Two coherent monochromatic light beams of intensities 4I and 9I are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is xI. The value of x is ________.

### 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

### 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 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