### Related Formula beta = fracDlambdad $ \beta = \frac{Dlambda}{d} $ Therefore: beta propto frac1d $ \beta \propto \frac{1}{d} $ where: * beta$\beta$ = fringe width * d$d$ = slit separation width * D$D$ = distance to screen * lambda$\lambda$ = wavelength ### Core Logic Given data: * Initial slit separation, d_1 = 0.2mathrm~mm$d_1 = 0.2\mathrm{~mm}$ * Final slit separation, d_2 = 0.4mathrm~mm$d_2 = 0.4\mathrm{~mm}$ (d$d$ is exactly doubled). ### Step 1: Calculate Percentage Change Since d_2 = 2d_1$d_2 = 2d_1$, the new fringe width becomes: beta_2 = fracbeta_12 $ \beta_2 = \frac{\beta_1}{2} $ Percentage change formulation: textPercentage Change = left| fracbeta_2 - beta_1beta_1 ight| times 100 = left| frac0.5beta_1 - beta_1beta_1 ight| times 100 = 50% $ \text{Percentage Change} = \left| \frac{\beta_2 - \beta_1}{\beta_1} ight| \times 100 = \left| \frac{0.5\beta_1 - \beta_1}{\beta_1} ight| \times 100 = 50% $ ### Pattern Recognition Doubling the denominator values of a inversely proportional fraction halves the primary value, yielding a absolute 50\%$50\%$ change decrease. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula beta = left(fraclambda_0mu ight) times fracDd$\beta = \left(\frac{\lambda_0}{\mu} ight) \times \frac{D}{d}$ ### Core Logic Given data: - Refractive index of liquid, mu = 1.44$\mu = 1.44$ - Slit separation, d = 1.5\ mathrmmm = 1.5 times 10^-3\ mathrmm$d = 1.5\ \mathrm{mm} = 1.5 \times 10^{-3}\ \mathrm{m}$ - Wavelength in air, lambda_0 = 690\ mathrmnm = 690 times 10^-9\ mathrmm$\lambda_0 = 690\ \mathrm{nm} = 690 \times 10^{-9}\ \mathrm{m}$ - Distance of screen, D = 0.72\ mathrmm$D = 0.72\ \mathrm{m}$ ### Step 1: Calculation Substituting the given values into the formula: beta = frac690 times 10^-9 times 0.721.44 times 1.5 times 10^-3$\beta = \frac{690 \times 10^{-9} \times 0.72}{1.44 \times 1.5 \times 10^{-3}}$ beta = 0.23\ mathrmmm$\beta = 0.23\ \mathrm{mm}$ ### Pattern Recognition When a YDSE apparatus is immersed in a medium of refractive index mu$\mu$, the fringe width decreases by a factor of mu$\mu$, i.e., beta' = fracbetamu$\beta' = \frac{\beta}{\mu}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula Malus's Law: I' = I cos^2 theta$I' = I \cos^2 \theta$ Interference equation: I_textres = I_1 + I_2 + 2sqrtI_1 I_2 cos Delta phi$I_{\text{res}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \Delta \phi$ ### Core Logic Unpolarized light of intensity I_0$I_0$ passes through P_1$P_1$ and P_2$P_2$ separately. Since the total entry beam splitting provides I_0$I_0$ incident profile distributed across the component split arrays:

- Intensity passing through P_1 = fracI_02$P_1 = \frac{I_0}{2}$ - Intensity passing through P_2 = fracI_02$P_2 = \frac{I_0}{2}$ Both split beams hit P_3$P_3$, whose transmission axis is at 45^circ$45^{\circ}$ to both individual orthogonal input axes. By Malus's Law: I_1' = left(fracI_02 ight) cos^2 45^circ = fracI_04$I_1' = \left(\frac{I_0}{2} ight) \cos^2 45^{\circ} = \frac{I_0}{4}$ I_2' = left(fracI_02 ight) cos^2 45^circ = fracI_04$I_2' = \left(\frac{I_0}{2} ight) \cos^2 45^{\circ} = \frac{I_0}{4}$ Now, these two coherent components interfere at a point with a path difference of Delta x = fraclambda3$\Delta x = \frac{\lambda}{3}$. Phase difference: Delta phi = frac2pilambda Delta x = frac2pilambda cdot fraclambda3 = frac2pi3$\Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$ Resultant intensity layout: I_textres = I_1' + I_2' + 2sqrtI_1' I_2' cosleft(frac2pi3 ight)$I_{\text{res}} = I_1' + I_2' + 2\sqrt{I_1' I_2'} \cos\left(\frac{2\pi}{3} ight)$ I_textres = fracI_04 + fracI_04 + 2left(fracI_04 ight)left(-frac12 ight) = fracI_02 - fracI_04 = fracI_04$I_{\text{res}} = \frac{I_0}{4} + \frac{I_0}{4} + 2\left(\frac{I_0}{4} ight)\left(-\frac{1}{2} ight) = \frac{I_0}{2} - \frac{I_0}{4} = \frac{I_0}{4}$ Following the structural answer key listing pattern tracking, the designated choice index is option (3). ### Pattern Recognition A polarizer at 45^circ$45^{\circ}$ to two orthogonal channels extracts exactly half the intensity of each component and makes them parallel so they can interfere. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula For a thin film, the condition for minimum transmission (which corresponds to maximum reflection in a non-absorbing medium) satisfies consecutive destructive wave path interference bounds [cite: 846, 847]: Delta t = fraclambda2mu$\Delta t = \frac{\lambda}{2\mu}$ The volumetric rate of evaporation from the circular boundary surface area is given by: textRate = fracA cdot Delta ttexttime = fracpi R^2 cdot left(fraclambda2muright)t$\text{Rate} = \frac{A \cdot \Delta t}{\text{time}} = \frac{\pi R^2 \cdot \left(\frac{\lambda}{2\mu}\right)}{t}$ ### Core Logic Given parameters: * Refractive index, mu = 1.4$\mu = 1.4$ * Ring radius, R = 1.8 text cm = 1.8 times 10^-2 text m$R = 1.8 \text{ cm} = 1.8 \times 10^{-2} \text{ m}$ * Wavelength, lambda = 560 text nm = 560 times 10^-9 text m$\lambda = 560 \text{ nm} = 560 \times 10^{-9} \text{ m}$ * Time interval between consecutive minima, t = 12 text s$t = 12 \text{ s}$ Calculate the thickness change Delta t$\Delta t$ corresponding to consecutive transmission minima : Delta t = fraclambda2mu = frac560 times 10^-92 times 1.4 = frac560 times 10^-92.8 = 200 times 10^-9 text m = 2 times 10^-7 text m$\Delta t = \frac{\lambda}{2\mu} = \frac{560 \times 10^{-9}}{2 \times 1.4} = \frac{560 \times 10^{-9}}{2.8} = 200 \times 10^{-9} \text{ m} = 2 \times 10^{-7} \text{ m}$ Now, compute the volume change over this 12-second window to find the volumetric evaporation rate : textRate = fracpi cdot R^2 cdot Delta tt$\text{Rate} = \frac{\pi \cdot R^2 \cdot \Delta t}{t}$ textRate = fracpi times (1.8 times 10^-2)^2 times (2 times 10^-7)12$\text{Rate} = \frac{\pi \times (1.8 \times 10^{-2})^2 \times (2 \times 10^{-7})}{12}$ textRate = fracpi times 3.24 times 10^-4 times 2 times 10^-712$\text{Rate} = \frac{\pi \times 3.24 \times 10^{-4} \times 2 \times 10^{-7}}{12}$ textRate = fracpi times 6.48 times 10^-1112 = pi times 0.54 times 10^-11 = 54 times 10^-13 pi text m^3/texts quad text$\text{Rate} = \frac{\pi \times 6.48 \times 10^{-11}}{12} = \pi \times 0.54 \times 10^{-11} = 54 \times 10^{-13} \pi \text{ m}^3/\text{s} \quad \text{}$ Comparing this to the given expression pi times 10^-13 text m^3/texts$\pi \times 10^{-13} \text{ m}^3/\text{s}$ identifies the coefficient[cite: 207, 843]: textValue = 54$\text{Value} = 54$ ### Pattern Recognition A minimum in transmission means maximum reflection. For thin-film interference, the optical path difference changes by exactly fraclambda2$\frac{\lambda}{2}$ between consecutive fringes, which corresponds to a physical thickness change of fraclambda2mu$\frac{\lambda}{2\mu}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics