Problem Set

1. A string of length $L = 1.20$ m is fixed at both ends and has wave speed $v = 240$ m/s. Find the fundamental frequency and the frequencies of the first three harmonics.

2. Show that $u(x,t) = A e^{i(kx-\omega t)} + B e^{-i(kx+\omega t)}$ satisfies the 1D wave Equation $\partial^2 u/\partial x^2 = (1/v^2)\partial^2 u/\partial t^2$. Identify the physical Meaning of each term and find the condition on $\omega$ and $k$.

3. A wave packet in a dispersive medium has central angular frequency $\omega_0 = 10^{15}$ rad/s And bandwidth $\Delta\omega = 10^{12}$ rad/s. The group velocity dispersion is $\alpha = d^2\omega/dk^2 = 2.0 \times 10^6$ m$^2$/s. Estimate the time required for the packet To double in spatial width after travelling a distance of 1.0 m.

4. The electric field of a plane wave is $\mathbf{E} = (20\hat{\mathbf{x}} - 15\hat{\mathbf{y}})\cos(kz - \omega t)$ V/m in vacuum. Find the amplitude, the polarisation state (including the angle and handedness), and the time-averaged Intensity.

5. Show that for normal incidence on a dielectric interface, the amplitude reflection and Transmission coefficients satisfy $t = 1 + r$. Prove this from the boundary conditions.

6. Unpolarised light is incident from water ($n = 1.33$) onto glass ($n = 1.50$). Calculate the Reflectance for (a) normal incidence, (b) $\theta_i = 45°$And (c) Brewster”s angle. At which Angle is the reflected light most strongly polarised?

7. An optical fibre has core index $n_1 = 1.48$ and cladding index $n_2 = 1.46$. Find the Critical angle for total internal reflection and the numerical aperture. What is the maximum Acceptance angle in air?

8. In a Young’s double-slit experiment, the slit separation is $d = 0.50$ mm and the screen is $L = 1.5$ m away. The fifth bright fringe is 8.2 mm from the central maximum. Find the wavelength.

9. A magnesium fluoride ($n = 1.38$) anti-reflection coating is deposited on a glass lens ($n = 1.52$). Find the minimum coating thickness for minimum reflection at $\lambda = 550$ nm. What is the reflectance of the uncoated lens at normal incidence?

10. A Michelson interferometer uses light from a sodium lamp ($\lambda = 589$ nm, $\Delta\lambda = 0.6$ nm). (a) If 1000 fringes are counted when one mirror moves, how far did it move? (b) Over what range of mirror displacement will interference fringes remain visible?

11. A Fabry-Perot etalon with $R = 0.85$ and plate separation $d = 0.50$ mm is illuminated at Normal incidence with $\lambda = 500$ nm. Calculate the finesse, the free spectral range (in Hz), And the resolving power. What is the minimum wavelength difference that can be resolved?

12. Monochromatic light of wavelength $\lambda = 633$ nm passes through a slit of width $a = 0.050$ mm onto a screen at distance $L = 3.0$ m. (a) Find the width of the central maximum. (b) Calculate the intensity at the position of the second secondary maximum relative to $I_0$.

13. A diffraction grating with 1200 lines/mm is illuminated at normal incidence by light containing Two wavelengths $\lambda_1 = 500.0$ nm and $\lambda_2 = 500.5$ nm. What minimum grating width is Needed to resolve these lines in the second order?

14. The Hubble Space Telescope has a primary mirror of diameter $D = 2.4$ m. Calculate its Angular resolution at $\lambda = 500$ nm in both radians and arcseconds. A ground-based telescope With $D = 8$ m operates under atmospheric seeing of $1.0''$. Which telescope achieves better Resolution, and why?

15. Unpolarised light of intensity $I_0$ passes through two ideal linear polarisers whose Transmission axes are at angle $\theta$ to each other. For what value of $\theta$ is the transmitted Intensity equal to $I_0/8$?

16. Linearly polarised light at $30°$ to the fast axis passes through a quarter-wave plate, Then through a half-wave plate whose fast axis is aligned with the quarter-wave plate’s fast axis. Describe the polarisation state after each element. What is the final polarisation state?

17. Light is incident from air onto a glass surface ($n = 1.50$) at Brewster’s angle. (a) Calculate the Brewster angle. (b) Find the angle of refraction and verify that the reflected and refracted beams are Perpendicular. (c) If the incident light is unpolarised with intensity $I_0$What is the intensity and Polarisation state of the reflected light?

18. An object is placed 30 cm from a converging lens ($f_1 = 20$ cm). A diverging lens ($f_2 = -15$ cm) is placed 60 cm from the converging lens on the opposite side. Using the ray Transfer matrix method, find the position and magnification of the final image. Verify your result Using the thin lens equation applied twice.

Selected Solutions

Solution 1. $f_1 = v/(2L) = 240/(2 \times 1.20) = 100$ Hz. The harmonics are $f_n = nf_1$: $f_1 = 100$ Hz, $f_2 = 200$ Hz, $f_3 = 300$ Hz.

Solution 2. $\partial^2 u/\partial x^2 = -k^2(Ae^{i(kx-\omega t)} + Be^{-i(kx+\omega t)})$ and $\partial^2 u/\partial t^2 = -\omega^2(Ae^{i(kx-\omega t)} + Be^{-i(kx+\omega t)})$. The wave Equation requires $k^2 = \omega^2/v^2$I.e., $\omega = \pm vk$. The first term is a wave Travelling in the $+x$ direction; the second is a wave travelling in the $-x$ direction.

Solution 3. Group velocity: $v_g = d\omega/dk \approx \Delta\omega/\Delta k$. Initial spatial Width: $\sigma_0 \approx 1/\Delta k = 1/(\Delta\omega/v_g)$. The packet doubles when $1 + (\alpha t/(2\sigma_0^2))^2 = 4$Giving $t = 2\sigma_0\sqrt{3}/\alpha$. Using $\sigma_0 = v_g/\Delta\omega$ and $v_g = \omega_0/k_0$ with $k_0 = \omega_0/c$ (assuming $v_g \approx c$ for Estimation): $\sigma_0 \approx 3 \times 10^8/10^{12} = 3 \times 10^{-4}$ m. $t = 2(3 \times 10^{-4})(\sqrt{3})/(2 \times 10^6) = 5.2 \times 10^{-10}$ s. Time to travel 1 m: $\sim 3.3 \times 10^{-9}$ s $\gg t$So the packet doubles well before reaching 1 m.

Solution 5. At normal incidence, $\theta_i = \theta_t = 0$ and $r_s = r_p = (n_1 - n_2)/(n_1 + n_2)$. Boundary condition on tangential $E$: $E_0(1 + r) = E_0 t$So $t = 1 + r$. $\blacksquare$

Solution 8. $y_5 = 5\lambda L/d \implies \lambda = y_5 d/(5L) = (8.2 \times 10^{-3})(0.50 \times 10^{-3})/(5 \times 1.5) = 5.47 \times 10^{-7}$ m $= 547$ nm.

Solution 9. Thickness: $t = \lambda/(4n) = 550/(4 \times 1.38) = 99.6$ nm. Uncoated reflectance: $R = [(1 - 1.52)/(1 + 1.52)]^2 = (0.52/2.52)^2 = 0.0426 = 4.26\%$.

Solution 10. (a) $\Delta d = m\lambda/2 = 1000 \times 589 \times 10^{-9}/2 = 2.945 \times 10^{-4}$ m $= 0.295$ mm. (b) Fringes are visible for path difference $\Delta x \lt L_c = \lambda^2/\Delta\lambda = (589 \times 10^{-9})^2/(0.6 \times 10^{-9}) = 5.78 \times 10^{-4}$ m $= 0.578$ mm. Since the path difference is $2\Delta d$The mirror can move up to $\Delta d = L_c/2 = 0.289$ mm before fringes wash out. Note that 1000 fringes correspond to $\Delta d = 0.295$ mm, which slightly exceeds $L_c/2$ — the outermost fringes would already be fading.

Solution 11. $\mathcal{F} = \pi\sqrt{0.85}/(1 - 0.85) = 19.3$. $\Delta\nu_{\mathrm{FSR} = c/(2nd) = 3 \times 10^8/(2 \times 0.5 \times 10^{-3}) = 3 \times 10^{11}}$ Hz. $m = 2nd/\lambda = 2000$. $\mathcal{R} = m\mathcal{F} = 38600$. $\delta\lambda = \lambda/\mathcal{R} = 500/38600 = 0.0130$ nm.

Solution 14. $\theta_{\min} = 1.22\lambda/D = 1.22(500 \times 10^{-9})/2.4 = 2.54 \times 10^{-7}$ rad $= 0.0527''$. The ground-based $D = 8$ m telescope has a diffraction limit of $1.22(500 \times 10^{-9})/8 = 7.63 \times 10^{-8}$ rad $= 0.0158''$But atmospheric seeing of $1.0''$ degrades this by a factor of $\sim 63$. Hubble, being above the atmosphere, achieves its diffraction-limited $0.053''$ resolution, far surpassing the ground-based telescope’s effective resolution.

Solution 17. (a) $\theta_B = \arctan(1.50) = 56.3°$. (b) $\theta_t = 90° - \theta_B = 33.7°$. The reflected and refracted beams are separated by $90°$. $\checkmark$ (c) $r_s = (n_1\cos\theta_B - n_2\cos\theta_t)/(n_1\cos\theta_B + n_2\cos\theta_t) = (0.555 - 1.248)/(0.555 + 1.248) = -0.384$. $R_s = 0.148$. Reflected intensity: $I_r = R_s(I_0/2) = 0.074\,I_0$. The reflected light is 100% s-polarised.

Solution 15. $I = (I_0/2)\cos^2\theta = I_0/8 \implies \cos^2\theta = 1/4 \implies \theta = 60°$.

Solution 4. $E_0 = \sqrt{20^2 + (-15)^2} = 25$ V/m. Polarisation angle from $x$-axis: $\theta = \arctan(-15/20) = -36.9°$ (below the $x$-axis). The field has $\delta = 0$ (no phase difference between components), so it is linearly polarised. $I = \frac{1}{2}c\varepsilon_0 E_0^2 = \frac{1}{2}(3 \times 10^8)(8.854 \times 10^{-12})(625) = 0.830$ W/m$^2$.

Solution 6. (a) $R = [(1.33 - 1.50)/(1.33 + 1.50)]^2 = (0.17/2.83)^2 = 0.00361 = 0.361\%$. (b) $\sin\theta_t = 1.33\sin 45°/1.50 = 0.627$, $\theta_t = 38.8°$. $r_s = (1.33\cos 45° - 1.50\cos 38.8°)/(1.33\cos 45° + 1.50\cos 38.8°) = (0.940 - 1.170)/(0.940 + 1.170) = -0.109$. $R_s = 0.0119$. $r_p = (1.50\cos 45° - 1.33\cos 38.8°)/(1.50\cos 45° + 1.33\cos 38.8°) = (1.061 - 1.038)/(1.061 + 1.038) = 0.011$. $R_p = 1.2 \times 10^{-4}$. (c) $\theta_B = \arctan(1.50/1.33) = 48.4°$. The reflected light is most strongly polarised at $\theta_B$.

Solution 7. $\theta_c = \arcsin(1.46/1.48) = 80.6°$. $\mathrm{NA} = \sqrt{1.48^2 - 1.46^2} = \sqrt{2.190 - 2.132} = \sqrt{0.0588} = 0.242$. $\theta_{\max} = \arcsin(0.242) = 14.0°$.

Solution 12. (a) First minimum at $\sin\theta_1 = \lambda/a = 633 \times 10^{-9}/(5.0 \times 10^{-5}) = 1.266 \times 10^{-2}$, $\theta_1 = 0.726°$. Central maximum width on screen: $2y_1 \approx 2L\theta_1 = 2(3.0)(1.266 \times 10^{-2}) = 7.60$ cm. (b) Second secondary maximum near $\alpha \approx 5\pi/2 = 7.854$. $I/I_0 = (\sin 7.854/7.854)^2 = (1/7.854)^2 = 0.0162$About 1.6% of $I_0$.

Solution 16. After the QWP: fast-axis component $E_f = E_0\cos 30° = 0.866\,E_0$Slow-axis component $E_s = E_0\sin 30° = 0.500\,E_0$ with a $\pi/2$ phase delay. Since $E_f \neq E_s$The output is elliptically polarised (not circular). After the HWP (same fast axis), the phase difference doubles to $\pi$ and the slow-axis component is negated: the output is linearly polarised at $-30°$ to the fast axis (reflected about the fast axis).

Solution 18. First lens: $1/s_1' = 1/20 - 1/30 = 1/60$So $s_1' = 60$ cm. The image forms at The position of the second lens. Object distance for second lens: $s_2 = 60 - 60 = 0$ cm (object at Infinity for the second lens). $1/s_2' = 1/f_2 - 1/s_2$: since $s_2 = 0$ (parallel rays enter the Second lens), $s_2' = f_2 = -15$ cm. The final image is virtual, 15 cm to the left of the diverging Lens.

Matrix method: $M_{\mathrm{sys} = M_{\mathrm{lens_2} \cdot M_{\mathrm{prop} \cdot M_{\mathrm{lens_1}}}}}$ $= \begin{pmatrix} 1 & 0 \\ 1/15 & 1 \end{pmatrix} \begin{pmatrix} 1 & 60 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1/20 & 1 \end{pmatrix}$ $= \begin{pmatrix} 1 & 0 \\ 1/15 & 1 \end{pmatrix} \begin{pmatrix} 1 & 60 \\ -1/20 & 1 \end{pmatrix}$ $= \begin{pmatrix} 1 & 60 \\ 1/15 - 1/20 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 60 \\ 1/60 & 5 \end{pmatrix}$.

From the matrix $\begin{pmatrix} A & B \\ C & D \end{pmatrix}$: $s' = -A/C = -1/(1/60) = -60$ cm (measured From the second lens, so 60 cm to the left, but this is the object distance for a virtual object). The Effective focal length: $-1/C = -60$ cm. The image forms where parallel output rays converge: at $f_2 = -15$ cm. Total magnification: $M = A - s'C = 1 - (-15)(1/60) = 1 + 1/4 = 5/4 = 1.25$ (upright, slightly magnified).