### Related Formula cos^2alpha + cos^2beta + cos^2gamma = 1$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ ### Core Logic The direction of propagation of a light wave is always perpendicular to its plane wavefronts. For wavefronts of the form x + y + z = textconstant$x + y + z = \text{constant}$, the normal vector representing the direction of wave propagation is given by: vecn = hati + hatj + hatk$\vec{n} = \hat{i} + \hat{j} + \hat{k}$ Since the coefficients of x$x$, y$y$, and z$z$ are all equal, the propagation direction is symmetric with respect to all three axes: cosalpha = cosbeta = cosgamma$\cos\alpha = \cos\beta = \cos\gamma$ Substituting this into the direction cosine identity: cos^2alpha + cos^2alpha + cos^2alpha = 1 implies 3cos^2alpha = 1$\cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \implies 3\cos^2\alpha = 1$ cosalpha = frac1sqrt3$\cos\alpha = \frac{1}{\sqrt{3}}$ alpha = cos^-1left(frac1sqrt3right)$\alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$ ### Step 1: Final Conclusion The angle made by the direction of wave propagation with the x$x$-axis is cos^-1left(frac1sqrt3right)$\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$. ### Pattern Recognition Sees: plane wavefront Ax + By + Cz = textd$Ax + By + Cz = \text{d}$ → propagation vector is parallel to the normal veck = Ahati + Bhatj + Chatk$\vec{k} = A\hat{i} + B\hat{j} + C\hat{k}$. Use standard direction cosines to find the angle with any axis. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics Class 12 Mathematics: Three Dimensional Geometry

### Related Formula a sintheta_n = n lambda$a \sin\theta_n = n \lambda$ theta_textapprox = n fraclambdaa quad text(for small angles)$\theta_{\text{approx}} = n \frac{\lambda}{a} \quad \text{(for small angles)}$ ### Core Logic Let the width of the slit be a$a$, and the light wavelength be lambda = 628mathrm~nm = 628 times 10^-9mathrm~m$\lambda = 628\mathrm{~nm} = 628 \times 10^{-9}\mathrm{~m}$. The second minimum (n = 2$n = 2$) is located at angular position: sintheta_1 = frac2lambdaa$\sin\theta_1 = \frac{2\lambda}{a}$ The third minimum (n = 3$n = 3$) is located at angular position: sintheta_2 = frac3lambdaa$\sin\theta_2 = \frac{3\lambda}{a}$ The total angular separation is 30^circ$30^{circ}$: theta_1 + theta_2 = 30^circ = fracpi6mathrm~rad$\theta_1 + \theta_2 = 30^{circ} = \frac{\pi}{6}\mathrm{~rad}$ Using the small-angle approximation (where theta approx sintheta$\theta \approx \sin\theta$): theta_1 + theta_2 approx frac2lambdaa + frac3lambdaa = frac5lambdaa$\theta_1 + \theta_2 \approx \frac{2\lambda}{a} + \frac{3\lambda}{a} = \frac{5\lambda}{a}$ Equating to the given separation: frac5lambdaa = fracpi6 implies a = frac30lambdapi$\frac{5\lambda}{a} = \frac{\pi}{6} \implies a = \frac{30\lambda}{\pi}$ Substitute the given values (using pi approx 3.14$\pi \approx 3.14$): a = frac30 times 628 times 10^-9mathrm~m3.14 = 30 times 200 times 10^-9mathrm~m = 6 times 10^-6mathrm~m = 6mathrm~mu m$a = \frac{30 \times 628 \times 10^{-9}\mathrm{~m}}{3.14} = 30 \times 200 \times 10^{-9}\mathrm{~m} = 6 \times 10^{-6}\mathrm{~m} = 6\mathrm{~\mu m}$ ### Step 1: Final Conclusion The width of the slit is $ ### Step 1: Final Conclusion The width of the slit is $6\mathrm{~\mu m}. ### Pattern Recognition In single slit diffraction, the position of minima is $. ### Pattern Recognition In single slit diffraction, the position of minima is $a \sin\theta = n \lambda. The angular spread from the $. The angular spread from the $n_1-th minimum on one side to the $-th minimum on one side to the $n_2-th minimum on the other is $-th minimum on the other is $(n_1 + n_2) \frac{\lambda}{a}. Since $. Since $\pi \approx 3.14, note how $, note how $628 / 3.14 = 200$, resolving to a neat integer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula The intensity of a slit is proportional to its width: I propto w$I \propto w$ The ratio of maximum to minimum intensity in an interference pattern is given by: fracI_maxI_min = fracleft(sqrtI_1 + sqrtI_2right)^2left(sqrtI_1 - sqrtI_2right)^2$\frac{I_{\max}}{I_{\min}} = \frac{\left(\sqrt{I_1} + \sqrt{I_2}\right)^2}{\left(\sqrt{I_1} - \sqrt{I_2}\right)^2}$ ### Core Logic Let the width of the larger slit be w_2 = w$w_2 = w$ and the smaller slit be w_1 = w/2$w_1 = w/2$. Thus: I_2 = I_0, quad I_1 = fracI_02$I_2 = I_0, \quad I_1 = \frac{I_0}{2}$ sqrtI_2 = sqrtI_0, quad sqrtI_1 = fracsqrtI_0sqrt2$\sqrt{I_2} = \sqrt{I_0}, \quad \sqrt{I_1} = \frac{\sqrt{I_0}}{\sqrt{2}}$ ### Step 1: Substitute values into intensity ratio fracI_maxI_min = fracleft(sqrtI_0 + fracsqrtI_0sqrt2right)^2left(sqrtI_0 - fracsqrtI_0sqrt2right)^2$\frac{I_{\max}}{I_{\min}} = \frac{\left(\sqrt{I_0} + \frac{\sqrt{I_0}}{\sqrt{2}}\right)^2}{\left(\sqrt{I_0} - \frac{\sqrt{I_0}}{\sqrt{2}}\right)^2}$ fracI_maxI_min = fracleft(1 + frac1sqrt2right)^2left(1 - frac1sqrt2right)^2 = fracleft(fracsqrt2 + 1sqrt2right)^2left(fracsqrt2 - 1sqrt2right)^2 = fracleft(sqrt2 + 1right)^2left(sqrt2 - 1right)^2$\frac{I_{\max}}{I_{\min}} = \frac{\left(1 + \frac{1}{\sqrt{2}}\right)^2}{\left(1 - \frac{1}{\sqrt{2}}\right)^2} = \frac{\left(\frac{\sqrt{2} + 1}{\sqrt{2}}\right)^2}{\left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right)^2} = \frac{\left(\sqrt{2} + 1\right)^2}{\left(\sqrt{2} - 1\right)^2}$ ### Step 2: Expand the terms fracI_maxI_min = frac2 + 1 + 2sqrt22 + 1 - 2sqrt2 = frac3 + 2sqrt23 - 2sqrt2$\frac{I_{\max}}{I_{\min}} = \frac{2 + 1 + 2\sqrt{2}}{2 + 1 - 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}$ ### Pattern Recognition When dealing with fractional width ratio beta = w_2/w_1$\beta = w_2/w_1$, the ratio sqrtI_2/I_1 = sqrtbeta$\sqrt{I_2/I_1} = \sqrt{\beta}$. The intensity ratio formula can be rewritten as (sqrtbeta+1)^2 / (sqrtbeta-1)^2$(\sqrt{\beta}+1)^2 / (\sqrt{\beta}-1)^2$. For beta = 2$\beta = 2$, this directly resolves to (sqrt2+1)^2/(sqrt2-1)^2 = (3+2sqrt2)/(3-2sqrt2)$(\sqrt{2}+1)^2/(\sqrt{2}-1)^2 = (3+2\sqrt{2})/(3-2\sqrt{2})$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula The intensity limits for two overlapping coherent light sources of intensities I_1$I_1$ and I_2$I_2$ are: fracI_maxI_min = left( fracsqrtI_2 + sqrtI_1sqrtI_2 - sqrtI_1 right)^2$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_2} + \sqrt{I_1}}{\sqrt{I_2} - \sqrt{I_1}} \right)^2$ ### Core Logic Given ratio: fracI_1I_2 = frac19$\frac{I_1}{I_2} = \frac{1}{9}$ If we let I_1 = I_0$I_1 = I_0$, then I_2 = 9I_0$I_2 = 9I_0$. Taking square roots: sqrtI_1 = sqrtI_0, quad sqrtI_2 = 3sqrtI_0$\sqrt{I_1} = \sqrt{I_0}, \quad \sqrt{I_2} = 3\sqrt{I_0}$ ### Step 1: Calculate the ratio fracI_maxI_min = left( frac3sqrtI_0 + sqrtI_03sqrtI_0 - sqrtI_0 right)^2 = left( frac42 right)^2 = (2)^2 = 4$\frac{I_{\max}}{I_{\min}} = \left( \frac{3\sqrt{I_0} + \sqrt{I_0}}{3\sqrt{I_0} - \sqrt{I_0}} \right)^2 = \left( \frac{4}{2} \right)^2 = (2)^2 = 4$ Thus, the ratio is 4:1$4:1$. ### Pattern Recognition A classic shortcut is using amplitude ratios. If intensities are in ratio 1:9$1:9$, amplitudes a_1:a_2$a_1:a_2$ are in ratio 1:3$1:3$. fraca_maxa_min = frac3+13-1 = 2 Rightarrow I_max:I_min = 4:1$\frac{a_{\max}}{a_{\min}} = \frac{3+1}{3-1} = 2 \Rightarrow I_{\max}:I_{\min} = 4:1$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula For two waves of identical direction and frequency superimposing with phase difference phi$\phi$: E_textres = sqrtE_1^2 + E_2^2 + 2E_1E_2cosphi$E_{\text{res}} = \sqrt{E_1^2 + E_2^2 + 2E_1E_2\cos\phi}$ ### Core Logic The amplitudes of the two waves are E_1 = E_0$E_1 = E_0$ and E_2 = E_0$E_2 = E_0$. The phase difference is: phi = fracpi3$\phi = \frac{\pi}{3}$ Substitute these values into the resultant amplitude equation: E_textres = sqrtE_0^2 + E_0^2 + 2E_0^2cosleft(fracpi3right)$E_{\text{res}} = \sqrt{E_0^2 + E_0^2 + 2E_0^2\cos\left(\frac{\pi}{3}\right)}$ ### Step 1: Simplify the calculation Since cosleft(fracpi3right) = 0.5$\cos\left(\frac{\pi}{3}\right) = 0.5$: E_textres = sqrt2E_0^2 + 2E_0^2(0.5) = sqrt3E_0^2 = sqrt3E_0 approx 1.732E_0$E_{\text{res}} = \sqrt{2E_0^2 + 2E_0^2(0.5)} = \sqrt{3E_0^2} = \sqrt{3}E_0 \approx 1.732E_0$ This is closest to 1.7mathrmE_0$1.7\mathrm{E}_0$. ### Pattern Recognition Sees: Equal amplitudes (A$A$) with a phase angle of 60^circ$60^\circ$ (pi/3${\pi}/{3}$). Shortcut: The vector sum of two vectors of equal magnitude A$A$ separated by 60^circ$60^\circ$ is always sqrt3A approx 1.73A$\sqrt{3}A \approx 1.73A$. If separated by 120^circ$120^\circ$, it is A$A$. If 90^circ$90^\circ$, it is sqrt2A$\sqrt{2}A$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics