Solution: When a mirror moves with velocity $V$, the effective path for secondary wavelets changes. If the source and observer are on the same side, the apparent frequency $f' = f \frac{c+V}{c-V}$ for reflection. The wavefront construction must account for the shift in wavelet centers during time $\Delta t$.
Solution: $\delta = (i-r_1) + (e-r_2) = i+e - (r_1+r_2)$. For minimum deviation, construction shows $i=e$ and the wavefront inside is parallel to the base. Total deviation $\delta_m = 2i - A$.
Solution: Only light within the critical angle $C$ emerges. $\sin C = 1/n$. The radius of the emerging light disc is $R = h \tan C = h \frac{\sin C}{\cos C} = \frac{h}{\sqrt{n^2-1}}$. The emerging wavefront part is the spherical cap bounded by this cone.
Solution: $n \lambda_1 = m \lambda_2 \Rightarrow n(400) = m(560) \Rightarrow n/m = 560/400 = 7/5$. Thus, the 7th bright fringe of $400 \text{ nm}$ coincides with the 5th bright fringe of $560 \text{ nm}$. Distance $y = \frac{7 \cdot \lambda_1 D}{d}$.
Solution: Path difference introduced $\Delta x = (\mu - 1)t = (1.5 - 1) \times 2 \mu\text{m} = 1 \mu\text{m} = 1000 \text{ nm}$. Number of fringes shifted $N = \Delta x / \lambda = 1000 / 500 = 2$. The central maximum shifts by 2 fringe widths.
Solution: $A_2/A_1 = \sqrt{2}$. $I_{max}/I_{min} = \frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2} \approx \frac{5.83}{0.17} \approx 34:1$. Contrast (visibility) $V = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2} = \frac{2\sqrt{2}}{3} \approx 94\%$.
Solution: For reflection from film on water (low-high-higher $n$? No, soap $n=1.33$ is less than glass but more than air). Usually air-soap-air. Phase change of $\pi$ at first surface. Condition for constructive: $2\mu t = (n + 1/2)\lambda$. For minimum thickness $n=0$, $2(1.33)t = \lambda/2 \Rightarrow t = \lambda/5.32 = 600 / 5.32 \approx 112.8 \text{ nm}$.
Solution: Phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{3}$. Resultant intensity $I = 4I_0 \cos^2(\phi/2) = 4I_0 \cos^2(\pi/3) = 4I_0 (1/4) = I_0$. Ratio to central maximum ($4I_0$) is $I/I_{max} = 1/4$.
Solution: $a \sin \theta = \lambda \Rightarrow a \sin 30^\circ = 650 \text{ nm} \Rightarrow a \cdot (0.5) = 650 \Rightarrow a = 1300 \text{ nm} = 1.3 \mu\text{m}$.
Solution: $RP \propto n \sin \theta$. $RP_{oil} / RP_{air} = n_{oil} / n_{air} = 1.5 / 1 = 1.5$. Resolving power increases by $50\%$.
Solution: $\Delta \theta = 1.22 \lambda / D = 1.22 \cdot 500 \times 10^{-9} / 2.5 = 2.44 \times 10^{-7} \text{ rad}$. Minimum distance $x = r \cdot \Delta \theta = 10^{15} \times 2.44 \times 10^{-7} = 2.44 \times 10^8 \text{ m}$.
Solution: Intensity $I \approx I_0 \left[ \frac{\sin \beta}{\beta} \right]^2$ where $\beta = \frac{\pi a \sin \theta}{\lambda}$. For first secondary maximum, $\beta \approx 3\pi/2$. $I \approx I_0 [1 / (3\pi/2)]^2 = 4 I_0 / 9\pi^2 \approx I_0 / 22 \approx 4.5\% I_0$.
Solution: $y_n = n \lambda D / a$. $y_1 = 500\text{n} \cdot 1 / 0.1\text{m} = 5 \text{ mm}$. $y_2 = 2 y_1 = 10 \text{ mm}$. Distance between them is $5 \text{ mm}$.
Solution: After first Polaroid, $I_1 = I_0/2$. After each subsequent Polaroid, intensity scales by $\cos^2 \theta$. Final intensity $I_N = \frac{I_0}{2} (\cos^2 \theta)^{N-1}$.
Solution: $I/I_0 = \cos^2 \theta$. (i) $30^\circ \to 3/4$, (ii) $45^\circ \to 1/2$, (iii) $60^\circ \to 1/4$, (iv) $90^\circ \to 0$.
Solution: $I = I_p \cos^2 \theta \Rightarrow 1/3 = \cos^2 \theta \Rightarrow \cos \theta = 1/\sqrt{3}$. $\theta = \cos^{-1}(0.577) \approx 54.7^\circ$.
Solution: $\tan i_p = 1.5 \Rightarrow i_p = 56.3^\circ$. Since $i_p + r = 90^\circ$ at Brewster angle, $r = 90 - 56.3 = 33.7^\circ$.
