of intersection, which is impossible.
4. Draw a clear and labeled diagram of an alternating current generator.
प्रत्यावर्ती धारा जनित्र का स्पष्ट एवं नामांकित चित्र बनाइए।
Diagram description: An AC generator consists of a rectangular coil (armature) placed between the poles of a strong magnet (field magnet). The coil is rotated about an axis. The ends of the coil are connected to slip rings, which are in contact with carbon brushes. The diagram shows the axis of rotation, the armature coil, the field magnet, and the slip rings with brushes.
(Note: A labeled diagram would show: Axis, Armature coil, Field magnet, Slip rings, Brushes, and the direction of motion.)
5. A 5 m long straight horizontal conducting wire situated in the east to west direction is falling with a speed of 2 m/s perpendicular to the horizontal component of the earth's magnetic field of 0.3 × 10-4 T. Find the instantaneous value of the emf induced between the ends of the wire.
पूर्व से पश्चिम दिशा में स्थित 5 m लम्बा सीधा क्षैतिज चालक तार 0.3 × 10-4 T के पृथ्वी के चुंबकीय क्षेत्र के क्षितिज घटक के लम्बवत् 2 m/s की चाल से गिर रहा है। तार के सिरों के मध्य प्रेरित विद्युत वाहक बल का तात्क्षणिक मान ज्ञात कीजिए।
Solution:
Given: Length of wire, l = 5 m
Speed of falling, v = 2 m/s
Horizontal component of Earth's magnetic field, B = 0.3 × 10-4 T
The induced emf is given by: ε = B l v
ε = (0.3 × 10-4) × 5 × 2
ε = 0.3 × 10-4 × 10
ε = 3 × 10-4 V
Therefore, the instantaneous induced emf is 3 × 10-4 V.
6. Write the names of any three waves (radiations) produced in the electromagnetic spectrum.
वैद्युत चुम्बकीय स्पेक्ट्रम में उत्पन्न किन्हीं तीन तरंगों (विकिरणों) के नाम लिखिए।
Three electromagnetic waves:
- X-rays
- Ultraviolet (UV) rays
- Gamma (γ) rays
70.
‘The magnifying power of a small telescope is 9 and the length of the tube is 700 cm. Find the focal lengths of the objective and eyepiece of the telescope.
किसी छोटी दूरबीन की आवर्धन क्षमता 9 तथा नली (ट्यूब) की लम्बाई 700 cm है। दूरबीन के अभिदृश्यक तथा नेत्रिका की फोकस दूरियाँ ज्ञात कीजिए।
Solution:
Given: Magnifying power (M) = 9, Length of tube (L) = 700 cm.
For a telescope in normal adjustment:
M = fo / fe ⇒ fo = M × fe = 9 fe
L = fo + fe ⇒ 700 = 9 fe + fe = 10 fe
⇒ fe = 700 / 10 = 70 cm
⇒ fo = 9 × 70 = 630 cm
Answer: Focal length of objective (fo) = 630 cm, Focal length of eyepiece (fe) = 70 cm.
44.
Derive Snell's law for refraction of light by Huygen's wave theory.
हाइगेंस के तरंग सिद्धान्त से प्रकाश के अपवर्तन हेतु स्नेल के नियम का निगमन कीजिए।
Solution:
Laws of Refraction (Snell’s Law) at a Plane Surface
Let 1, 2, 3 be the incident rays and 1', 2', 3' be the corresponding refracted rays.
If v1 and v2 are the speeds of light in the two media, and t is the time taken by light to go from B to C or A to D or E to G through F, then:
t = EF / v1 + FG / v2
In ΔAFE, sin i = EF / AF
In ΔFGC, sin r = FG / FC
⇒ t = (AF sin i) / v1 + (FC sin r) / v2
For rays from different parts of the incident wavefront, the values of AF are different. However, light from different points of the incident wavefront should take the same time to reach corresponding points on the refracted wavefront. So, t should not depend upon AF. This is possible only if:
sin i / v1 - sin r / v2 = 0
⇒ sin i / sin r = v1 / v2
⇒ sin i / sin r = c/v2 / c/v1 = μ2 / μ1 = 1μ2
This is known as Snell’s law of refraction.
प्रश्न 42 (5 अंक)
निम्न को परिभाषित कीजिए:
(a) प्रकाश का व्यतिकरण (b) प्रकाश का ध्रुवण
Define the following:
(a) Interference of light (b) Polarization of light
(a) प्रकाश का व्यतिकरण (Interference of light): जब दो संगत स्रोतों से निकली प्रकाश तरंगें एक-दूसरे पर अध्यारोपित होती हैं, तो ऊर्जा का पुनर्वितरण होता है, जिसके परिणामस्वरूप कुछ बिंदुओं पर प्रकाश की तीव्रता अधिकतम (दीप्त फ्रिंज) और कुछ पर न्यूनतम (अदीप्त फ्रिंज) हो जाती है। इस घटना को प्रकाश का व्यतिकरण कहते हैं।
(b) प्रकाश का ध्रुवण (Polarization of light): प्रकाश एक अनुप्रस्थ तरंग है। सामान्य प्रकाश में कंपन सभी दिशाओं में होते हैं। जब इन कंपनों को किसी एक ही तल में सीमित कर दिया जाता है, तो इस प्रकाश को ध्रुवित प्रकाश कहते हैं। अध्रुवित प्रकाश को ध्रुवित प्रकाश में बदलने की प्रक्रिया को प्रकाश का ध्रुवण कहते हैं।
प्रश्न 43 (5 अंक)
20 वॉट के बल्ब से 5 × 1019 फोटॉन प्रति सेकण्ड उत्सर्जित होते हैं। प्रत्येक फोटॉन की ऊर्जा ज्ञात कीजिए।
A 20 Watt bulb emits 5 × 1019 photons per second. Find the energy of each photon.
हल:
बल्ब की शक्ति (P) = 20 वॉट = 20 जूल/सेकण्ड
प्रति सेकण्ड उत्सर्जित फोटॉनों की संख्या (n) = 5 × 1019
माना प्रत्येक फोटॉन की ऊर्जा E है।
कुल ऊर्जा = n × E
⇒ 20 = (5 × 1019) × E
⇒ E = 20 / (5 × 1019)
⇒ E = 4 × 10-19 जूल
अतः प्रत्येक फोटॉन की ऊर्जा 4 × 10-19 जूल है।
प्रश्न 44 (5 अंक)
दे ब्रॉग्ली परिकल्पना से बोर के द्वितीय अभिगृहीत (क्वाण्टीकरण) की व्याख्या कीजिए।
Explain Bohr's second postulate of quantisation by de Broglie hypothesis.
व्याख्या:
दे ब्रॉग्ली के अनुसार, इलेक्ट्रॉन से जुड़ी एक तरंग होती है। बोर के परमाणु मॉडल में, इलेक्ट्रॉन नाभिक के चारों ओर वृत्ताकार कक्षाओं में घूमता है। दे ब्रॉग्ली ने प्रस्तावित किया कि स्थायी कक्षा वह होती है जिसमें इलेक्ट्रॉन से जुड़ी दे ब्रॉग्ली तरंग एक अपवाही (standing wave) बनाती है।
इसका अर्थ है कि कक्षा की परिधि, दे ब्रॉग्ली तरंगदैर्ध्य (λ) की पूर्णांक गुणज होनी चाहिए।
यदि nवीं कक्षा की त्रिज्या rn है, तो:
2πrn = nλ, जहाँ n = 1, 2, 3, ...
दे ब्रॉग्ली तरंगदैर्ध्य का सूत्र: λ = h / (mvn)
जहाँ vn इलेक्ट्रॉन की चाल है। मान रखने पर:
2πrn = n × (h / mvn)
⇒ mvnrn = n × (h / 2π)
यह बोर के द्वितीय अभिगृहीत (कोणीय संवेग क्वाण्टीकरण) का गणितीय रूप है, जो सिद्ध करता है कि इलेक्ट्रॉन का कोणीय संवेग h/2π का पूर्णांक गुणज होता है।
प्रश्न 45 (5 अंक)
परिभाषित कीजिए:
(a) नाभिकीय विखंडन (b) नाभिकीय संलयन
Define:
(a) Nuclear fission (b) Nuclear fusion
(a) नाभिकीय विखंडन (Nuclear fission): यह एक नाभिकीय अभिक्रिया है जिसमें किसी भारी नाभिक (जैसे यूरेनियम-235) को न्यूट्रॉन से बमबारी करने पर वह दो या दो से अधिक हल्के नाभिकों में टूट जाता है। इस प्रक्रिया में अपार ऊर्जा निकलती है।
(b) नाभिकीय संलयन (Nuclear fusion): यह वह प्रक्रिया है जिसमें दो या दो से अधिक हल्के नाभिक (जैसे हाइड्रोजन के समस्थानिक) आपस में संयोग करके एक भारी नाभिक (जैसे हीलियम) का निर्माण करते हैं। इस प्रक्रिया में भी भारी मात्रा में ऊर्जा मुक्त होती है।
खण्ड-स (SECTION - C)
प्रश्न 46.
Derive formula for the electric field due to electric dipole at any point on the equatorial plane. Draw necessary diagram.
अपने द्विध्रुव के कारण विषुवतीय तल पर स्थित किसी बिन्दु पर विद्युत क्षेत्र का सूत्र व्युत्पन्न कीजिए। आवश्यक चित्र बनाइए।
Electric field at Equatorial point of a dipole :
As shown in figure, consider an electric dipole consisting of charges –q and +q, separated by distance 2a and placed in vacuum. Let P be a point on the equatorial line of the dipole at distance r from it. OP = r.
(Diagram: Electric field at an equatorial point of a dipole)
Electric field at point P due to +q charge is:
E+q = 1⁄4πε₀ · q⁄(r² + a²) directed along BP
Electric field at point P due to –q charge is:
E–q = 1⁄4πε₀ · q⁄(r² + a²) directed along PA
Thus the magnitudes of E–q and E+q are equal i.e.,
E–q = E+q = 1⁄4πε₀ · q⁄(r² + a²)
Clearly, the components of E–q and E+q normal to the dipole axis will cancel out. The components parallel to the dipole axis add up. The total electric field E is opposite to p.
E = –(E+q cos θ + E–q cos θ) p̂ = –2E+q cos θ p̂ [∵ E+q = E–q]
Now, cos θ = a⁄√(r² + a²)
E = –2 · 1⁄4πε₀ · q⁄(r² + a²) · a⁄√(r² + a²) p̂
E = –1⁄4πε₀ · p⁄(r² + a²)3/2 p̂ = –k p⁄(r² + a²)3/2 p̂
where p = 2aq, is the electric dipole moment. If the point P is located far away from the dipole, r >> a, then
E = –1⁄4πε₀ · p⁄r³ p̂ = –k p⁄r³ p̂
Clearly, the direction of electric field at any point on the equatorial line of the dipole will be antiparallel to the dipole moment p.
OR / अथवा
Obtain an expression for the electric field at any point due to a uniformly charged infinite plane sheet with the help of Gauss's law. Draw necessary diagram.
गाउस नियम द्वारा एकसमान आवेशित अनंत समतल चादर के कारण किसी बिन्दु पर विद्युत क्षेत्र का व्यंजक प्राप्त कीजिए। आवश्यक चित्र बनाइए।
Electric field due to a uniformly charged infinite plane sheet :
Consider an infinite plane sheet of charge with uniform surface charge density σ (C/m²). To find the electric field at a point P at distance r from the sheet, we use Gauss's law.
(Diagram: Gaussian cylinder passing through the sheet)
We choose a cylindrical Gaussian surface of cross-sectional area A, with its axis perpendicular to the sheet and passing through point P. The cylinder extends equally on both sides of the sheet.
By symmetry, the electric field E is perpendicular to the sheet and has the same magnitude at all points equidistant from the sheet. The flux through the curved surface of the cylinder is zero because E is parallel to it.
Flux through each flat end of the cylinder = EA (since E is perpendicular to the surface).
Total electric flux through the Gaussian surface = 2EA
Charge enclosed by the Gaussian surface = σA
Applying Gauss's law:
2EA = σA⁄ε₀
E = σ⁄2ε₀
Thus, the electric field due to a uniformly charged infinite plane sheet is independent of distance from the sheet and is directed perpendicularly away from the sheet (for positive σ) or towards the sheet (for negative σ).
Electric field due to a uniformly charged infinite plane sheet
Consider a thin, infinite plane sheet of charge with uniform surface charge density σ. We wish to calculate its electric field at a point P at distance r from it.
Diagram: A plane sheet with cross-sectional area A, and a cylindrical Gaussian surface of cross-sectional area A and length 2r, with its axis perpendicular to the sheet. Points P and P' are equidistant from the sheet on opposite sides.
By symmetry, electric field E points outwards normal to the sheet. Also, it must have the same magnitude and opposite direction at two points P and P' equidistant from the sheet and on opposite sides. We choose a cylindrical Gaussian surface of cross-sectional area A and length 2r with its axis perpendicular to the sheet.
As the lines of force are parallel to the curved surface of the cylinder, the flux through the curved surface is zero. The flux through the plane-end faces of the cylinder is:
φE = EA + EA = 2EA (A = cross-sectional area of plane-end faces)
Charge enclosed by the Gaussian surface:
q = σA
According to Gauss's theorem:
φE = q / ε₀
2EA = σA / ε₀
E = σ / (2ε₀)
Result: The electric field due to a uniformly charged infinite plane sheet is E = σ / (2ε₀), directed perpendicularly away from the sheet (for positive charge).
Magnetic field on the axis of a current-carrying circular loop (Biot-Savart's law)
Consider a circular loop of wire of radius a carrying current I. Let the plane of the loop be perpendicular to the plane of paper. We wish to find the magnetic field B at an axial point P at a distance r from the centre C.
Diagram: A circular loop of radius a with centre C. Point P is on the axis at distance r from C. A small current element Idℓ on the loop produces magnetic field dB at P. The angle between dℓ and the line joining the element to P is 90°. The component of dB perpendicular to the axis cancels, while the axial component dB cos φ adds up.
Derivation:
From Biot-Savart's law, the magnetic field due to a small current element Idℓ at point P is:
dB = (μ₀ / 4π) × (I dℓ sin θ) / R²
Here, θ = 90° (since dℓ is perpendicular to the line joining element to P), so sin θ = 1. Also, R = √(a² + r²).
dB = (μ₀ / 4π) × (I dℓ) / (a² + r²)
The direction of dB is perpendicular to the plane containing dℓ and R. The component perpendicular to the axis cancels due to symmetry. The axial component is:
dBaxial = dB cos φ
From the geometry, cos φ = a / R = a / √(a² + r²)
So, dBaxial = (μ₀ / 4π) × (I dℓ) / (a² + r²) × a / √(a² + r²)
dBaxial = (μ₀ I a dℓ) / [4π (a² + r²)3/2]
Integrating over the entire loop (∫ dℓ = 2πa):
B = ∫ dBaxial = (μ₀ I a) / [4π (a² + r²)3/2] × 2πa
B = (μ₀ I a²) / [2 (a² + r²)3/2]
Result: The magnetic field at a point on the axis of a circular current loop of radius a carrying current I, at distance r from the centre, is:
B = (μ₀ I a²) / [2 (a² + r²)3/2]
The direction is along the axis, given by the right-hand rule.
OR: Force per unit length between two straight parallel current-carrying conductors
Consider two long, straight, parallel conductors separated by distance d, carrying currents I₁ and I₂ in the same direction.
Diagram: Two parallel wires separated by distance d. Wire 1 carries current I₁ (upward), wire 2 carries current I₂ (upward). The magnetic field B₁ due to wire 1 at the location of wire 2 is directed into the page (by right-hand rule). The force on wire 2 due to B₁ is towards wire 1 (attractive).
Derivation:
The magnetic field produced by conductor 1 at the location of conductor 2 is:
B₁ = (μ₀ I₁) / (2π d)
The force on a length L of conductor 2 due to this field is:
F = I₂ L B₁ sin θ
Here, θ = 90° (current in wire 2 is perpendicular to B₁), so sin θ = 1.
F = I₂ L × (μ₀ I₁) / (2π d)
F = (μ₀ I₁ I₂ L) / (2π d)
Force per unit length:
F/L = (μ₀ I₁ I₂) / (2π d)
Result: The force per unit length between two parallel current-carrying conductors is:
F/L = (μ₀ I₁ I₂) / (2π d)
The force is attractive if currents are in the same direction, and repulsive if currents are in opposite directions.
Magnetic Field on the Axis of a Circular Current Loop
Consider a current element dl at the top of the loop. It has an outward coming current.
If r be the position vector of point P relative to the element dl, then from Biot-Savart law, the field at point P due to the current element is:
dB = (μ₀ / 4π) × (I dl × r) / r³
Since θ = 90° (i.e., dl ⟂ r), therefore:
dB = (μ₀ I dl) / (4π r²)
The field dB lies in the plane of the paper and is perpendicular to r, as shown by PQ. Let φ be the angle between OP and CP. Then dB can be resolved into two rectangular components:
- dB sin φ along the axis
- dB cos φ perpendicular to the axis
For any two diametrically opposite elements of the loop, the components perpendicular to the axis of the loop will be equal and opposite and will cancel out. Their axial components will be in the same direction, i.e., along CP and get added up.
∴ Total magnetic field at point P in the direction CP is:
B = ∫ dB sin φ
But sin φ = a / r and dB = (μ₀ I dl) / (4π r²)
∴ B = ∫ (μ₀ I dl) / (4π r²) × (a / r) = (μ₀ I a) / (4π r³) ∫ dl
Since μ₀, I, and a are constant, and r and a are same for all points on the circular loop, we have:
B = (μ₀ I a) / (4π r³) × 2πa = (μ₀ I a²) / (2 r³)
[∵ ∫ dl = circumference = 2πa]
Or B = (μ₀ I a²) / [2 (r² + a²)³/²] [since r² = r² + a²]
As the direction of the field is along +ve X-direction, so we can write:
B = [μ₀ I a² / 2 (r² + a²)³/²] î
If the coil consists of N turns, then:
B = [μ₀ N I a² / 2 (r² + a²)³/²] î
Explanation: This derivation uses the Biot-Savart law to find the magnetic field at a point on the axis of a circular current loop. The perpendicular components cancel due to symmetry, leaving only the axial component. The final expression shows that the field depends on the current (I), radius (a), distance from the center (r), and number of turns (N).
8. Expression for the force between two parallel current-carrying wires
Consider two long parallel wires AB and CD carrying currents I1 and I2. Let r be the separation between them.
Diagram: Two parallel wires AB (left) and CD (right), separated by distance r. Current I1 flows upward in AB, current I2 flows upward in CD. Magnetic field B1 due to wire AB points into the page at wire CD.
The magnetic field produced by current I1 at any point on wire CD is:
B1 = μ0I1 / (2πr)
This field acts perpendicular to wire CD and points into the plane of the paper. It exerts a force on current-carrying wire CD:
F2 = I2 B1 sin 90° = I2 × (μ0I1 / 2πr) × 1 = μ0I1I2 / (2πr)
Force per unit length:
F / L = μ0I1I2 / (2πr)
According to Fleming's left-hand rule, this force acts at right angles to CD, towards AB in the plane of the paper. Similarly, an equal force is exerted on wire AB by the field of wire CD. Thus, when the currents in the two wires are in the same direction, the forces between them are attractive. It can be easily seen that:
F1 = –F2
As shown in the figure, when the current in the two parallel wires flow in opposite directions (anti-parallel), the forces between the two wires are repulsive.
(a) On the basis of energy band theory, write the difference between conductor, insulator and semiconductor.
(b) Draw energy band diagram of n-type semiconductor.
Solution (a): Differences based on energy band theory
| Property | Conductor | Insulator | Semiconductor |
|---|---|---|---|
| Energy gap (Eg) | No energy gap (valence and conduction bands overlap) | Large energy gap (Eg > 3 eV) | Small energy gap (Eg ≈ 0.1 – 1.5 eV) |
| Conduction band | Partially filled or overlapping with valence band | Empty at 0 K | Empty at 0 K |
| Valence band | Partially filled or completely filled (overlap) | Completely filled | Completely filled at 0 K |
| Conductivity | Very high (107 – 108 S/m) | Very low (10–10 – 10–20 S/m) | Intermediate (10–5 – 103 S/m) |
| Effect of temperature | Conductivity decreases with increase in temperature | Conductivity remains very low | Conductivity increases with increase in temperature |
| Examples | Copper, Silver, Aluminum | Rubber, Glass, Wood | Silicon, Germanium |
Solution (b): Energy band diagram of n-type semiconductor
Energy Band Diagram of n-type Semiconductor:
Energy ↑ | Conduction Band (CB) | --------------------- | ▲ | | Donor level (ED) — just below CB | ▼ | --------------------- | Energy gap (small) | --------------------- | Valence Band (VB) | --------------------- |___________________________→ Distance
Explanation: In an n-type semiconductor, pentavalent impurity atoms (donors) are added. These donor atoms create an extra energy level (donor level) just below the conduction band. Electrons from this donor level can easily jump to the conduction band, increasing the number of free electrons (negative charge carriers).
अथवा / OR
Write the name of device 'Y' in the following given diagram. Explain its working making with circuit diagram.
Given Diagram: A circuit with input signal applied to a device labeled 'Y', which has two outputs: one connected to +2V and the other to –2V. The output is taken from the device.
Solution:
Name of device 'Y': Full Wave Rectifier (or Bridge Rectifier)
Working: A full wave rectifier converts both halves of an AC input signal into a DC output. It uses four diodes arranged in a bridge configuration. During the positive half cycle of the input, two diodes conduct, allowing current to flow through the load in one direction. During the negative half cycle, the other two diodes conduct, again allowing current to flow through the load in the same direction. Thus, the output is a pulsating DC signal.
Circuit Diagram of Full Wave Bridge Rectifier:
D1 ▲ D2 ▲
/ \ / \
/ \ / \
AC --- ------ ---
Input | | | |
| | | |
| | | |
D3 ▼ D4 ▼
\ / \ /
\ /
\ /
\ /
\ /
\ /
\/
Load (RL)
|
GND
Explanation: D1, D2, D3, D4 are four diodes. During positive half cycle, D1 and D2 conduct; during negative half cycle, D3 and D4 conduct. The load resistor RL receives current in the same direction during both half cycles, producing a full-wave rectified output.
Physics (SS-40-2024) – Page 14
(a) Differences between Metals, Insulators, and Semiconductors
| Property | Metals | Insulators | Semiconductors |
|---|---|---|---|
| Definition | Substances naturally found below the earth; good conductors of heat and electricity. | Poor conductors of heat and electricity. | Substances that have properties between conductors and insulators. |
| Band Structure | Conduction band is either filled or partially filled; valence band is partially empty. | Valence band is completely filled; conduction band is partially filled. | Valence band is completely filled; conduction band is empty. |
| Forbidden Gap | No forbidden gap. | Large energy gap. | Small energy gap. |
Energy Band Diagrams
Metals: Conduction band (partially filled) and valence band (partially empty) overlap – no band gap.
Insulators: Large band gap (Eg > 3 eV) between filled valence band and empty conduction band.
Semiconductors: Small band gap (Eg ≈ 1 eV) between filled valence band and empty conduction band.
(b) Energy Band Diagram for n-type Semiconductor
In an n-type semiconductor, donor impurities introduce an extra energy level (donor level) just below the conduction band. The band gap is small (e.g., ~0.01 eV for some materials).
Diagram description:
- Valence band (completely filled) at the bottom.
- Conduction band (empty) at the top.
- A donor level (ED) is present very close to the conduction band edge (EC), with energy difference ~0.01 eV.
- Electrons from donor level easily jump to conduction band, increasing conductivity.
OR
Full Wave Rectifier
Answer: 'Y' is a full wave rectifier.
Explanation: A full wave rectifier uses two p-n junction diodes (D1 and D2) to convert both halves of the AC input into DC output.
Circuit Diagram Description:
- An AC input source is connected to the primary coil of a center-tapped transformer.
- The secondary coil has a center tap, providing two equal voltages opposite in phase.
- Diode D1 is connected to one end of the secondary coil; diode D2 is connected to the other end.
- The cathodes of both diodes are connected together to the load resistor (RL).
- The center tap is connected to the other end of the load resistor, completing the circuit.
Working: During the positive half-cycle, D1 is forward biased and conducts; D2 is reverse biased and does not conduct. During the negative half-cycle, D2 conducts and D1 does not. Thus, current flows through the load in the same direction for both half-cycles, producing a pulsating DC output.
खण्ड-द (SECTION - D)
(a) Prove that the peak value (Im) of an alternating current is √2 times its root mean square (rms) value.
Proof:
Let an alternating current be represented by: i = Im sin ωt
The root mean square (rms) value of current is defined as:
Irms = √[ (1/T) ∫0T i² dt ]
Substituting i² = Im² sin² ωt:
Irms² = (1/T) ∫0T Im² sin² ωt dt
Using sin² ωt = (1 - cos 2ωt)/2:
Irms² = (Im²/T) ∫0T (1 - cos 2ωt)/2 dt
Irms² = (Im²/2T) [ ∫0T dt - ∫0T cos 2ωt dt ]
The integral of cos 2ωt over a complete cycle is zero. Therefore:
Irms² = (Im²/2T) × T = Im²/2
Taking square root: Irms = Im/√2
Hence, Im = √2 × Irms (Proved)
(b) If alternating current I = 4 sin ωt and voltage V = 200 sin (ωt + π/3), then calculate the average power dissipated in the circuit.
Solution:
Given: I = 4 sin ωt, V = 200 sin (ωt + π/3)
Peak current, Im = 4 A
Peak voltage, Vm = 200 V
Phase difference between voltage and current, φ = π/3
Average power in an AC circuit is given by:
Pavg = Vrms × Irms × cos φ
Where Vrms = Vm/√2 = 200/√2 V
Irms = Im/√2 = 4/√2 A
cos φ = cos(π/3) = 1/2
Therefore:
Pavg = (200/√2) × (4/√2) × (1/2)
Pavg = (200 × 4)/(2 × 2) = 800/4 = 200 W
Average power dissipated = 200 W
अथवा / OR
(a) Prove that the average power supplied to an inductor over one complete cycle is zero.
Proof:
In a purely inductive AC circuit, current lags voltage by 90° (π/2).
Let voltage: V = Vm sin ωt
Current: I = Im sin (ωt - π/2) = -Im cos ωt
Instantaneous power: p = V × I = Vm sin ωt × (-Im cos ωt)
p = -VmIm sin ωt cos ωt = -(VmIm/2) sin 2ωt
Average power over one complete cycle:
Pavg = (1/T) ∫0T p dt = -(VmIm/2T) ∫0T sin 2ωt dt
The integral of sin 2ωt over a complete cycle is zero.
Hence, Pavg = 0 (Proved)
This shows that the inductor stores energy during one half-cycle and returns it during the next half-cycle, resulting in zero net power consumption.
(b) If in an LCR alternating current circuit R = 24 Ω, XL = 40 Ω and XC = 70 Ω, then find the impedance of the circuit.
Solution:
Given: R = 24 Ω, XL = 40 Ω, XC = 70 Ω
Net reactance, X = XL - XC = 40 - 70 = -30 Ω
Impedance of LCR circuit is given by:
Z = √(R² + X²) = √(24² + (-30)²)
Z = √(576 + 900) = √1476
Z = √(36 × 41) = 6√41 Ω
Impedance of the circuit = 6√41 Ω ≈ 38.42 Ω
Physics (SS-40-2024) – Page 16
Solution (8)
(a) RMS Value of AC:
We have:
\( I_{\text{rms}} = \sqrt{ \frac{1}{T} \int_0^T i^2 \, dt } \)
Let \( i = I_0 \sin \omega t \). Then:
\( I_{\text{rms}}^2 = \frac{1}{T} \int_0^T I_0^2 \sin^2 \omega t \, dt \)
\( = \frac{I_0^2}{T} \int_0^T \frac{1 - \cos 2\omega t}{2} \, dt \)
\( = \frac{I_0^2}{2T} \left[ \int_0^T dt - \int_0^T \cos 2\omega t \, dt \right] \)
\( = \frac{I_0^2}{2T} \left[ T - 0 \right] \) (since integral of cos over a full period is zero)
\( = \frac{I_0^2}{2} \)
Therefore:
\( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = 0.707 I_0 \)
Thus, the effective or rms value of an AC is \( \frac{1}{\sqrt{2}} \) times its peak value.
(b) Power in AC Circuit:
\( P = V_{\text{rms}} I_{\text{rms}} \cos \phi \)
OR
(a) Instantaneous Power in Inductive Circuit:
Instantaneous power is given by \( p = v \cdot i \).
For a purely inductive circuit, \( v = V_0 \sin \omega t \) and \( i = -I_0 \cos \omega t \).
\( p = (V_0 \sin \omega t)(-I_0 \cos \omega t) \)
\( = -V_0 I_0 \sin \omega t \cos \omega t \)
\( = -\frac{V_0 I_0}{2} \sin 2\omega t \)
The average power over one time period is:
\( P_{\text{av}} = \frac{1}{T} \int_0^T p \, dt = -\frac{V_0 I_0}{2} \langle \sin 2\omega t \rangle \)
Since the average value of \(\sin 2\omega t\) over a complete cycle is zero:
\( P_{\text{av}} = 0 \)
(b) Impedance Calculation:
Given: \( R = 24 \, \Omega \), \( X_L = 40 \, \Omega \), \( X_C = 440 \, \Omega \).
Impedance \( Z = \sqrt{R^2 + (X_L - X_C)^2} \)
\( Z = \sqrt{(24)^2 + (40 - 440)^2} \)
\( Z = \sqrt{576 + (-400)^2} \)
\( Z = \sqrt{576 + 160000} \)
\( Z = \sqrt{160576} \)
\( Z \approx 400.72 \, \Omega \)
20.
Define total internal reflection. Establish relation between \(u\), \(v\) and \(f\) for a spherical mirror. Draw necessary ray diagram.
पूर्ण आंतरिक परावर्तन को परिभाषित कीजिए | किसी गोलीय दर्पण के लिए \(u\), \(v\) तथा \(f\) में संबंध स्थापित कीजिए। आवश्यक किरण चित्र बनाइए। [4+2+4½]
अथवा / OR
Define lateral shift. Derive the lens maker's formula \(\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\). Draw necessary ray diagram. (where symbols carry usual meaning).
पार्श्व विस्थापन को परिभाषित कीजिए। लेंस मेकर सूत्र \(\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\) व्युत्पन्न कीजिए। आवश्यक किरण चित्र बनाइए (जहाँ संकेतों के सामान्य अर्थ हैं) [4+2+4½]
Solution (First Option): Total Internal Reflection and Mirror Formula
Total Internal Reflection
When a ray of light travels from a denser medium to a rarer medium, and the angle of incidence in the denser medium exceeds the critical angle, the ray is reflected back entirely into the denser medium. This phenomenon is called total internal reflection.
Mirror Formula for a Concave Mirror (Real Image)
Consider an object AB placed on the principal axis beyond the centre of curvature C of a concave mirror of small aperture. The mirror forms a real, inverted image A'B'.
Sign Conventions:
- Object distance, BP = \(-u\)
- Image distance, B'P = \(-v\)
- Focal length, FP = \(-f\)
- Radius of curvature, CP = \(-R = -2f\)
Derivation:
From similar triangles \(\triangle A'B'C\) and \(\triangle ABC\):
\[\frac{A'B'}{AB} = \frac{CB'}{CB} = \frac{CP - B'P}{BP - CP} = \frac{-R + v}{-u + R} \quad \text{(1)}\]
From similar triangles \(\triangle A'B'P\) and \(\triangle ABP\) (since \(\angle A'PB' = \angle ABP\)):
\[\frac{A'B'}{AB} = \frac{B'P}{BP} = \frac{-v}{-u} = \frac{v}{u} \quad \text{(2)}\]
Equating (1) and (2):
\[\frac{-R + v}{-u + R} = \frac{v}{u}\]
Cross-multiplying:
\[u(-R + v) = v(-u + R)\]
\[-uR + uv = -uv + vR\]
\[vR + uR = 2uv\]
Dividing both sides by \(uvR\):
\[\frac{1}{u} + \frac{1}{v} = \frac{2}{R}\]
But \(R = 2f\), therefore:
\[\frac{1}{u} + \frac{1}{v} = \frac{2}{2f} = \frac{1}{f}\]
Thus, the mirror formula is:
\[\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\]
Note: A ray diagram should be drawn showing object AB beyond C, image A'B' between C and F, with rays reflecting off the concave mirror.
Lateral Shift & Lens Maker's Formula
OR
The perpendicular distance between the incident ray and the emergent ray is defined as lateral shift.
Assumptions made in the derivation of lens maker's formula:
- The lens used is thin so that the distances measured from its optical centre.
- The object is a point object placed on the principal axis.
- The aperture of the lens is small.
- All the rays are paraxial, i.e., they make very small angles with the normal to the lens faces and with the principal axis.
Refraction through a double convex lens
Suppose a point object O is placed on the principal axis in the rarer medium of refractive index μ₁. The ray OM is incident on the first surface ABC. It is refracted along MN, bending towards the normal at this surface. If the second surface ADC were absent, the ray MN would have met the principal axis at I₁. So we can treat I₁ as the real image formed by first surface ABC in the medium of refractive index μ₂.
For refraction at surface ABC, we can write the relation between the object distance u, image distance v₁, and radius of curvature R₁ as:
Equation (1): μ₂/v₁ - μ₁/u = (μ₂ - μ₁)/R₁
For refraction at second surface, I₁ acts as a virtual object placed in the medium of refractive index μ₂ and I is the real image formed in the medium of refractive index μ₁. Therefore, the relation between the object distance v₁, image distance v and radius of curvature R₂ can be written as:
Equation (2): μ₁/v - μ₂/v₁ = (μ₁ - μ₂)/R₂
Adding equation (1) and (2), we get:
Equation (3): μ₁/v - μ₁/u = (μ₂ - μ₁)(1/R₁ - 1/R₂)
If the object is placed at infinity (u = ∞), the image will be formed at the focus, i.e. v = f. Therefore,
Lens Maker's Formula: 1/f = (μ₂/μ₁ - 1)(1/R₁ - 1/R₂)
Or, if the surrounding medium is air (μ₁ = 1), then: 1/f = (μ - 1)(1/R₁ - 1/R₂)