Solution: Each point on a wavefront acts as a source of secondary wavelets spreading in all directions with speed of light. The forward tangent to these wavelets gives the new wavefront.
Solution: Consider incident wavefront $AB$. In time $t$, point $B$ reaches $C$ ($BC=vt$). Simultaneously, wavelet from $A$ spreads to $D$ ($AD=vt$). $\triangle ABC \cong \triangle ADC$. Hence $\angle i = \angle r$.
Solution: $n = v_1 / v_2 = \sin i / \sin r$. Since $v = f\lambda$ and frequency $f$ is constant, $\lambda_1/\lambda_2 = v_1/v_2 = n$. Wavelength decreases because speed decreases.
Solution: (a) Reflection: $\lambda=589 \text{ nm}, v=3 \times 10^8 \text{ m/s}, f=c/\lambda \approx 5.09 \times 10^{14} \text{ Hz}$. (b) Refraction: $f$ unchanged. $v = c/n = 2.25 \times 10^8 \text{ m/s}$, $\lambda = \lambda_{air}/n \approx 443 \text{ nm}$.
Solution: $n = c/v = 3/2 = 1.5$. $\lambda_m = \lambda_v / n = 6000 / 1.5 = 4000 \text{ \AA}$.
Solution: $\sin r = \sin 30 / 1.5 = 0.5 / 1.5 = 1/3$. $r = \sin^{-1}(1/3) \approx 19.47^\circ$. The wavefront angle with interface is the same as the refraction angle.
Solution: $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$. If $a_1, a_2$ are amplitudes, $I = a_1^2 + a_2^2 + 2a_1 a_2\cos\phi$.
Solution: $I_{res} = I + 9I + 2\sqrt{9I^2}\cos\phi = 10I + 6I\cos\phi$. (a) $\phi=0, I=16I$. (b) $\phi=\pi, I=4I$. (c) $\phi=\pi/2, I=10I$.
Solution: $x_4 = 4 \lambda D / d \Rightarrow \lambda = x_4 \cdot d / 4D = (0.012 \cdot 0.00028) / (4 \cdot 1.4) = 600 \text{ nm}$.
Solution: $I_{max} = (\sqrt{4I}+\sqrt{I})^2 = 9I$. $I_{min} = (\sqrt{4I}-\sqrt{I})^2 = I$. Visibility $V = (I_{max}-I_{min})/(I_{max}+I_{min}) = 8/10 = 0.8$.
Solution: $I \propto \cos^2(\phi/2)$. (a) $\Delta x=0 \Rightarrow \phi=0, I=I_0$. (b) $\Delta x=\lambda/4 \Rightarrow \phi=\pi/2, I = I_0 \cos^2(\pi/4) = I_0/2$. Ratio $2:1$.
Solution: $n_1 \lambda_1 = n_2 \lambda_2 \Rightarrow n_1/n_2 = 520/650 = 4/5$. So 4th of $\lambda_1$ coincides with 5th of $\lambda_2$. $x = 4 \cdot \lambda_1 D/d = 1.56 \text{ mm}$.
Solution: $\beta_{air} = 18/9 = 2 \text{ mm}$. $\beta_{liq} = 12.6/9 = 1.4 \text{ mm}$. $n = \beta_{air}/\beta_{liq} = 2/1.4 \approx 1.43$.
Solution: $a \sin \theta = \lambda$. $\sin 30^\circ = 0.5$. $a = \lambda/0.5 = 2 \times 700 \text{ nm} = 1.4 \mu\text{m}$.
Solution: $2\theta = 2\lambda/a = 2 \cdot (5 \times 10^{-7}) / (2 \times 10^{-4}) = 5 \times 10^{-3} \text{ rad}$.
Solution: $y_1 = \lambda D/a \Rightarrow \lambda = y_1 a / D = (0.005 \cdot 0.0001) / 2 = 250 \text{ nm}$.
Solution: $Z_F = a^2/\lambda = (3 \times 10^{-3})^2 / (5 \times 10^{-7}) = 18 \text{ m}$.
Solution: $RP = D / 1.22 \lambda = 5 / (1.22 \cdot 5.5 \times 10^{-7}) \approx 7.45 \times 10^6$.
Solution: $I_{max}/2 = I_{max} \cos^2\theta \Rightarrow \cos^2\theta = 1/2 \Rightarrow \theta = 45^\circ$.
Solution: $\tan i_p = 1.55 \Rightarrow i_p \approx 57.2^\circ$. $r = 90 - 57.2 = 32.8^\circ$.
Solution: $I_1 = I_0/2$. $I_2 = I_1 \cos^2 45 = I_0/4$. $I_3 = I_2 \cos^2 45 = I_0/8$.
Solution: $\sin C = \sin 45 = 1/\sqrt{2} \Rightarrow n = \sqrt{2}$. $\tan i_p = \sqrt{2} \Rightarrow i_p = 54.7^\circ$.
Solution: $I = (I_{un}/2) \cos^2\theta = I_{un}/8 \Rightarrow \cos^2\theta = 1/4 \Rightarrow \theta = 60^\circ$.
