### Related Formula For a stable, visible interference pattern to form, the light sources passing through the two slits must be **coherent**: - They must have the same wavelength (or frequency). - They must maintain a constant phase difference over time. ### Core Logic If one slit is covered by a red filter and the other by a green filter: - Only red light (lambda_textred approx 700mathrm~nm$\lambda_{\text{red}} \approx 700\mathrm{~nm}$) passes through the first slit. - Only green light (lambda_textgreen approx 500mathrm~nm$\lambda_{\text{green}} \approx 500\mathrm{~nm}$) passes through the second slit. Since the two passing waves have completely different wavelengths and frequencies, they are **incoherent**. ### Step 1: Resulting Pattern Analysis Coherent sources are a prerequisite for producing stable bright and dark interference fringes. Incoherent waves of different frequencies merely superimpose to create a general background illumination without any distinct, observable spatial fringe lines. Thus, **there shall be no interference fringes**. ### Pattern Recognition Sees: YDSE + opposite colored filters (red and green) on slits. Trap: Don't get confused thinking separate patterns will overlay. Since the slits emit different colors, the sources are incoherent, so the interference term langle cosphi rangle = 0$\langle \cos\phi \rangle = 0$. Shortcut: Different colors = different wavelengths = incoherent sources = NO fringes. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula mathrmY = fracmathrmnlambda mathrmDmathrmd implies mathrmY propto lambda$\mathrm{Y} = \frac{\mathrm{n}\lambda \mathrm{D}}{\mathrm{d}} \implies \mathrm{Y} \propto \lambda$ ### Core Logic Since the fringe index mathrmn$\mathrm{n}$ and apparatus parameters mathrmD, mathrmd$\mathrm{D}, \mathrm{d}$ remain constant across both runs: fracmathrmy_2mathrmy_1 = fraclambda_2lambda_1$\frac{\mathrm{y}_2}{\mathrm{y}_1} = \frac{\lambda_2}{\lambda_1}$ Substituting the values into the proportionality equation: fracmathrmy_210 mathrm~mm = frac660 mathrm~nm600 mathrm~nm$\frac{\mathrm{y}_2}{10 \mathrm{~mm}} = \frac{660 \mathrm{~nm}}{600 \mathrm{~nm}}$ mathrmy_2 = 10 times 1.1 = 11 mathrm~mm$\mathrm{y}_2 = 10 \times 1.1 = 11 \mathrm{~mm}$ ### Step 1: Final Numerical Value The distance of the tenth bright fringe shifts to exactly 11 mathrm~mm$11 \mathrm{~mm}$. ### Pattern Recognition Fringe position scales linearly with wavelength in standard Young's setups. Increasing the wavelength by 10\%$10\%$ shifts the entire pattern outward by exactly 10\%$10\%$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula For superposition of two coherent beams of intensities I_1$I_1$ and I_2$I_2$: I_textmax = left(sqrtI_1 + sqrtI_2right)^2$I_{\text{max}} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2$ I_textmin = left(sqrtI_1 - sqrtI_2right)^2$I_{\text{min}} = \left(\sqrt{I_1} - \sqrt{I_2}\right)^2$ ### Core Logic Given values: - I_1 = 4I$I_1 = 4I$ - I_2 = 9I$I_2 = 9I$ Let's calculate the square roots of the intensities: - sqrtI_1 = sqrt4I = 2sqrtI$\sqrt{I_1} = \sqrt{4I} = 2\sqrt{I}$ - sqrtI_2 = sqrt9I = 3sqrtI$\sqrt{I_2} = \sqrt{9I} = 3\sqrt{I}$ ### Step 1: Calculating Max and Min Intensities Substitute these values into the intensity formulas: I_textmax = (2sqrtI + 3sqrtI)^2 = (5sqrtI)^2 = 25I$I_{\text{max}} = (2\sqrt{I} + 3\sqrt{I})^2 = (5\sqrt{I})^2 = 25I$ I_textmin = (3sqrtI - 2sqrtI)^2 = (1sqrtI)^2 = I$I_{\text{min}} = (3\sqrt{I} - 2\sqrt{I})^2 = (1\sqrt{I})^2 = I$ ### Step 2: Finding the Difference The difference between the maximum and minimum intensities is: I_textmax - I_textmin = 25I - I = 24I$I_{\text{max}} - I_{\text{min}} = 25I - I = 24I$ Since this difference is given as xI$xI$: x = 24$x = 24$ ### Pattern Recognition Algebraic Shortcut: I_textmax - I_textmin = left(sqrtI_1 + sqrtI_2right)^2 - left(sqrtI_1 - sqrtI_2right)^2 = 4sqrtI_1 I_2$I_{\text{max}} - I_{\text{min}} = \left(\sqrt{I_1} + \sqrt{I_2}\right)^2 - \left(\sqrt{I_1} - \sqrt{I_2}\right)^2 = 4\sqrt{I_1 I_2}$ Substitute I_1 = 4I$I_1 = 4I$ and I_2 = 9I$I_2 = 9I$: 4sqrt4I cdot 9I = 4sqrt36 I^2 = 4 times 6I = 24I$4\sqrt{4I \cdot 9I} = 4\sqrt{36 I^2} = 4 \times 6I = 24I$. This beautiful identity (4ab$4ab$ formula) lets you solve the problem instantly without separately calculating maximum and minimum values! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula Malus's Law: I = I_0 cos^2theta$I = I_0 \cos^2\theta$ ### Core Logic Since P_1$P_1$ and P_2$P_2$ are crossed, the angle between their transmission axes is 90^circ$90^\circ$. Let the angle between P_1$P_1$ and P_3$P_3$ be theta$\theta$. Then the angle between P_3$P_3$ and P_2$P_2$ is left(90^circ - thetaright)$\left(90^\circ - \theta\right)$. Intensity after passing through P_3$P_3$: I_1 = I_0 cos^2theta$I_1 = I_0 \cos^2\theta$. Intensity after passing through P_2$P_2$: I_textnet = I_1 cos^2(90^circ - theta) = I_0 cos^2theta sin^2theta$I_{\text{net}} = I_1 \cos^2(90^\circ - \theta) = I_0 \cos^2\theta \sin^2\theta$. ### Step 1: Maximize Net Intensity Rewrite the expression: I_textnet = fracI_04 [2sinthetacostheta]^2 = fracI_04 [sin(2theta)]^2$I_{\text{net}} = \frac{I_0}{4} [2\sin\theta\cos\theta]^2 = \frac{I_0}{4} [\sin(2\theta)]^2$ For maximum transmitted intensity, sin(2theta) = 1 implies 2theta = 90^circ implies theta = 45^circ = fracpi4$\sin(2\theta) = 1 \implies 2\theta = 90^\circ \implies \theta = 45^\circ = \frac{\pi}{4}$. The angle between P_2$P_2$ and P_3$P_3$ is 90^circ - 45^circ = 45^circ = fracpi4$90^\circ - 45^\circ = 45^\circ = \frac{\pi}{4}$.

### Pattern Recognition Inserting a polarization filter at exactly 45^circ$45^\circ$ ( pi/4$ \pi/4$) between crossed polarizers symmetrically splits up components, maximizing overall transmission throughput. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula Width of central maximum in single-slit diffraction: Delta y_textdiff = frac2lambda Da$\Delta y_{\text{diff}} = \frac{2\lambda D}{a}$ Fringe width in double-slit interference: beta = fraclambda Dd$\beta = \frac{\lambda D}{d}$ ### Core Logic Given condition: 20$20$ interference fringes fit inside the central diffraction envelope: 20 times beta = Delta y_textdiff$20 \times \beta = \Delta y_{\text{diff}}$ 20 times fraclambda Dd = frac2lambda Da$20 \times \frac{\lambda D}{d} = \frac{2\lambda D}{a}$ Cancel common parameters: frac10d = frac1a implies a = fracd10$\frac{10}{d} = \frac{1}{a} \implies a = \frac{d}{10}$ ### Step 1: Substitute Given Parameters Slit separation d = 1.5text mm = 0.15text cm$d = 1.5\text{ mm} = 0.15\text{ cm}$. a = frac0.15text cm10 = 0.015text cm = 15 times 10^-3text cm$a = \frac{0.15\text{ cm}}{10} = 0.015\text{ cm} = 15 \times 10^{-3}\text{ cm}$ Comparing with x times 10^-3text cm$x \times 10^{-3}\text{ cm}$, the value of x$x$ is **15**. ### Pattern Recognition Envelope matching conditions rely strictly on the geometric ratio of slit separation (d$d$) to individual slit width (a$a$). Wavelength (lambda$\lambda$) and screen distance (D$D$) cancel out completely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics