A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is:

### Core Logic Refractive index represents the ratio of the speed of light in vacuum to its speed in a given medium[cite: 719]. Glass slows light down more than air does, hence mu_textglass approx 1.5 > mu_textair approx 1.0$\mu_{\text{glass}} \approx 1.5 > \mu_{\text{air}} \approx 1.0$, which makes Assertion (A) correct[cite: 130]. However, optical density is defined by a medium's capacity to refract light and is completely conceptually distinct from inertial mass density (mass per unit volume)[cite: 719]. For example, turpentine has a lower mass density than water but possesses a higher optical density and refractive index. Therefore, Reason (R) is fundamentally incorrect[cite: 722]. ### Pattern Recognition Optical density vs mass density is a signature conceptual trick in refraction theory[cite: 719]. They share the word 'density' but have entirely different physical meanings and no fixed mathematical proportionality[cite: 719, 722]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula sintheta_c = fracmu_1mu_2$$\sin\theta_c = \frac{\mu_1}{\mu_2}$$ [cite: 810] mu_1 sintheta = mu_2 sin r$$\mu_1 \sin\theta = \mu_2 \sin r$$ [cite: 807] ### Core Logic From the boundary geometry at the top interface, the angle of refraction r$r$ at the first surface satisfies: [cite: 806] r + theta_c = 90^circ implies r = 90^circ - theta_c$$r + \theta_c = 90^{\circ} \implies r = 90^{\circ} - \theta_c$$ [cite: 806] Applying Snell's law at the first entry interface: [cite: 170, 807] mu_1 sintheta = mu_2 sin r = mu_2 sin(90^circ - theta_c) = mu_2 costheta_c$$\mu_1 \sin\theta = \mu_2 \sin r = \mu_2 \sin(90^{\circ} - \theta_c) = \mu_2 \cos\theta_c$$ [cite: 172, 173, 807, 809] Since sintheta_c = fracmu_1mu_2 = frac1.01.25 = frac45$\sin\theta_c = \frac{\mu_1}{\mu_2} = \frac{1.0}{1.25} = \frac{4}{5}$, we have costheta_c = sqrt1 - left(frac45right)^2 = frac35$\cos\theta_c = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$[cite: 169, 810, 811]. Substituting back into the expression: [cite: 811] 1.0 cdot sintheta = 1.25 times frac35 = frac54 times frac35 = frac34$$1.0 \cdot \sin\theta = 1.25 \times \frac{3}{5} = \frac{5}{4} \times \frac{3}{5} = \frac{3}{4}$$ [cite: 169, 811] theta = sin^-1left(frac34 ight)$$\theta = \sin^{-1}\left(\frac{3}{4} ight)$$ [cite: 811] ### Pattern Recognition Maximum angle at the entry face ensures minimum angle of incidence at the subsequent wall[cite: 806, 807]. Setting that exact internal angle equal to the critical threshold condition values solves for the operational scanning range edge directly[cite: 808]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula Areal Magnification: m^2 = fracA_textimageA_textobject = left(fracff+u ight)^2 approx left(fracfu ight)^2$$m^2 = \frac{A_{\text{image}}}{A_{\text{object}}} = \left(\frac{f}{f+u} ight)^2 \approx \left(\frac{f}{u} ight)^2$$ since object distance u = -18\ mathrmkm$u = -18\ \mathrm{km}$ is vastly larger than f$f$. ### Core Logic Given parameters:

- Object height distance, H = 18\ mathrmkm = 18 times 10^3\ mathrmm$H = 18\ \mathrm{km} = 18 \times 10^3\ \mathrm{m}$ - Film size area, A_textimage = 2\ mathrmcm times 2\ mathrmcm = 4\ mathrmcm^2 = 4 times 10^-4\ mathrmm^2$A_{\text{image}} = 2\ \mathrm{cm} \times 2\ \mathrm{cm} = 4\ \mathrm{cm^2} = 4 \times 10^{-4}\ \mathrm{m^2}$ - Landscape area, A_textobject = 400\ mathrmkm^2 = 400 times 10^6\ mathrmm^2$A_{\text{object}} = 400\ \mathrm{km^2} = 400 \times 10^6\ \mathrm{m^2}$ Linear magnification factor: fracyx = sqrtfracA_textimageA_textobject = sqrtfrac4 times 10^-4400 times 10^6 = sqrt10^-12 = 10^-6$$\frac{y}{x} = \sqrt{\frac{A_{\text{image}}}{A_{\text{object}}}} = \sqrt{\frac{4 \times 10^{-4}}{400 \times 10^6}} = \sqrt{10^{-12}} = 10^{-6}$$ Using the simple pinhole/thin lens perspective ratio: fracfH = 10^-6 implies f = 18 times 10^3 times 10^-6 = 18 times 10^-3\ mathrmm = 1.8\ mathrmcm$$\frac{f}{H} = 10^{-6} \implies f = 18 \times 10^3 \times 10^{-6} = 18 \times 10^{-3}\ \mathrm{m} = 1.8\ \mathrm{cm}$$ ### Pattern Recognition For aerial satellite imaging contexts where u gg f$u \gg f$, linear sizing scales directly as fractextfilm sidetextground side = fracfH$\frac{\text{film side}}{\text{ground side}} = \frac{f}{H}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula The relationship between optical power P$P$ and focal length F$F$ is given by: F = frac1P$$F = \frac{1}{P}$$ Relative decrease in focal length is defined as: fracDelta FF = fracF - F'F$$\frac{\Delta F}{F} = \frac{F - F'}{F}$$ ### Core Logic Given initial power P = 2.5text D$P = 2.5\text{ D}$[cite: 19, 600]. After an increase of 0.1text D$0.1\text{ D}$, the new power is[cite: 19, 603]: P' = 2.5 + 0.1 = 2.6text D$$P' = 2.5 + 0.1 = 2.6\text{ D}$$ ### Step 1: Calculate Focal Length Change Find the initial and final focal lengths [cite: 602, 604]: F = frac12.5 = frac25$$F = \frac{1}{2.5} = \frac{2}{5}$$ F' = frac12.6 = frac513$$F' = \frac{1}{2.6} = \frac{5}{13}$$ Now, calculate the relative decrease: fracF - F'F = 1 - fracF'F = 1 - fracPP' = 1 - frac2.52.6 = frac0.12.6 = frac126 approx 0.04$$\frac{F - F'}{F} = 1 - \frac{F'}{F} = 1 - \frac{P}{P'} = 1 - \frac{2.5}{2.6} = \frac{0.1}{2.6} = \frac{1}{26} \approx 0.04$$ ### Pattern Recognition For a small change, we can approximate using differentiation: P = frac1F implies dP = -fracdFF^2 implies fracdFF = -fracdPP$P = \frac{1}{F} \implies dP = -\frac{dF}{F^2} \implies \frac{dF}{F} = -\frac{dP}{P}$. Thus, the relative change magnitude is frac0.12.5 = frac125 = 0.04$\frac{0.1}{2.5} = \frac{1}{25} = 0.04$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula The net focal power of a silvered tracking lens system is given by: P = 2P_L + P_M$$P = 2P_{L} + P_{M}$$ frac1f = frac2f_L + frac1f_M$$\frac{1}{f} = \frac{2}{f_{L}} + \frac{1}{f_{M}}$$ ### Core Logic As shown in diagram

, the plane flat side boundary interface has an infinite radius of curvature (R_2 = infty$R_{2} = \infty$), meaning its mirror focal component is f_M = infty implies P_M = 0$f_{M} = \infty \implies P_{M} = 0$. The power depends entirely on the refraction step: frac1f = frac2f_L $$\frac{1}{f} = \frac{2}{f_{L}} $$ ### Step 1: Lens Maker Formulation Find the focal expression of the immersed lens element [cite: 91, 679]: frac1f_L = left(fracmu_textglassmu_textliquid - 1 ight)left(frac1R ight) = left(frac1.51.2 - 1 ight)frac1R = frac0.31.2frac1R = frac14R$$\frac{1}{f_{L}} = \left(\frac{\mu_{\text{glass}}}{\mu_{\text{liquid}}} - 1 ight)\left(\frac{1}{R} ight) = \left(\frac{1.5}{1.2} - 1 ight)\frac{1}{R} = \frac{0.3}{1.2}\frac{1}{R} = \frac{1}{4R}$$ Now insert this into the total system tracking balance relation : frac1f = 2 left(frac14R ight) = frac12R $$\frac{1}{f} = 2 \left(\frac{1}{4R} ight) = \frac{1}{2R} $$ Given the final effective concave configuration matches f = 0.2text m$f = 0.2\text{ m}$ : frac10.2 = frac12R implies 2R = 0.2 implies R = 0.10text m$$\frac{1}{0.2} = \frac{1}{2R} \implies 2R = 0.2 \implies R = 0.10\text{ m}$$ ### Pattern Recognition Silvering a plano-flat back boundary means light traverses the initial curved face interface exactly twice, mapping to R = 2 cdot f cdot (mu_textrel - 1)$R = 2 \cdot f \cdot (\mu_{\text{rel}} - 1)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments