Which of the following are true for a single slit diffraction? (A) Width of central maxima increases with increase in wavelength keeping slit width constant. (B) Width of central maxima increases with decrease in wavelength keeping slit width constant. (C) Width of central maxima increases with decrease in slit width at constant wavelength. (D) Width of central maxima increases with increase in slit width at constant wavelength. (E) Brightness of central maxima increases for decrease in wavelength at constant slit width. Options :

### Related Formula textShift (x) = fracDd (mu - 1)t$$\text{Shift } (x) = \frac{D}{d} (\mu - 1)t$$ Alternatively written: d left(fracxDright) = (mu - 1)t$d \left(\frac{x}{D}\right) = (\mu - 1)t$ ### Core Logic Given parameters: Distance between slits, d = 0.1text cm$d = 0.1\text{ cm}$ Distance to screen, D = 50text cm$D = 50\text{ cm}$ Fringe shift, x = 0.2text cm$x = 0.2\text{ cm}$ Refractive index, mu = 1.5$\mu = 1.5$ ### Step 1: Calculate Thickness Rearranging the formula for thickness t$t$: t = fracx cdot dD(mu - 1)$$t = \frac{x \cdot d}{D(\mu - 1)}$$ t = frac(0.2)(0.1)50(1.5 - 1)$$t = \frac{(0.2)(0.1)}{50(1.5 - 1)}$$ t = frac0.0250(0.5) = frac0.0225$$t = \frac{0.02}{50(0.5)} = \frac{0.02}{25}$$ t = 8 times 10^-4text cm$$t = 8 \times 10^{-4}\text{ cm}$$ ### Pattern Recognition Direct plug-and-play into the slab shift formula. Working completely in cm avoids unit conversion errors as long as all given lengths (x$x$, d$d$, D$D$) are consistently in cm and the answer is requested in cm. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula textAngular fringe width theta = fraclambdad$$\text{Angular fringe width } \theta = \frac{\lambda}{d}$$ ### Core Logic The angular separation (angular fringe width) theta$\theta$ is given by fraclambdad$\frac{\lambda}{d}$, where lambda$\lambda$ is the wavelength and d$d$ is the separation between slits. It is completely independent of the distance to the screen D$D$. Statement I claims moving the screen away (increasing D$D$) increases angular separation. Since theta$\theta$ doesn't depend on D$D$, Statement I is false. Statement II claims using a higher wavelength lambda$\lambda$ increases angular separation. Since theta propto lambda$\theta \propto \lambda$, increasing lambda$\lambda$ increases theta$\theta$. Statement II is true. ### Step 1: Final Conclusion Statement I is false but Statement II is true. ### Pattern Recognition Linear fringe width y = lambda D / d$y = \lambda D / d$ depends on D$D$, but angular fringe width theta = lambda / d$\theta = \lambda / d$ isolates only the slit separation and wavelength. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula y_textshift = frac(mu - 1)t cdot Dd$$y_{\text{shift}} = \frac{(\mu - 1)t \cdot D}{d}$$ ### Core Logic Introducing a transparent sheet of thickness t$t$ and refractive index mu$\mu$ adds an extra optical path length of (mu - 1)t$(\mu - 1)t$. This translates to a shift of the entire fringe pattern on the screen. Given values: d = 0.4 text mm = 0.4 times 10^-3 text m$d = 0.4 \text{ mm} = 0.4 \times 10^{-3} \text{ m}$ D = 1 text m$D = 1 \text{ m}$ t = 20 \,mutextm = 20 times 10^-6 text m$t = 20 \,\mu\text{m} = 20 \times 10^{-6} \text{ m}$ y_textshift = 20 text mm = 20 times 10^-3 text m$y_{\text{shift}} = 20 \text{ mm} = 20 \times 10^{-3} \text{ m}$ ### Step 1: Substitution and Solving 20 times 10^-3 = frac(mu - 1) times 20 times 10^-6 times 10.4 times 10^-3$$20 \times 10^{-3} = \frac{(\mu - 1) \times 20 \times 10^{-6} \times 1}{0.4 \times 10^{-3}}$$ 20 times 10^-3 = frac(mu - 1) times 20 times 10^-64 times 10^-4$$20 \times 10^{-3} = \frac{(\mu - 1) \times 20 \times 10^{-6}}{4 \times 10^{-4}}$$ 20 times 10^-3 = (mu - 1) times 5 times 10^-2$$20 \times 10^{-3} = (\mu - 1) \times 5 \times 10^{-2}$$ (mu - 1) = frac20 times 10^-35 times 10^-2 = 4 times 10^-1 = 0.4$$(\mu - 1) = \frac{20 \times 10^{-3}}{5 \times 10^{-2}} = 4 \times 10^{-1} = 0.4$$ ### Step 2: Final Conclusion mu = 1 + 0.4 = 1.4$$\mu = 1 + 0.4 = 1.4$$ We are given mu = fracalpha10$\mu = \frac{\alpha}{10}$. fracalpha10 = 1.4 implies alpha = 14$$\frac{\alpha}{10} = 1.4 \implies \alpha = 14$$ ### Pattern Recognition The central fringe always shifts towards the slit covered by the slab. The linear shift y_0 = fracDd(mu-1)t$y_0 = \frac{D}{d}(\mu-1)t$ is independent of the wavelength used. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula mu = fraccv = fraclambda_0lambda$$\mu = \frac{c}{v} = \frac{\lambda_0}{\lambda}$$ mu_1 lambda_1 = mu_2 lambda_2$$\mu_1 \lambda_1 = \mu_2 \lambda_2$$ ### Core Logic Since frequency f$f$ remains constant across different media: fracmu_1mu_2 = fraclambda_2lambda_1$$\frac{\mu_1}{\mu_2} = \frac{\lambda_2}{\lambda_1}$$ Given mu_1 = frac43, lambda_1 = 540 mathrm~nm$\mu_1 = \frac{4}{3}, \lambda_1 = 540 \mathrm{~nm}$ and mu_2 = frac32$\mu_2 = \frac{3}{2}$: frac4/33/2 = fraclambda_2540$$\frac{4/3}{3/2} = \frac{\lambda_2}{540}$$ lambda_2 = left(frac4 times 23 times 3right) times 540 = frac89 times 540 = 480 mathrm~nm$$\lambda_2 = \left(\frac{4 \times 2}{3 \times 3}\right) \times 540 = \frac{8}{9} \times 540 = 480 \mathrm{~nm}$$ ### Step 1: Final Conclusion The wavelength in the second medium is 480 mathrm~nm$480 \mathrm{~nm}$. ### Pattern Recognition Product rule: mu cdot lambda = textconstant$\mu \cdot \lambda = \text{constant}$. (4/3) times 540 = (3/2) times lambda_2 implies lambda_2 = 480 mathrm~nm$(4/3) \times 540 = (3/2) \times \lambda_2 \implies \lambda_2 = 480 \mathrm{~nm}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics