Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

### Related Formula I = I_textincident cos^2 theta quad text(Malus's Law)$$I = I_{\text{incident}} \cos^2 \theta \quad \text{(Malus's Law)}$$ ### Core Logic When unpolarised light of intensity I_0$I_0$ passes through the first polaroid mathrmA$\mathrm{A}$, it becomes plane-polarised, and its intensity drops by exactly half. I_1 = fracI_02$$I_1 = \frac{I_0}{2}$$ When this polarised light passes through the second polaroid mathrmB$\mathrm{B}$, the transmitted intensity is determined by Malus's Law. ### Step 1: Apply Malus's Law The angle between the principal planes of polaroids mathrmA$\mathrm{A}$ and mathrmB$\mathrm{B}$ is theta = 45^circ$\theta = 45^\circ$. I_2 = I_1 cos^2(45^circ)$$I_2 = I_1 \cos^2(45^\circ)$$ I_2 = left(fracI_02right) left(frac1sqrt2right)^2$$I_2 = \left(\frac{I_0}{2}\right) \left(\frac{1}{\sqrt{2}}\right)^2$$ I_2 = fracI_02 times frac12 = fracI_04$$I_2 = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$$ ### Pattern Recognition Unpolarised to Polarised rightarrow I_0/2$\rightarrow I_0/2$. Polarised to Polarised rightarrow I cos^2theta$\rightarrow I \cos^2\theta$. At theta = 45^circ$\theta = 45^\circ$, cos^2theta = 1/2$\cos^2\theta = 1/2$, resulting in a final intensity of I_0/4$I_0/4$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula textWidth of secondary maxima = fraclambda Da$$\text{Width of secondary maxima} = \frac{\lambda D}{a}$$ ### Core Logic In a single slit diffraction pattern, the linear width of any secondary maxima (fringe width of secondary bright bands) is given by W = fraclambda Da$W = \frac{\lambda D}{a}$, whereas the central maximum is double this width (2fraclambda Da$2\frac{\lambda D}{a}$). ### Step 1: Parameter Identification Given values: Slit width, a = 0.2 times 10^-3 mathrm~m$a = 0.2 \times 10^{-3} \mathrm{~m}$ Wavelength, lambda = 400 times 10^-9 mathrm~m$\lambda = 400 \times 10^{-9} \mathrm{~m}$ Distance to screen (focal length of the lens), D = 100 times 10^-2 mathrm~m = 1 mathrm~m$D = 100 \times 10^{-2} \mathrm{~m} = 1 \mathrm{~m}$ ### Step 2: Execution Substitute these into the formula: textWidth = frac400 times 10^-90.2 times 10^-3 times 1$$\text{Width} = \frac{400 \times 10^{-9}}{0.2 \times 10^{-3}} \times 1$$ textWidth = frac4000.2 times 10^-6 mathrm~m$$\text{Width} = \frac{400}{0.2} \times 10^{-6} \mathrm{~m}$$ textWidth = 2000 times 10^-6 mathrm~m = 2 times 10^-3 mathrm~m$$\text{Width} = 2000 \times 10^{-6} \mathrm{~m} = 2 \times 10^{-3} \mathrm{~m}$$ textWidth = 2 mathrm~mm$$\text{Width} = 2 \mathrm{~mm}$$ ### Pattern Recognition Remember to strictly distinguish between central maximum (2lambda D/a$2\lambda D/a$) and secondary maxima (lambda D/a$\lambda D/a$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics

### Related Formula Brewster's Law states that at complete polarization upon reflection, the reflected and refracted rays are perpendicular to each other: i_p + r = 90^circ$$i_p + r = 90^{\circ}$$ ### Core Logic The incident angle is given as i_p = 60^circ$i_p = 60^{\circ}$. At this angle, since the reflected ray is completely polarized, the geometry of Brewster's angle applies.

### Step 1: Calculate Refraction Angle 60^circ + r = 90^circ$$60^{\circ} + r = 90^{\circ}$$ r = 90^circ - 60^circ = 30^circ$$r = 90^{\circ} - 60^{\circ} = 30^{\circ}$$ ### Pattern Recognition The condition "reflected ray is completely polarized" is a direct trigger for Brewster's Law (i_p + r = 90^circ$i_p + r = 90^{\circ}$). No refractive index (mu$\mu$) calculation is needed if only the geometric angle is asked. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula I_textincoherent = I_A + I_B$$I_{\text{incoherent}} = I_A + I_B$$ I_textcoherent = I_A + I_B + 2sqrtI_A I_B cosphi$$I_{\text{coherent}} = I_A + I_B + 2\sqrt{I_A I_B} \cos\phi$$ ### Core Logic Let the individual intensities be I_A = I_0$I_A = I_0$ and I_B = 9I_0$I_B = 9I_0$. For incoherent waves, the net intensity is simply the algebraic sum: I_1 = I_A + I_B = I_0 + 9I_0$$I_1 = I_A + I_B = I_0 + 9I_0$$ I_1 = 10I_0$I_1 = 10I_0$ ### Step 2: Coherent Waves Interference For coherent waves with a phase difference of phi = 60^circ$\phi = 60^\circ$: I_2 = I_A + I_B + 2sqrtI_A I_B cos(60^circ)$$I_2 = I_A + I_B + 2\sqrt{I_A I_B} \cos(60^\circ)$$ I_2 = I_0 + 9I_0 + 2sqrt(I_0)(9I_0) left(frac12right)$$I_2 = I_0 + 9I_0 + 2\sqrt{(I_0)(9I_0)} \left(\frac{1}{2}\right)$$ I_2 = 10I_0 + 2(3I_0) left(frac12right)$$I_2 = 10I_0 + 2(3I_0) \left(\frac{1}{2}\right)$$ I_2 = 10I_0 + 3I_0 = 13I_0$$I_2 = 10I_0 + 3I_0 = 13I_0$$ ### Step 3: Finding x Taking the ratio: fracI_1I_2 = frac10I_013I_0 = frac1013$$\frac{I_1}{I_2} = \frac{10I_0}{13I_0} = \frac{10}{13}$$ Comparing with the given expression frac10x$\frac{10}{x}$, we get: x = 13$x = 13$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Wave Optics