Answer: Zero (since they are in the same phase). Note: Wavelength is the distance between consecutive wavefronts.
Answer: (b) Perpendicular.
Answer: Small circular wavelets originating from every point on a primary wavefront.
Answer: The path of energy flow perpendicular to the wavefront.
Answer: Due to the factor $(1+\cos\theta)$ in wave propagation; for backwave $\theta=180^\circ$, making intensity zero.
Answer: Even multiple of $\pi$ or integral multiple of $2\pi$.
Answer: Odd multiple of $\lambda/2$.
Answer: Decrease (since $\lambda$ decreases in a medium).
Answer: Hyperbolic (or straight near the axis).
Answer: (b) $2:1$. Explanation: $(A_1+A_2)^2 / (A_1-A_2)^2 = 9/1 \Rightarrow (A_1+A_2)/(A_1-A_2) = 3/1$. Solving gives $A_1/A_2 = 2/1$.
Answer: (b) Fringes disappear (you get single slit diffraction instead).
Answer: $\beta = \lambda D / d = (500 \times 10^{-9} \times 2) / (1 \times 10^{-3}) = 10^{-3} \text{ m} = 1 \text{ mm}$.
Answer: $\Delta x = (2n-1)\lambda/2$. For $n=3$, $\Delta x = 5\lambda/2$.
Answer: $I_{max}/I_{min} = (\sqrt{9}+\sqrt{1})^2 / (\sqrt{9}-\sqrt{1})^2 = (3+1)^2 / (3-1)^2 = 16/4 = 4:1$.
Answer: $y_n = n\beta$. $10 \text{ mm} = 5\beta \Rightarrow \beta = 2 \text{ mm}$. For $n=2$, $y_2 = 2 \times 2 = 4 \text{ mm}$.
Answer: No, because they are incoherent (phase changes randomly and rapidly).
Answer: The resultant displacement is the vector sum of individual displacements: $y = y_1 + y_2$.
Answer: $\beta$ is doubled ($\beta \propto D$).
Answer: Wavelength ($\lambda$).
Answer: $a$ (Slit width). Formula: $2\lambda D/a$.
Answer: (c) $3\lambda/2$.
Answer: False (sound waves also diffract).
Answer: Sound wavelength is large ($\sim$m), comparable to daily objects like doors/windows. Light wavelength is tiny ($\sim$nm).
Answer: Diffraction pattern disappears; you see a clear image of the slit (rectilinear propagation).
Answer: The minimum distance/angle between two objects so they can just be seen as separate.
Answer: Interference is from two separate wavefronts; diffraction is from different parts of the same wavefront.
Answer: The distance $Z_F = a^2/\lambda$ up to which ray optics is a good approximation.
Answer: Pass-axis (it transmits parallel and absorbs perpendicular).
Answer: Perpendicular ($90^\circ$).
Answer: (d) Polarization.
Answer: (b) $I_0/4$. Explanation: $I = I_0 \cos^2(60) = I_0 (1/2)^2 = I_0/4$.
Answer: (c) $60^\circ$. ($\tan 60^\circ = \sqrt{3}$).
Answer: $\mu = \tan 57^\circ \approx 1.5$.
Answer: $0.25 = \cos^2\theta \Rightarrow \cos\theta = 0.5 \Rightarrow \theta = 60^\circ$.
Answer: An optical filter that lets light waves of a specific polarization pass and blocks others.
Answer: Intensity of transmitted polarized light varies as square of cosine of the angle between polarizer and analyzer.
Answer: The plane perpendicular to the plane of vibration which contains no electric vectors.
Answer: Rotate a Polaroid. If intensity remains constant, it's unpolarized. If it varies from max to zero, it's plane polarized.
Answer: Sunglasses to reduce glare, 3D movies, car windshields, LCD screens.
Answer: At Brewster's angle, the reflected ray has vibrations only parallel to the surface, hence it's plane polarized.
