Solution: A wavefront is defined as the locus of all points in a medium that vibrate in the same phase at a given instant. According to Huygens Principle, each point on a primary wavefront acts as a source of secondary wavelets. The envelope of these wavelets in the forward direction gives the position of the new wavefront. For a plane wavefront, the secondary wavelets are spheres of radius $v \cdot t$; their forward tangential plane forms the new plane wavefront.
Solution: 1. **Spherical:** Produced by a point source at finite distance. Amplitude $A \propto 1/r$. Intensity $I \propto 1/r^2$. 2. **Plane:** Produced by source at infinity. Amplitude and intensity remain constant along the direction of propagation.
Solution: Speed in glass $v = c/n = 3 \times 10^8 / 1.5 = 2 \times 10^8 \text{ m/s}$. Wavelength in glass $\lambda' = \lambda/n = 600 / 1.5 = 400 \text{ nm}$.
Solution: Wavelength in medium $\lambda' = \lambda/n = 5000 / 1.33 \approx 3759 \text{ \AA}$. Path diff $\Delta x = 2 \text{ cm} = 2 \times 10^8 \text{ \AA}$. Phase diff $\Delta \phi = (2\pi / \lambda') \times \Delta x$. Calculation leads to a very large value, representing many full cycles of phase.
Solution: Path difference $\Delta x = d \sin \theta \approx d \cdot y/D$. For bright fringes, $\Delta x = n\lambda$. So $y_n = n \lambda D / d$. Fringe width $\beta = y_{n+1} - y_n = \frac{\lambda D}{d}$.
Solution: $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{I} + \sqrt{4I})^2 = (1+2)^2 I = 9I$.
$I_{min} = (\sqrt{4I} - \sqrt{I})^2 = (2-1)^2 I = I$. Ratio $9:1$.
Solution: $y_4 = 4 \lambda D / d$. $\lambda = \frac{y_4 \cdot d}{4D} = \frac{0.012 \times 0.0002}{4 \times 1.5} = \frac{2.4 \times 10^{-6}}{6} = 4 \times 10^{-7} \text{ m} = 400 \text{ nm}$.
Solution: $n_1 \lambda_1 = n_2 \lambda_2$. $n_1/n_2 = 520/650 = 4/5$. So 4th bright of $650 \text{ nm}$ coincides with 5th bright of $520 \text{ nm}$. Distance $y = 4 \cdot \frac{650 \times 10^{-9} \cdot 1.2}{2 \times 10^{-3}} = 1.56 \text{ mm}$.
Solution: Width ratio $w_1/w_2 = 4/1 \Rightarrow I_1/I_2 = 4/1 \Rightarrow A_1/A_2 = 2/1$.
$I_{max}/I_{min} = (2+1)^2 / (2-1)^2 = 9:1$.
Solution: $I = I_0 \cos^2(45) = I_0 (1/\sqrt{2})^2 = I_0/2$. Ratio $1:2$.
Solution: $\tan i_p = 1.732 = \sqrt{3} \Rightarrow i_p = 60^\circ$. $r = 90 - 60 = 30^\circ$.
Solution: After 1st: $I_1 = I_0/2$. After 2nd: $I_2 = I_1 \cos^2(30) = (I_0/2) \cdot (3/4) = 3I_0/8$.
Solution: $i_p = \tan^{-1}(1.33) = \tan^{-1}(4/3) \approx 53.1^\circ$.
Solution: Total $I = (I_{un}/2) \cos^2\theta = I_{un}/8 \Rightarrow \cos^2\theta = 1/4 \Rightarrow \cos\theta = 1/2 \Rightarrow \theta = 60^\circ$.
