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Using the double-slit formula: \( w = \frac{\lambda D}{d} \). For the first setup: \( w_1 = \frac{\lambda_1 D}{d} \). For the second setup: \( w_2 = \frac{\lambda_2 D}{2d} \). Dividing the second equation by the first: \( \frac{w_2}{w_1} = \frac{\lambda_2 D / (2d)}{\lambda_1 D / d} = \frac{\lambda_2}{2\lambda_1} \). Substituting the wavelengths: \( \frac{w_2}{w_1} = \frac{540 \text{ nm}}{2 \times 660 \text{ nm}} = \frac{540}{1320} \approx 0.41 \).

1 mark for the correct option A. Reject all other options.

The resistance of a wire is given by \( R = \rho \frac{L}{A} = \rho \frac{L}{\pi (d/2)^2} = \frac{4\rho L}{\pi d^2} \). Therefore, resistance is proportional to \( \frac{L}{d^2} \). Comparing Y to X: \( R_Y = R_X \times \frac{L_Y}{L_X} \times \left(\frac{d_X}{d_Y}\right)^2 = R_X \times 2 \times \left(\frac{1}{2}\right)^2 = 0.5 R_X \). Since the wires are in parallel, the potential difference \( V \) is the same across both. The power dissipated is \( P = \frac{V^2}{R} \), which means power is inversely proportional to resistance: \( \frac{P_X}{P_Y} = \frac{R_Y}{R_X} = \frac{0.5 R_X}{R_X} = 0.5 \). Thus, the ratio of power in X to Y is \( 1 : 2 \).

1 mark for the correct option B. Reject all other options.

According to the photoelectric equation, \( E_k = \frac{hc}{\lambda} - \phi \), where \( \phi \) is the work function of the metal. If the wavelength is halved to \( \frac{\lambda}{2} \), the energy of each incident photon doubles. The new maximum kinetic energy of the photoelectrons is: \( E_{k,\text{new}} = \frac{hc}{\lambda/2} - \phi = 2\left(\frac{hc}{\lambda}\right) - \phi \). Substituting \( \frac{hc}{\lambda} = E_k + \phi \) gives: \( E_{k,\text{new}} = 2(E_k + \phi) - \phi = 2E_k + \phi \). Since the work function \( \phi \) must be positive for photoemission to occur, \( E_{k,\text{new}} \) must be greater than \( 2E_k \).

1 mark for the correct option B. Reject all other options.

For a negative temperature coefficient (NTC) thermistor, a decrease in temperature causes its resistance to increase. In a series potential divider, \( V_{\text{out}} = V_{\text{supply}} \times \frac{R_{\text{thermistor}}}{R + R_{\text{thermistor}}} \). As the resistance of the thermistor increases, it takes a greater share of the total resistance, and therefore the potential difference across it, \( V_{\text{out}} \), increases.

1 mark for the correct option D. Reject all other options.

When unpolarised light passes through the first polarizing filter, its intensity is reduced by half, so \( I_1 = \frac{1}{2} I_0 \). When this polarized light passes through the second polarizing filter, Malus's Law applies: \( I_2 = I_1 \cos^2(\theta) \). Substituting the given values: \( I_2 = \left(\frac{1}{2} I_0\right) \cos^2(30^\circ) \). Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), we have \( \cos^2(30^\circ) = \frac{3}{4} \). Thus, \( I_2 = \frac{1}{2} I_0 \times \frac{3}{4} = \frac{3}{8} I_0 \).

1 mark for the correct option C. Reject all other options.

The critical angle \( \theta_c \) at a glass-air boundary is given by \( \sin(\theta_c) = \frac{1}{n} \), where \( n \) is the refractive index of the glass. An increase in refractive index \( n \) leads to a smaller value of \( \sin(\theta_c) \), and thus a smaller critical angle. The speed of light in glass is \( v = \frac{c}{n} \). An increase in \( n \) results in a lower speed of light. Therefore, both the critical angle and the speed of light decrease.

1 mark for the correct option A. Reject all other options.

The terminal potential difference \( V \) is related to the current \( I \) by the equation \( V = E - Ir \). Rearranging this in the form of a straight line equation \( y = mx + c \) gives \( V = -rI + E \). Thus, the vertical axis intercept (y-intercept) represents the e.m.f. \( E \), and the gradient of the graph is \( -r \). The magnitude of this gradient is therefore equal to the internal resistance \( r \).

1 mark for the correct option A. Reject all other options.

For a string of length \( L \) fixed at both ends, the condition for stationary waves is \( L = n \frac{\lambda}{2} \), where \( n \) is the harmonic number. For the third harmonic, \( n = 3 \), which gives \( L = \frac{3\lambda}{2} \) or \( \lambda = \frac{2}{3}L \). A vibration in the third harmonic consists of three loops, which are bounded by 4 nodes in total (one at each end and two along the length of the string).

1 mark for the correct option B. Reject all other options.

According to Einstein's photoelectric equation: \(E_k = hf - \Phi\), where \(h\) is Planck's constant. Rearranging this gives \(hf = E_k + \Phi\). When the frequency of the incident light is doubled to \(2f\), the new maximum kinetic energy \(E_k'\) is given by: \(E_k' = h(2f) - \Phi = 2(hf) - \Phi\). Substituting the first expression into the second: \(E_k' = 2(E_k + \Phi) - \Phi = 2E_k + 2\Phi - \Phi = 2E_k + \Phi\). Thus, the correct option is B.

[1] B - Correctly derives the new maximum kinetic energy as 2E_k + \Phi

The total resistance of the circuit is \(R + r\). As the resistance of the variable resistor \(R\) increases, the total resistance of the circuit increases. Since \(I = \frac{\mathcal{E}}{R+r}\) and the e.m.f. remains constant, the current \(I\) decreases. The terminal potential difference is given by \(V = \mathcal{E} - Ir\). Since the current \(I\) decreases, the 'lost volts' \(Ir\) also decreases. Therefore, the terminal potential difference \(V\) increases. This corresponds to option B.

[1] B - Correctly identifies that current decreases and terminal potential difference increases

(a) Energy of photon \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{2.40 \times 10^{-7}\text{ m}} = 8.29 \times 10^{-19}\text{ J}\). In eV: \(\frac{8.29 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 5.18\text{ eV}\). (b) Since the photon energy (\(5.18\text{ eV}\)) is greater than the work function (\(4.30\text{ eV}\)), photoelectric emission will occur. The maximum kinetic energy in eV is \(5.18\text{ eV} - 4.30\text{ eV} = 0.88\text{ eV}\). In Joules: \(0.88 \times 1.60 \times 10^{-19}\text{ J} = 1.41 \times 10^{-19}\text{ J}\). (c) Increasing intensity increases the number of photons incident per second, but does not change the energy of individual photons. Therefore, the maximum kinetic energy of the emitted photoelectrons remains unchanged.

(a) [1 mark] Use of \(E = hc/\lambda\). [1 mark] Calculation of energy in J (\(8.29 \times 10^{-19}\text{ J}\)). [1 mark] Conversion to eV (\(5.18\text{ eV}\)). (b) [1 mark] Clear comparison of photon energy with work function to conclude emission occurs. [1 mark] Calculation of max KE in eV (\(0.88\text{ eV}\)). [1 mark] Conversion of max KE to J (\(1.41 \times 10^{-19}\text{ J}\)). (c) [1 mark] State that max KE is unchanged. [1 mark] Explain that intensity affects photon arrival rate, not individual photon energy.

(a) The total resistance of the circuit is \(R_{\text{total}} = 1.5\text{ k}\Omega + 3.5\text{ k}\Omega = 5.0\text{ k}\Omega\). The potential difference across the fixed resistor is \(V = E \times \frac{R_{\text{fixed}}}{R_{\text{total}}} = 9.0\text{ V} \times \frac{1.5\text{ k}\Omega}{5.0\text{ k}\Omega} = 2.7\text{ V}\). (b) As temperature increases, the resistance of the NTC thermistor decreases. This reduces the total resistance of the circuit, which increases the current. Since \(V = IR\) and the resistance of the fixed resistor is constant, the potential difference across it increases. (c) When the potential difference across the thermistor is \(3.0\text{ V}\), the potential difference across the fixed resistor is \(9.0\text{ V} - 3.0\text{ V} = 6.0\text{ V}\). Using the ratio of voltages: \(\frac{R_{\text{thermistor}}}{R_{\text{fixed}}} = \frac{V_{\text{thermistor}}}{V_{\text{fixed}}} \Rightarrow R_{\text{thermistor}} = 1.5\text{ k}\Omega \times \frac{3.0\text{ V}}{6.0\text{ V}} = 0.75\text{ k}\Omega = 750\ \Omega\).

(a) [1 mark] Calculation of total resistance (\(5.0\text{ k}\Omega\)). [1 mark] Use of potential divider formula. [1 mark] Correct value of \(2.7\text{ V}\). (b) [1 mark] State that thermistor resistance decreases. [1 mark] State that total circuit current increases (or thermistor takes a smaller fraction of the voltage). [1 mark] Conclude that the voltage across the fixed resistor increases. (c) [1 mark] State that the voltage across the fixed resistor is \(6.0\text{ V}\). [1 mark] Use of correct voltage-resistance ratio. [1 mark] Correct calculation of thermistor resistance (\(750\ \Omega\) or \(0.75\text{ k}\Omega\)).

(a) Using the fringe spacing equation \(\Delta y = \frac{\lambda D}{d}\): \(\Delta y = \frac{6.33 \times 10^{-7}\text{ m} \times 1.80\text{ m}}{0.25 \times 10^{-3}\text{ m}} = 4.56 \times 10^{-3}\text{ m} = 4.56\text{ mm}\), which is approximately \(4.6\text{ mm}\). (b) Since the wavelength of green light (\(5.20 \times 10^{-7}\text{ m}\)) is shorter than that of red light, and fringe spacing is directly proportional to wavelength (\(\Delta y \propto \lambda\)), the fringe spacing will decrease. The fringes will appear closer together. (c) If one slit is covered, two-source interference can no longer occur, so the alternating bright and dark fringes of equal width disappear. Instead, a single-slit diffraction pattern is observed, featuring a broad, bright central maximum with much fainter, narrower secondary maxima on either side.

(a) [1 mark] Identification of variables: \(d = 0.25 \times 10^{-3}\text{ m}\), \(D = 1.80\text{ m}\). [1 mark] Substitution into \(\Delta y = \frac{\lambda D}{d}\). [1 mark] Correct calculation to yield \(4.56\text{ mm}\). (b) [1 mark] State that green light has a shorter wavelength. [1 mark] Reference to \(\Delta y \propto \lambda\) or equivalent reasoning. [1 mark] Conclude that the fringe spacing decreases. (c) [1 mark] State that the double-slit interference pattern (or equal-width fringes) disappears. [1 mark] State that a single-slit diffraction pattern is formed. [1 mark] Describe the single-slit pattern (e.g., wide central maximum).

(a) From conservation of energy, the total electrical energy per unit charge supplied by the source (EMF, \(E\)) equals the electrical energy per unit charge delivered to the external load (\(V\)) plus the energy per unit charge wasted as heat in the internal resistance (\(Ir\)). Thus, \(E = V + Ir\), which rearranges to \(V = -Ir + E\). (b) Comparing \(V = -rI + E\) to the straight-line equation \(y = mx + c\): the y-intercept is the EMF, so \(E = 6.20\text{ V}\). The gradient is \(-r\), so the internal resistance \(r = 1.50\ \Omega\). (c) (i) Terminal potential difference \(V = 6.20\text{ V} - (1.80\text{ A} \times 1.50\ \Omega) = 6.20 - 2.70 = 3.50\text{ V}\). (ii) Power dissipated in the internal resistance is \(P = I^2 r = (1.80\text{ A})^2 \times 1.50\ \Omega = 3.24 \times 1.50 = 4.86\text{ W}\).

(a) [1 mark] State that \(E = V + V_{\text{lost}}
\) or \(E = I(R + r)\) based on conservation of energy. [1 mark] Show clear rearrangement to \(V = -Ir + E\). (b) [1 mark] EMF = \(6.20\text{ V}\). [1 mark] Internal resistance = \(1.50\ \Omega\) (must be positive). (c) [1 mark] Use of \(V = E - Ir\) with correct substitution. [1 mark] Terminal PD = \(3.50\text{ V}\). [1 mark] Use of \(P = I^2 r\) or \(P = I V_{\text{lost}}\). [1 mark] Power dissipated = \(4.86\text{ W}\).

(a) Grating spacing \(d = \frac{1}{N} = \frac{1}{5.00 \times 10^5\text{ m}^{-1}} = 2.00 \times 10^{-6}\text{ m}\). (b) Using \(d \sin \theta = n \lambda\) with \(n = 2\): For \(\lambda_A = 4.50 \times 10^{-7}\text{ m}\): \(\sin \theta_A = \frac{2 \times 4.50 \times 10^{-7}}{2.00 \times 10^{-6}} = 0.450 \Rightarrow \theta_A = \arcsin(0.450) = 26.74^\circ\). For \(\lambda_B = 6.00 \times 10^{-7}\text{ m}\): \(\sin \theta_B = \frac{2 \times 6.00 \times 10^{-7}}{2.00 \times 10^{-6}} = 0.600 \Rightarrow \theta_B = \arcsin(0.600) = 36.87^\circ\). The angular separation is \(\Delta \theta = 36.87^\circ - 26.74^\circ = 10.13^\circ\) (or \(10.1^\circ\)). (c) For \(n = 4\) and \(\lambda_B = 6.00 \times 10^{-7}\text{ m}\): \(\sin \theta = \frac{4 \times 6.00 \times 10^{-7}}{2.00 \times 10^{-6}} = 1.20\). Since \(\sin \theta\) cannot exceed \(1\), the fourth-order maximum is not physically possible and cannot be observed.

(a) [1 mark] Recall of \(d = 1/N\). [1 mark] Correct calculation of \(d = 2.00 \times 10^{-6}\text{ m}\). (b) [1 mark] Use of \(d \sin \theta = n \lambda\). [1 mark] Calculation of \(\theta_A = 26.74^\circ\) (or \(26.7^\circ\)). [1 mark] Calculation of \(\theta_B = 36.87^\circ\) (or \(36.9^\circ\)). [1 mark] Correct subtraction to yield \(10.1^\circ\) (accept \(10.1^\circ\) to \(10.2^\circ\)). (c) [1 mark] Substitute \(n=4\) and \(\lambda_B\) into diffraction formula. [1 mark] Calculate \(\sin \theta = 1.20\). [1 mark] State that \(\sin \theta > 1\) is impossible, hence the fourth order cannot exist.

(a) Polarized light has its electric field vector oscillations restricted to a single plane, which is perpendicular to the direction of wave propagation. (b) When unpolarized light passes through the first polarizing filter, its intensity is halved: \(I_1 = 0.5 I_0\). Applying Malus's law for the second filter: \(I_2 = I_1 \cos^2 \theta = 0.5 I_0 \cos^2(35.0^\circ)\). Since \(\cos(35.0^\circ) \approx 0.8192\), we have \(\cos^2(35.0^\circ) \approx 0.6710\). Thus, \(I_2 = 0.5 I_0 \times 0.6710 = 0.336 I_0\). The ratio \(I_2 / I_0 = 0.336\). (c) As the second filter is rotated through \(360^\circ\): when aligned (\(0^\circ, 180^\circ, 360^\circ\)), the transmitted intensity is at a maximum value of \(0.5 I_0\). When perpendicular (\(90^\circ, 270^\circ\)), the transmitted intensity is zero. The variation is smooth and periodic, following a \(\cos^2 \theta\) relationship.

(a) [1 mark] Reference to oscillations of the electric field vector. [1 mark] Restricted to a single plane perpendicular to the direction of propagation (or energy transfer). (b) [1 mark] State that intensity is halved after the first filter (\(I_1 = 0.5 I_0\)). [1 mark] Use of Malus's Law \(I = I_{\text{max}} \cos^2 \theta\). [1 mark] Substitute \(\theta = 35.0^\circ\). [1 mark] Correct calculation of ratio as \(0.336\) (accept \(0.34\)). (c) [1 mark] Maxima occur at \(0^\circ, 180^\circ, 360^\circ\) (with value \(0.5 I_0\)). [1 mark] Minima (zero intensity) occur at \(90^\circ, 270^\circ\). [1 mark] Smooth/periodic variation following \(\cos^2 \theta\).

(a) The cross-sectional area \(A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(0.180 \times 10^{-3}\text{ m}\right)^2 = 1.018 \times 10^{-7}\text{ m}^2\). Using \(R = \frac{\rho L}{A}\): \(R = \frac{1.70 \times 10^{-8}\ \Omega\text{ m} \times 2.40\text{ m}}{1.018 \times 10^{-7}\text{ m}^2} = 0.401\ \Omega\), which is approximately \(0.40\ \Omega\). (b) Using the drift velocity formula \(I = nAvq\): \(v = \frac{I}{nAe} = \frac{1.50\text{ A}}{8.50 \times 10^{28}\text{ m}^{-3} \times 1.018 \times 10^{-7}\text{ m}^2 \times 1.60 \times 10^{-19}\text{ C}}\). Denominator \(= 1384.48\). \(v = \frac{1.50}{1384.48} = 1.08 \times 10^{-3}\text{ m s}^{-1}\). (c) Since the mass and density are constant, the volume \(V = A \times L\) remains constant. If length is doubled (\(L' = 2L\)), the cross-sectional area must be halved (\(A' = A/2\)). The new resistance \(R' = \frac{\rho (2L)}{A/2} = 4 \left(\frac{\rho L}{A}\right) = 4R\). Thus, the resistance increases by a factor of 4.

(a) [1 mark] Calculation of cross-sectional area \(A = 1.02 \times 10^{-7}\text{ m}^2\). [1 mark] Substitution of values into \(R = \rho L / A\). [1 mark] Correct calculation of \(0.401\ \Omega\). (b) [1 mark] Recall of \(I = nAvq\). [1 mark] Rearrangement to make \(v\) the subject. [1 mark] Substitution of electronic charge \(e = 1.60 \times 10^{-19}\text{ C}\). [1 mark] Correct calculation of \(1.08 \times 10^{-3}\text{ m s}^{-1}\) (accept \(1.1 \times 10^{-3}\)). (c) [1 mark] Explain that doubling length halves the cross-sectional area because volume is constant. [1 mark] Deduce that \(R\) increases by a factor of 4.

(a) Total internal reflection requires: 1. The light must be traveling in a medium of higher refractive index towards a medium of lower refractive index (i.e., \(n_{\text{core}} > n_{\text{cladding}}\)). 2. The angle of incidence at the boundary must be greater than the critical angle. (b) Using Snell's law for the critical angle: \(\sin C = \frac{n_2}{n_1} = \frac{1.46}{1.52} = 0.9605\). Therefore, critical angle \(C = \arcsin(0.9605) = 73.84^\circ\) (or \(73.8^\circ\)). (c) The angle of incidence (\(\theta_i = 75.0^\circ\)) is greater than the critical angle (\(C = 73.8^\circ\)). Additionally, the light is traveling from a higher refractive index medium (core, \(n = 1.52\)) to a lower refractive index medium (cladding, \(n = 1.46\)). Therefore, total internal reflection occurs. The ray does not refract into the cladding but is completely reflected back into the core, obeying the law of reflection with an angle of reflection equal to \(75.0^\circ\).

(a) [1 mark] Light must travel from higher refractive index to lower refractive index. [1 mark] Angle of incidence must be greater than the critical angle. (b) [1 mark] Use of \(\sin C = n_2 / n_1\). [1 mark] Substitution of \(1.46\) and \(1.52\). [1 mark] Correct calculation of \(73.8^\circ\) (or \(73.9^\circ\)). (c) [1 mark] Identify that \(\theta_i = 75.0^\circ\) is greater than the critical angle. [1 mark] Conclude that total internal reflection (TIR) occurs. [1 mark] State that no light escapes into the cladding (or all light is reflected). [1 mark] State that the angle of reflection is \(75.0^\circ\).