If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30^circ$30^{circ}$ in a single slit diffraction pattern recorded using 628mathrm~nm$628\mathrm{~nm}$ light, then the width of the slit is mumathrmm$\mu\mathrm{m}$.

### 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

### 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 The position y$y$ of the n$n$-th bright fringe from the central maximum in a YDSE setup is: y = fracnlambda Dd$$y = \frac{n\lambda D}{d}$$ For two wavelengths to overlap, their linear coordinates must match identically: n_1lambda_1 = n_2lambda_2$$n_{1}\lambda_{1} = n_{2}\lambda_{2}$$ ### Core Logic Equating the respective path positions: n_1 times 480text nm = n_2 times 600text nm$$n_{1} \times 480\text{ nm} = n_{2} \times 600\text{ nm}$$ fracn_1n_2 = frac600480 = frac54$$\frac{n_{1}}{n_{2}} = \frac{600}{480} = \frac{5}{4}$$ ### Step 1: Evaluating the Least Count To satisfy the smallest integer ratio requirement for first spatial coincidence, the numerator must scale up to its base irreducible integer divisor: n_1,min = 5$n_{1,\min} = 5$ ### Pattern Recognition Overlap occurs whenever indices inversely mimic their wavelength factors. The shorter wavelength always maps to a higher fringe index count. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics