ISC Class 12 Physics Board Exam Question Paper 2017 with Solutions

ISC Class 12 Physics Board Exam Question Paper 2017 with Solutions

Part - I (20 Marks)

Question 1.

(a) Choose the correct alternative (a), (b), (c) or (d) for each of the questions given below: [5]

(i) The electrostatic potential energy of two point charges, \( 1\,\mu\text{C} \) each, placed 1 metre apart in air is : [1 Mark]
(a) \( 9 \times 10^{3}\,\text{J} \)
(b) \( P \times 10^{9}\,\text{J} \)
(c) \( 9 \times 10^{-3}\,\text{J} \)
(d) \( 9 \times 10^{-3}\,\text{eV} \)

Answer: (c) \( 9 \times 10^{-3}\,\text{J} \)

\( U = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{10^{-6} \times 10^{-6}}{1} = 9 \times 10^{-3}\,\text{J} \).

Teacher's Note:
a) Use the standard formula for electrostatic potential energy between two point charges in vacuum or air.
b) Pay close attention to powers of ten and units of microcoulombs during conversion.

(ii) A wire of resistance \( R \) is cut into \( n \) equal parts. These parts are then connected in parallel with each other. The equivalent resistance of the combination is : [1 Mark]
(a) \( nR \)
(b) \( R/n \)
(c) \( n/R^2 \)
(d) \( R/n^2 \)

Answer: (d) \( R/n^2 \)

Resistance of each part is \( R' = R/n \). When \( n \) such identical resistors are connected in parallel, equivalent resistance \( R_{eq} = \frac{R'}{n} = \frac{R}{n^2} \).

Teacher's Note:
a) Resistance is directly proportional to length, so each cut piece has a resistance of \( R/n \).
b) A common student mistake is to write \( R/n \), which applies when resistors are connected in series.

(iii) Magnetic susceptibility of platinum is 0.0001. Its relative permeability is : [1 Mark]
(a) 1.0000
(b) 0.9999
(c) 1.0001
(d) 0

Answer: (c) 1.0001

Relative permeability \( \mu_r = 1 + \chi_m = 1 + 0.0001 = 1.0001 \).

Teacher's Note:
a) Recall the fundamental relation between relative permeability and magnetic susceptibility: \( \mu_r = 1 + \chi_m \).
b) Ensure proper addition of decimals; platinum is a paramagnetic substance having a small positive susceptibility.

(iv) When a light wave travels from air to glass : [1 Mark]
(a) its wavelength decreases
(b) its wavelength increases
(c) there is no change in wavelength
(d) its frequency decreases.

Answer: (a) its wavelength decreases

When light enters from a rarer medium (air) to a denser medium (glass), speed and wavelength decrease while frequency remains constant.

Teacher's Note:
a) Frequency is a characteristic of the source and does not change when light changes medium.
b) Since \( v = f\lambda \) and speed \( v \) decreases in glass, wavelength \( \lambda \) must decrease proportionally.

(v) A radioactive substance decays to \( 1/16^{\text{th}} \) of its initial mass in 40 days. The half life of the substance, in days, is: [1 Mark]
(a) 20
(b) 10
(c) 5
(d) 2.5

Answer: (b) 10

\( N = N_0 \left(\frac{1}{2}\right)^n \implies \frac{N}{N_0} = \frac{1}{16} = \left(\frac{1}{2}\right)^4 \), so \( n = 4 \) half-lives. Half-life \( T_{1/2} = \frac{40}{4} = 10 \) days.

Teacher's Note:
a) Relate the fraction remaining to powers of two to find the total number of half-lives elapsed.
b) Divide the total given time by the number of half-lives to determine the duration of one half-life.

(b) Answer all questions given below briefly and to the point: [15]

(i) Maximum torque acting on the electric dipole of moment \( 3 \times 10^{-29}\,\text{C m} \) in a uniform electric field \( E \) is \( 6 \times 10^{-25}\,\text{N m} \). Find \( E \). [1 Mark]

Answer:
\( \tau_{\max} = pE \sin(90^{\circ}) = pE \)
\( E = \frac{\tau_{\max}}{p} = \frac{6 \times 10^{-25}}{3 \times 10^{-29}} = 2 \times 10^{4}\,\text{N C}^{-1} \)

Teacher's Note:
a) Maximum torque on a dipole occurs when the dipole moment vector is perpendicular to the electric field direction.
b) Ensure standard SI units are maintained and powers of ten are simplified correctly.

(ii) What is meant by drift speed of free electrons ? [1 Mark]

Answer:
The average velocity with which free electrons get drifted towards the positive terminal under the influence of an applied external electric field is called drift speed.

Teacher's Note:
a) Mention that it is an average velocity of electrons superimposed over their random thermal motion.
b) Always specify the presence of an external electric field responsible for this directional drift.

(iii) On which conservation principle is Kirchhoff's Second Law of electrical networks based ? [1 Mark]

Answer:
Kirchhoff's Second Law (Loop rule) is based on the law of conservation of energy.

Teacher's Note:
a) Recall that Kirchhoff's First Law (junction rule) is based on conservation of charge.
b) Kirchhoff's Second Law states that the sum of potential changes around any closed loop is zero, signifying energy conservation.

(iv) Calculate magnetic flux density of the magnetic field at the center of a circular coil of 50 turns, having radius of \( 0.5\,\text{m} \) and carrying a current of \( 5\,\text{A} \). [1 Mark]

Answer:
\( B = \frac{\mu_0 NI}{2r} = \frac{4\pi \times 10^{-7} \times 50 \times 5}{2 \times 0.5} = 3.14 \times 10^{-4}\,\text{T} \)

Teacher's Note:
a) Apply the standard magnetic field formula at the centre of a circular current-carrying coil.
b) Substitute \( \pi \approx 3.14 \) carefully to arrive at the final numerical value with correct units.

(v) An a.c. generator generates an emf \( e \) where \( e = 314 \sin(50\pi t) \) volt. Calculate the frequency of the emf. [1 Mark]

Answer:
Comparing with standard equation \( e = e_0 \sin(\omega t) \), we get \( \omega = 50\pi \).
\( 2\pi f = 50\pi \implies f = \frac{50\pi}{2\pi} = 25\,\text{Hz} \).

Teacher's Note:
a) Compare the given sinusoidal equation with the standard equation to extract angular frequency \( \omega \).
b) Use the relation \( \omega = 2\pi f \) to determine the linear frequency.

(vi) With what type of source of light are cylindrical wave fronts associated ? [1 Mark]

Answer:
A line source of light (or a linear source).

Teacher's Note:
a) Point sources produce spherical wavefronts.
b) Line sources produce cylindrical wavefronts, and distant point sources produce plane wavefronts.

(vii) How is fringe width of an interference pattern in Young's double slit experiment affected if the two slits are brought closer to each other ? [1 Mark]

Answer:
Fringe width increases because fringe width \( \beta = \frac{\lambda D}{d} \), which is inversely proportional to the slit separation \( d \).

Teacher's Note:
a) State the fringe width formula clearly before concluding the effect.
b) Emphasize the inverse proportionality between slit separation \( d \) and fringe width \( \beta \).

(viii) In a regular prism, what is the relation between angle of incidence and angle of emergence when it is in the minimum deviation position ? [1 Mark]

Answer:
Angle of incidence equals angle of emergence (\( i = e \)).

Teacher's Note:
a) At minimum deviation, the ray passes symmetrically through the prism.
b) This symmetrical property implies that internal angles of refraction are also equal (\( r_1 = r_2 \)).

(ix) A converging lens of focal length \( 40\,\text{cm} \) is kept in contact with a diverging lens of focal length \( 30\,\text{cm} \). Find the focal length of the combination. [1 Mark]

Answer:
\( f_1 = +40\,\text{cm},\,f_2 = -30\,\text{cm} \)
\( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{40} + \frac{1}{-30} = \frac{3 - 4}{120} = -\frac{1}{120} \)
\( F = -120\,\text{cm} \) (Diverging combination)

Teacher's Note:
a) Apply proper sign conventions: positive for converging lens and negative for diverging lens.
b) Take reciprocal carefully after finding the net inverse focal length.

(x) How can the spherical aberration produced by a lens be minimised ? [1 Mark]

Answer:
By using a circular stop (or aperture) to block marginal rays, or by using a combination of suitable lenses.

Teacher's Note:
a) Spherical aberration occurs because marginal rays focus closer to the lens than paraxial rays.
b) Blocking marginal rays with an opaque stop or using crossed lenses minimizes this defect.

(xi) Calculate the momentum of a photon of energy \( 6 \times 10^{-19}\,\text{J} \). [1 Mark]

Answer:
\( E = pc \implies p = \frac{E}{c} = \frac{6 \times 10^{-19}}{3 \times 10^8} = 2 \times 10^{-27}\,\text{kg m s}^{-1} \)

Teacher's Note:
a) Use the relation between energy and momentum of a photon: \( E = pc \).

b) Substitute speed of light \( c = 3 \times 10^8\,\text{m s}^{-1} \) to obtain momentum in correct units.

(xii) According to Bohr, Angular momentum of an orbiting electron is quantised. What is meant by this statement ? [1 Mark]

Answer:
It means that the angular momentum of an orbiting electron can only take specific discrete values which are integral multiples of \( \frac{h}{2\pi} \), i.e., \( mvr = \frac{nh}{2\pi} \).

Teacher's Note:
a) Mention the quantization condition formula clearly along with the statement.
b) State that orbits where this condition is not satisfied are forbidden for electrons.

(xiii) Why nuclear fusion reaction is also called thermo-nuclear reaction ? [1 Mark]

Answer:
Because extremely high temperatures (of the order of \( 10^7\,\text{K} \)) are required to provide sufficient kinetic energy to nuclei to overcome electrostatic repulsion between them.

Teacher's Note:
a) Nuclear fusion requires overcoming high Coulomb barrier forces.
b) High thermal energy is mandatory to initiate fusion reactions, hence the term thermonuclear.

(xiv) What is the minimum energy which a gamma ray photon must possess in order to produce electron-positron pair ? [1 Mark]

Answer:
The minimum energy required is \( 1.02\,\text{MeV} \) (or equal to the rest mass energy of an electron-positron pair).

Teacher's Note:
a) Pair production requires energy equal to at least twice the rest mass energy of an electron (\( 2 \times 0.51\,\text{MeV} \)).
b) Specify \( 1.02\,\text{MeV} \) as the threshold energy for pair production.

(xv) Show the variation of voltage with time, for a digital signal. [1 Mark]

Answer:

[Figure: A square wave pulse showing abrupt transitions between two discrete voltage levels, typically 0V and 5V, plotted against time t.]

Teacher's Note:
a) Digital signals have discrete voltage levels (binary 0 and 1) rather than continuous variation.
b) Draw clear rectangular pulses showing sharp transitions between high and low states.

Part - II (20 Marks)

Section - A

(Answer any four questions)

Question 2.

(a) Show that electric potential at a point P, at a distance \( r \) from a fixed point charge \( Q \), is given by: [4 Marks]
\( V = \left(\frac{1}{4\pi\varepsilon_0}\right) \frac{Q}{r} \)

(b) Intensity of electric field at a perpendicular distance of \( 0.5\,\text{m} \) from an infinitely long line charge having linear charge density \( \lambda \) is \( 3.6 \times 10^3\,\text{V m}^{-1} \). Find the value of \( \lambda \). [1 Mark]

Answer:
(a) Let a point charge \( +Q \) be placed at origin A. Consider a test charge \( q_0 \) at point P at a distance \( r \) from A. The electrostatic force between \( Q \) and \( q_0 \) is \( F = \frac{1}{4\pi\varepsilon_0}\frac{Qq_0}{x^2} \).
Work done in moving test charge through a small displacement \( dx \) against electric force is \( dW = -F\,dx = -\frac{1}{4\pi\varepsilon_0}\frac{Qq_0}{x^2}\,dx \).
Total work done in moving test charge from infinity to distance \( r \) is:
\( W = \int_{\infty}^{r} -\frac{1}{4\pi\varepsilon_0}\frac{Qq_0}{x^2}\,dx = \frac{Q q_0}{4\pi\varepsilon_0 r} \)
Electric potential \( V = \frac{W}{q_0} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r} \).

(b) Electric field intensity due to infinite line charge is \( E = \frac{\lambda}{2\pi\varepsilon_0 r} \).
Given \( E = 3.6 \times 10^3\,\text{V m}^{-1},\,r = 0.5\,\text{m} \).
\( \lambda = E \cdot 2\pi\varepsilon_0 r = 3.6 \times 10^3 \times \frac{0.5}{9 \times 10^9} = 2 \times 10^{-7}\,\text{C m}^{-1} \) (or \( 10^{-7}\,\text{C m}^{-1} \) based on calculation values).

Teacher's Note:
a) For derivation part, proper integration limits from infinity to \( r \) must be clearly shown.
b) For numerical part, substitute values using \( \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \) for clean calculation.

Question 3.

(a) Three capacitors \( C_1 = 3\,\mu\text{F} \), \( C_2 = 6\,\mu\text{F} \) and \( C_3 = 10\,\mu\text{F} \) are connected to a 50 V battery as shown in the Figure 1 below : [3 Marks]
Calculate :
(i) The equivalent capacitance of the circuit between points A and B.
(ii) The charge on \( C_1 \)
[Figure: Capacitors C1 and C2 are in series across points A and B, and this combination is in parallel with C3 of 10 microfarad. A 50V battery is connected across the circuit terminals.]
(b) Two resistors \( R_1 = 60\,\Omega \) and \( R_2 = 90\,\Omega \) are connected in parallel. If electric power consumed by the resistor \( R_1 \) is \( 15\,\text{W} \), calculate the power consumed by the resistor \( R_2 \). [2 Marks]

Answer:
(a) (i) Capacitors \( C_1 \) and \( C_2 \) are in series, so their equivalent capacitance \( C' \) is:
\( \frac{1}{C'} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} \implies C' = 2\,\mu\text{F} \)
Now \( C' \) and \( C_3 \) are in parallel, so total equivalent capacitance \( C_{eq} = C' + C_3 = 2 + 10 = 12\,\mu\text{F} \).
(ii) Since \( C_1 \) and \( C_2 \) are in series across 50 V, charge on \( C_1 \) equals charge on combination \( C' \):
\( q_1 = C' \times V = 2\,\mu\text{F} \times 50\,\text{V} = 100\,\mu\text{C} = 10^{-4}\,\text{C} \).

(b) For parallel combination, potential difference across both resistors is same.
Power \( P_1 = \frac{V^2}{R_1} \implies 15 = \frac{V^2}{60} \implies V^2 = 900\,\text{V}^2 \implies V = 30\,\text{V} \).
Power consumed by \( R_2 \) is \( P_2 = \frac{V^2}{R_2} = \frac{900}{90} = 10\,\text{W} \).

Teacher's Note:
a) Remember that capacitors combine in reverse of resistors (series formula for parallel and vice-versa).
b) In parallel resistor circuits, voltage remains constant across each branch.

Question 4.

(a) Figure 2 below shows two resistors \( R_1 \) and \( R_2 \) connected to a battery having an emf of 40 V and negligible internal resistance. A voltmeter having a resistance of \( 300\,\Omega \) is used to measure potential difference across \( R_1 \). Find the reading of the voltmeter. [3 Marks]
[Figure: Resistor R1 (200 ohms) and R2 (880 ohms) are in series across a 40V source. A voltmeter of resistance 300 ohms is connected in parallel across R1.]
(b) A moving coil galvanometer has a coil of resistance \( 59\,\Omega \). It shows a full scale deflection for a current of \( 50\,\text{mA} \). How will you convert it to an ammeter having a range of 0 to 3A ? [2 Marks]

Answer:
(a) Equivalent resistance of \( R_1 \) (\( 200\,\Omega \)) connected in parallel with voltmeter (\( 300\,\Omega \)):
\( R' = \frac{200 \times 300}{200 + 300} = \frac{60000}{500} = 120\,\Omega \)
Total resistance of the circuit = \( R' + R_2 = 120 + 880 = 1000\,\Omega \).
Total current in circuit \( I = \frac{V}{R_{total}} = \frac{40}{1000} = 0.04\,\text{A} \).
Voltmeter reading = potential difference across \( R' \) = \( I \times R' = 0.04 \times 120 = 4.8\,\text{V} \).

(b) Given \( G = 59\,\Omega,\,I_g = 50\,\text{mA} = 50 \times 10^{-3}\,\text{A},\,I = 3\,\text{A} \).
Shunt resistance \( S = \frac{I_g G}{I - I_g} = \frac{50 \times 10^{-3} \times 59}{3 - 50 \times 10^{-3}} = \frac{2.95}{2.95} = 1\,\Omega \).
A shunt of \( 1\,\Omega \) must be connected in parallel with the galvanometer.

Teacher's Note:
a) A voltmeter connected across a resistor alters the circuit resistance, so calculate the parallel combination first.
b) For galvanometer conversion to ammeter, a small shunt resistance is always connected in parallel.

Question 5.

(a) In a meter bridge circuit, resistance in the left hand gap is \( 2\,\Omega \) and an unknown resistance \( X \) is in the right hand gap as shown in Figure 3 below. The null point is found to be 40 cm from the left end of the wire. What resistance should be connected to \( X \) so that the new null point is 50 cm from the left end of the wire ? [3 Marks]
[Figure: Standard meter bridge setup with 2 ohm resistance in left gap, unknown resistance X in right gap, and a null point at 40 cm.]
(b) The horizontal component of earth's magnetic field at a place is \( 1/\sqrt{3} \) times the vertical component. Determine the angle of dip at that place. [2 Marks]

Answer:
(a) Initially, using meter bridge formula: \( \frac{R}{X} = \frac{l_1}{100 - l_1} \implies \frac{2}{X} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3} \implies X = 3\,\Omega \).
For new null point at \( 50\,\text{cm} \), let equivalent resistance in right gap be \( S \):
\( \frac{2}{S} = \frac{50}{100 - 50} = 1 \implies S = 2\,\Omega \).\br />Since \( S \) is formed by connecting resistance \( R \) in parallel with \( X \) (\( 3\,\Omega \)):
\( \frac{1}{S} = \frac{1}{X} + \frac{1}{R} \implies \frac{1}{2} = \frac{1}{3} + \frac{1}{R} \implies \frac{1}{R} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \implies R = 6\,\Omega \).

(b) Given \( B_h = \frac{1}{\sqrt{3}} B_v \).
Angle of dip \( \theta \) is given by \( \tan\theta = \frac{B_v}{B_h} = \frac{B_v}{(1/\sqrt{3})B_v} = \sqrt{3} \).
\( \theta = \tan^{-1}(\sqrt{3}) = 60^{\circ} \).

Teacher's Note:
a) Remember to solve for the unknown resistance \( X \) first before finding the parallel resistance required for the new null point.
b) Angle of dip is defined as the angle made by the total earth's magnetic field vector with the horizontal.

Question 6.

(a) Using Ampere's circuital law, obtain an expression for the magnetic flux density 'B' at a point X at a perpendicular distance 'r' from a long current carrying conductor. (Statement of the law is not required). [3 Marks]
(b) PQ is a long straight conductor carrying a current of 3A as shown in Figure 4 below. An electron moves with a velocity of \( 2 \times 10^7\,\text{m s}^{-1} \) parallel to it. Find the force acting on the electron. [2 Marks]
[Figure: Long straight wire PQ carrying 3A current. An electron moves parallel to it at a perpendicular distance of 0.6 m with velocity \( 2 \times 10^7\,\text{m s}^{-1} \).]

Answer:
(a) Consider a long straight wire carrying current \( I \). By Ampere's circuital law, \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I \).
For a circular Amperian loop of radius \( r \), \( B \cdot 2\pi r = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r} = \frac{\mu_0}{4\pi}\frac{2I}{r} \).

(b) Given \( I = 3\,\text{A},\,r = 0.6\,\text{m},\,v = 2 \times 10^7\,\text{m s}^{-1},\,q = 1.6 \times 10^{-19}\,\text{C} \).
Magnetic field at distance \( r \) due to conductor: \( B = \frac{\mu_0 I}{2\pi r} = \frac{2 \times 10^{-7} \times 3}{0.6} = 10^{-6}\,\text{T} \).
Force on electron \( F = qvB\sin(90^{\circ}) = 1.6 \times 10^{-19} \times 2 \times 10^7 \times 10^{-6} = 3.2 \times 10^{-18}\,\text{N} \).

Teacher's Note:
a) State Ampere's law clearly and choose a symmetrical circular loop centred on the wire.
b) Ensure proper substitution of magnetic field and Lorentz force formula.

Question 7.

(a) (i) AB and CD are two parallel conductors kept 1 m apart and connected by a resistance R of \( 6\,\Omega \) as shown in Figure 5 below. They are placed in a magnetic field \( B = 3 \times 10^{-2}\,\text{T} \) which is perpendicular to the plane of the conductors and directed into the paper. A wire MN is placed over AB and CD and then made to slide with a velocity \( 2\,\text{m s}^{-1} \). (Neglect the resistance of AB, CD, and MN). [3 Marks]
Calculate the induced current flowing through the resistor R.
(ii) In an ideal transformer, an output of 66 kV is required when an input voltage of 220 V is available. If the primary has 300 turns, how many turns should the secondary have ?
[Figure: Parallel rails AB and CD connected by a resistor R, with a sliding wire MN across them in a uniform magnetic field directed into the page.]
(b) In a series LCR circuit, obtain an expression for the resonant frequency. [2 Marks]

Answer:
(a) (i) Induced emf \( e = Bvl = 3 \times 10^{-2} \times 2 \times 1 = 6 \times 10^{-2}\,\text{V} \).
Induced current \( I = \frac{e}{R} = \frac{6 \times 10^{-2}}{6} = 10^{-2}\,\text{A} = 0.01\,\text{A} \).
(ii) \( E_p = 220\,\text{V},\,E_s = 66\,\text{kV} = 66 \times 10^3\,\text{V},\,N_p = 300 \).
\( \frac{E_p}{E_s} = \frac{N_p}{N_s} \implies \frac{220}{66 \times 10^3} = \frac{300}{N_s} \implies N_s = \frac{300 \times 66 \times 10^3}{220} = 90000\,\text{turns} \).

(b) In a series LCR circuit, impedance is given by \( Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \).
Condition for resonance is when inductive reactance equals capacitive reactance: \( \omega L = \frac{1}{\omega C} \).
\( \omega_r^2 = \frac{1}{LC} \implies \omega_r = \frac{1}{\sqrt{LC}} \implies f_r = \frac{1}{2\pi\sqrt{LC}} \).

Teacher's Note:
a) Use motional emf formula \( e = Bvl \) for the sliding rod problem.
b) At resonance, impedance is minimum and purely resistive, leading to maximum current.

Section - B

(Answer any three questions)

Question 8.

(a) (i) State any one property which is common to all electromagnetic waves. [3 Marks]
(ii) Arrange the following electromagnetic waves in increasing order of their frequencies (i.e., begin with the lowest frequency) :
Visible light, \( \gamma \)-rays, X-rays, microwaves, radio waves, infrared radiations and ultraviolet radiations.
(b) (i) What is meant by diffraction of light ? [2 Marks]
(ii) In Fraunhofer diffraction, what kind of source of light is used and where is it situated?

Answer:
(a) (i) All electromagnetic waves travel through vacuum with the same speed (\( c = 3 \times 10^8\,\text{m s}^{-1} \)) and are uncharged (do not deflect in electric and magnetic fields).
(ii) Radio waves < Microwaves < Infrared radiations < Visible light < Ultraviolet radiations < X-rays < \( \gamma \)-rays.
(b) (i) The bending of light around the corners of an obstacle or aperture of size comparable to the wavelength of light into the region of geometrical shadow is called diffraction.
(ii) A monochromatic point or slit source placed at the focal point of a convex lens is used to obtain a parallel beam of light.

Teacher's Note:
a) Memorize the electromagnetic spectrum order either in terms of increasing frequency or increasing wavelength.
b) Clearly distinguish diffraction from refraction and reflection by mentioning obstacle dimensions.

Question 9.

(a) In Young's double slit experiment using monochromatic light of wavelength 600 nm, 5th bright fringe is at a distance of 0.48 mm from the center of the pattern. If the screen is at a distance of 80 cm from the plane of the two slits, calculate : [3 Marks]
(i) Distance between the two slits.
(ii) Fringe width i.e., fringe separation.
(b) (i) State Brewster's law. [2 Marks]
(ii) Find Brewster's angle for glass of refractive index 1.5.

Answer:
(a) Given \( \lambda = 600\,\text{nm} = 6 \times 10^{-7}\,\text{m},\,x_5 = 0.48\,\text{mm} = 0.48 \times 10^{-3}\,\text{m},\,D = 80\,\text{cm} = 0.8\,\text{m},\,n = 5 \).
(i) Position of \( n^{\text{th}} \) bright fringe: \( x_n = \frac{n\lambda D}{d} \)
\( d = \frac{n\lambda D}{x_5} = \frac{5 \times 6 \times 10^{-7} \times 0.8}{0.48 \times 10^{-3}} = 5 \times 10^{-3}\,\text{m} = 5\,\text{mm} \).
(ii) Fringe width \( \beta = \frac{\lambda D}{d} = \frac{6 \times 10^{-7} \times 0.8}{5 \times 10^{-3}} = 9.6 \times 10^{-5}\,\text{m} = 0.096\,\text{mm} \).

(b) (i) Brewster's law states that the tangent of the polarising angle (Brewster's angle) is equal to the refractive index of the reflecting medium (\( \mu = \tan i_p \)).
(ii) Given \( \mu = 1.5 \implies i_p = \tan^{-1}(1.5) \approx 56.3^{\circ} \).

Teacher's Note:
a) Ensure all quantities are converted to standard SI units before substitution.
b) Brewster's law also implies that at polarising angle, reflected and refracted rays are mutually perpendicular.

Question 10.

(a) Find critical angle for glass and water pair, given refractive index of glass is 1.62 and that of water is 1.33. [2 Marks]
(b) Starting with an expression for refraction at a single spherical surface, obtain Lens Maker's Formula. [3 Marks]

Answer:
(a) Relative refractive index of glass with respect to water: \( ^w\mu_g = \frac{\mu_g}{\mu_w} = \frac{1.62}{1.33} \approx 1.218 \).
Critical angle \( c \) is given by \( \sin c = \frac{1}{^w\mu_g} = \frac{1.33}{1.62} \approx 0.821\, \implies c = \sin^{-1}(0.821) \approx 55^{\circ} \).

(b) For refraction at a single spherical surface of radius \( R \) separating media of refractive indices \( \mu_1 \) and \( \mu_2 \):
\( \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \).
For a thin lens of refractive index \( \mu \) placed in air (\( \mu_1 = 1 \)), applying refraction twice for both surfaces:
First surface: \( \frac{\mu}{v_1} - \frac{1}{u} = \frac{\mu - 1}{R_1} \).
Second surface: \( \frac{1}{v} - \frac{\mu}{v_1} = \frac{1 - \mu}{R_2} \).
Adding both equations: \( \frac{1}{v} - \frac{1}{u} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \).
Since object is at infinity (\( u = \infty \)) and image is at focus (\( v = f \)), we get Lens Maker's Formula:
\( \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \).

Teacher's Note:
a) For critical angle between two media, the denser medium refractive index goes in the denominator relative to rarer medium.
b) Clearly state the assumptions made for thin lenses during derivation.

Question 11.

(a) A compound microscope consists of two convex lenses of focal length 2 cm and 5 cm. When an object is kept at a distance of 2.1 cm from the objective, a virtual and magnified image is formed 25 cm from the eye piece. Calculate the magnifying power of the microscope. [3 Marks]
(b) (i) What is meant by resolving power of a telescope ?
(ii) State any one method of increasing the resolving power of an astronomical telescope. [2 Marks]

Answer:
(a) Given \( f_o = 2\,\text{cm},\,f_e = 5\,\text{cm},\,u_o = -2.1\,\text{cm},\,v_e = -25\,\text{cm} \).
For objective lens: \( \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \implies \frac{1}{2} = \frac{1}{v_o} - \frac{1}{-2.1} \)
\( \frac{1}{v_o} = \frac{1}{2} - \frac{1}{2.1} = \frac{2.1 - 2}{4.2} = \frac{0.1}{4.2} \implies v_o = 42\,\text{cm} \).
Magnifying power \( M = -\frac{v_o}{u_o}(1 + \frac{D}{f_e}) = -\frac{42}{-2.1}(1 + \frac{25}{5}) = 20 \times (1 + 5) = 20 \times 6 = -120 \).

(b) (i) Resolving power of a telescope is defined as the reciprocal of the smallest angular separation between two distant objects whose images can be distinctly resolved by the telescope.
(ii) By increasing the diameter (aperture) of the objective lens.

Teacher's Note:
a) Pay close attention to sign conventions for microscope lens formulas.
b) Resolving power increases with larger objective lens aperture due to decreased diffraction limits.

Section - C

(Answer any three questions)

Question 12.

(a) (i) Plot a labelled graph of \( V_s \) where \( V_s \) is stopping potential versus frequency \( f \) of the incident radiation.
(ii) State how will you use this graph to determine the value of Planck's constant. [3 Marks]
(b) (i) Find the de Broglie wavelength of electrons moving with a speed of \( 7 \times 10^6\,\text{m s}^{-1} \). [2 Marks]
(ii) Describe in brief what is observed when moving electrons are allowed to fall on a thin graphite film and the emergent beam falls on a fluorescent screen.

Answer:
(a) (i) [Figure: Graph of stopping potential \( V_s \) on y-axis versus frequency \( f \) on x-axis. A straight line starting from threshold frequency \( f_0 \) with a positive slope.]
(ii) The slope of the stopping potential versus frequency graph gives \( \frac{h}{e} \). Multiplying the slope by electronic charge \( e \) gives Planck's constant \( h \).

(b) (i) \( \lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 7 \times 10^6} \approx 0.1\,\text{nm} \) (or \( 1.03 \times 10^{-10}\,\text{m} \)).
(ii) Concentric rings are observed on the fluorescent screen, which demonstrates the wave nature (diffraction) of electrons.

Teacher's Note:
a) Einstein's photoelectric equation \( eV_s = hf - W_0 \) forms the basis of the \( V_s \) vs \( f \) graph.
b) Davisson-Germer experiment or electron diffraction confirms de Broglie's hypothesis of matter waves.

Question 13.

(a) Draw energy level diagram for hydrogen atom, showing first four energy levels corresponding to \( n = 1, 2, 3 \) and 4. Show transitions responsible for : [3 Marks]
(i) Absorption spectrum of Lyman series.
(ii) Emission spectrum of Balmer series.
(b) (i) Find maximum frequency of X-rays produced by an X-ray tube operating at a tube potential of 66 kV. [2 Marks]
(ii) State any one difference between characteristic X-rays and continuous X-rays.

Answer:
(a) (i) [Figure: Energy level diagram showing upward transitions from ground state \( n=1 \) to higher states \( n=2, 3, 4 \) representing Lyman absorption.]
(ii) [Figure: Energy level diagram showing downward transitions ending at \( n=2 \) from higher states \( n=3, 4, \dots \) representing Balmer emission.]

(b) (i) Maximum frequency \( \nu_{\max} = \frac{eV}{h} = \frac{1.6 \times 10^{-19} \times 66 \times 10^3}{6.6 \times 10^{-34}} = 1.6 \times 10^{19}\,\text{Hz} \).
(ii) Continuous X-rays have a continuous range of wavelengths with a sharp minimum wavelength limit, whereas characteristic X-rays have definite discrete wavelengths.

Teacher's Note:
a) Lyman series transitions always terminate at \( n=1 \) (ultraviolet region), while Balmer series terminate at \( n=2 \) (visible region).
b) Duane-Hunt law relates maximum frequency of X-rays directly to operating accelerating voltage.

Question 14.

(a) Obtain a relation between half life of a radioactive substance and decay constant \( \lambda \). [2 Marks]
(b) Calculate mass defect and binding energy per nucleon of \( _{10}^{20}\text{Ne} \), given : [3 Marks]
Mass of \( _{10}^{20}\text{Ne} = 19.992397\,\text{u} \)
Mass of \( _1^1\text{H} = 1.007825\,\text{u} \)
Mass of \( _0^1\text{n} = 1.008665\,\text{u} \)

Answer:
(a) Radioactive decay law: \( N = N_0 e^{-\lambda t} \).
At \( t = T_{1/2} \), \( N = \frac{N_0}{2} \implies \frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \implies e^{\lambda T_{1/2}} = 2 \).
Taking natural logarithm on both sides: \( \lambda T_{1/2} = \ln(2) \implies T_{1/2} = \frac{0.693}{\lambda} \).

(b) Neon nucleus \( _{10}^{20}\text{Ne} \) has 10 protons and 10 neutrons.
Mass of constituents = \( (10 \times 1.007825) + (10 \times 1.008665) = 10.07825 + 10.08665 = 20.1649\,\text{u} \).
Mass defect \( \Delta m = 20.1649 - 19.992397 = 0.172503\,\text{u} \).
Binding energy \( BE = 0.172503 \times 931\,\text{MeV} \approx 160.6\,\text{MeV} \).
Binding energy per nucleon = \( \frac{160.6}{20} \approx 8.03\,\text{MeV/nucleon} \).

Teacher's Note:
a) Always memorize the half-life formula \( T_{1/2} = \frac{0.693}{\lambda} \) for quick verification.
b) When calculating mass defect, use proton and neutron masses accurately and multiply by 931 MeV to convert atomic mass units to energy.

Question 15.

(a) With reference to a semiconductor diode, what is meant by :
(i) Forward bias
(ii) Reverse bias
(iii) Depletion region
(b) Draw a diagram to show how NAND gates can be combined to obtain an OR gate (Truth table is not required). [2 Marks]

Answer:
(a) (i) Forward bias: When the p-type region of a diode is connected to the positive terminal and n-type to the negative terminal of a battery, reducing potential barrier.
(ii) Reverse bias: When the p-type region is connected to the negative terminal and n-type to the positive terminal, widening the depletion region.
(iii) Depletion region: The region near the p-n junction devoid of free mobile charge carriers due to electron-hole recombination.

(b) [Figure: Logic circuit diagram showing two NOT gates (made from single-input NAND gates) connected to the two inputs of a final NAND gate to form an OR gate.]

Teacher's Note:
a) Clearly mention polarity connections for both forward and reverse biasing.
b) De Morgan's laws help explain how universal gates like NAND can construct basic logic gates like OR.