### Related Formula Mirror Formula: frac1v + frac1u = frac1f$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ ### Core Logic For the convex mirror, u = -30text cm$u = -30\text{ cm}$ and f = +30text cm$f = +30\text{ cm}$. frac1v - frac130 = frac130 implies frac1v = frac230 implies v = +15text cm$\frac{1}{v} - \frac{1}{30} = \frac{1}{30} \implies \frac{1}{v} = \frac{2}{30} \implies v = +15\text{ cm}$ So, the convex mirror forms a virtual image 15 cm behind its surface. ### Step 1: Align Plane Mirror Image The total distance from the object to the image location is 30 + 15 = 45text cm$30 + 15 = 45\text{ cm}$. For a plane mirror to create an image at this same exact location, it must be placed precisely midway between the object and the image. Distance from object to plane mirror: d = frac452 = 22.5text cm$d = \frac{45}{2} = 22.5\text{ cm}$ Therefore, the clearance distance between the convex mirror and the plane mirror surface is: textDistance = 30 - 22.5 = 7.5text cm$\text{Distance} = 30 - 22.5 = 7.5\text{ cm}$ ### Pattern Recognition Coinciding images imply identical coordinate endpoints. Calculate the convex position explicitly, find the total path length from the real source object, and slice it in half for the plane mirror location. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula Magnification relation for spherical mirrors: m = -fracvu$m = -\frac{v}{u}$ Mirror equation: frac1f = frac1v + frac1u$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ ### Core Logic Given m = -frac13$m = -\frac{1}{3}$: -fracvu = -frac13 implies u = 3v$-\frac{v}{u} = -\frac{1}{3} implies u = 3v$ Since magnification is negative, a real inverted image is formed on the same side as the object in a concave mirror layout configuration.

### Step 1: Formulate Position Distances The distance between the object and image is given as 30mathrm~cm$30\mathrm{~cm}$: |u| - |v| = 30 implies 3v - v = 30 implies 2v = 30 implies v = 15mathrm~cm$|u| - |v| = 30 implies 3v - v = 30 implies 2v = 30 implies v = 15\mathrm{~cm}$ This gives u = 3(15) = 45mathrm~cm$u = 3(15) = 45\mathrm{~cm}$. ### Step 2: Solve for Focal Length and x Apply standard sign conventions (u = -45mathrm~cm$u = -45\mathrm{~cm}$, v = -15mathrm~cm$v = -15\mathrm{~cm}$): frac1f = -frac115 - frac145 = frac-3 - 145 = -frac445$\frac{1}{f} = -\frac{1}{15} - \frac{1}{45} = \frac{-3 - 1}{45} = -\frac{4}{45}$ |f| = frac454mathrm~cm$|f| = \frac{45}{4}\mathrm{~cm}$ Matching this with the prompt pattern form fracx4$\frac{x}{4}$ yields: x = 45$x = 45$ ### Pattern Recognition Real inverted diminished images (|m| < 1$|m| < 1$) mean the image forms closer to the mirror surface than the object, situated between the focal center F$F$ and center of curvature C$C$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula m = -fracvu$m = -\frac{v}{u}$ [cite: 700] frac1v + frac1u = frac1f$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ [cite: 707] ### Core Logic Given magnification magnitude |m| = frac14$|m| = \frac{1}{4}$[cite: 113]. Assuming a real image formed by a concave mirror: [cite: 701] fracvu = frac14 implies u = 4v$\frac{v}{u} = \frac{1}{4} \implies u = 4v$ [cite: 701] The distance between the object and the image is given as 40\ textcm$40\ \text{cm}$ [cite: 114, 116]: u - v = 40 implies 4v - v = 40 implies 3v = 40 implies v = frac403\ textcm$u - v = 40 \implies 4v - v = 40 \implies 3v = 40 \implies v = \frac{40}{3}\ \text{cm}$ u = 4 times frac403 = frac1603\ textcm$u = 4 \times \frac{40}{3} = \frac{160}{3}\ \text{cm}$ Applying mirror sign conventions (u = -frac1603$u = -\frac{160}{3}$, v = -frac403$v = -\frac{40}{3}$): [cite: 701, 704] frac1f = -frac340 - frac3160 = -frac12 + 3160 = -frac15160$\frac{1}{f} = -\frac{3}{40} - \frac{3}{160} = -\frac{12 + 3}{160} = -\frac{15}{160}$ f = -frac16015 approx -10.67\ textcm$f = -\frac{160}{15} \approx -10.67\ \text{cm}$ [cite: 712] Rounding to the matching options choice gives 10.7\ textcm$10.7\ \text{cm}$[cite: 118, 712]. ### Pattern Recognition Pay attention to sign conventions in mirror systems. A smaller real image formed by a concave mirror sits between the focus and center of curvature, resulting in u > v$u > v$ configurations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### 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