Two identical symmetric double convex lenses of focal length f$f$ are cut into two equal parts L_1, L_2$L_1, L_2$ by AB plane and L_3, L_4$L_3, L_4$ by XY plane as shown in figure respectively. The ratio of focal lengths of lenses L_1$L_1$ and L_3$L_3$ 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