### Related Formula Magnification m$m$ of a mirror is given by: m = -fracvu$m = -\frac{v}{u}$ The mirror equation relates focal length to distance positions: frac1f = frac1v + frac1u implies f = fracuvu+v$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies f = \frac{uv}{u+v}$ Radius of curvature R = 2f$R = 2f$. ### Core Logic Given, magnification m = -3$m = -3$. This tells us the image is real and inverted[cite: 131, 757]: -3 = -fracvu implies v = 3u$-3 = -\frac{v}{u} \implies v = 3u$ Since both real objects and real images lie on the same side in front of a concave mirror, u$u$ and v$v$ are both negative fields. The physical distance separation between them is: |v| - |u| = 20 implies 3|u| - |u| = 20 implies 2|u| = 20 implies |u| = 10 text cm$|v| - |u| = 20 \implies 3|u| - |u| = 20 \implies 2|u| = 20 \implies |u| = 10 \text{ cm}$ Therefore, object distance u = -10 text cm$u = -10 \text{ cm}$ and image distance v = -30 text cm$v = -30 \text{ cm}$ . Substitute into the focal equation formula: f = frac(-10)(-30)-10 - 30 = frac300-40 = -7.5 text cm$f = \frac{(-10)(-30)}{-10 - 30} = \frac{300}{-40} = -7.5 \text{ cm}$ R = 2 times |f| = 2 times 7.5 = 15 text cm$R = 2 \times |f| = 2 \times 7.5 = 15 \text{ cm}$ ### Step 1: Visual Diagram The visual positioning profile tracking focal path boundaries is given below:

### Pattern Recognition A magnification of -3$-3$ tells you immediately that the object lies between the Focus (F$F$) and Center of Curvature (C$C$), while the image forms beyond C$C$. This geometric layout instantly verifies that the radius value must exceed 10text cm$10\text{ cm}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics

### Core Logic Critical angle setups dictate [cite: 613, 614]: sin theta_1C = fracn_1n_2, quad sin theta_2C = fracn_1n_3$\sin \theta_{1C} = \frac{n_1}{n_2}, \quad \sin \theta_{2C} = \frac{n_1}{n_3}$ [cite: 613, 614] Given sin theta_2C - sin theta_1C = frac12$\sin \theta_{2C} - \sin \theta_{1C} = \frac{1}{2}$ [cite: 2, 615]: fracn_1n_3 - fracn_1n_2 = frac12 implies n_1 left(frac25 - 1right) = fracn_22$\frac{n_1}{n_3} - \frac{n_1}{n_2} = \frac{1}{2} \implies n_1 \left(\frac{2}{5} - 1\right) = \frac{n_2}{2}$ [cite: 617, 619, 621] fracn_1n_2 = -frac56$\frac{n_1}{n_2} = -\frac{5}{6}$ This produces an impossible negative value for a refractive index ratio, meaning the data contains a contradiction. This question was dropped by NTA.

### Related Formula (u+f)(v-f) = f^2$(u+f)(v-f) = f^2$ ### Core Logic

By rewriting the lens equation coordinates with respect to the focal focuses, we track a rectangular hyperbola structure. In the quadrant satisfying |u| > f$|u| > f$, the matching curvature coordinates belong strictly to the profile shown in option (2). ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula n_1 sin i = n_2 sin r$n_1 \sin i = n_2 \sin r$ ### Core Logic

By Snell\'s law matching standard boundary normal configurations : n_1 sin(90^circ - theta_1) = n_2 sin theta_3 implies n_1 cos theta_1 = n_2 sin theta_3$n_1 \sin(90^{\circ} - \theta_1) = n_2 \sin \theta_3 \implies n_1 \cos \theta_1 = n_2 \sin \theta_3$ [cite: 711, 712] Substituting the angle identity macro given [cite: 2, 713]: n_1 left(fracn_22n_1right) = n_2 sin theta_3 implies sin theta_3 = frac12 implies theta_3 = 30^circ$n_1 \left(\frac{n_2}{2n_1}\right) = n_2 \sin \theta_3 \implies \sin \theta_3 = \frac{1}{2} \implies \theta_3 = 30^{\circ}$ ### Step 1: Geometrical Thickness Resolution From the block triangles geometry : tan 30^circ = fracd/2t implies frac1sqrt3 = fracd2t$\tan 30^{\circ} = \frac{d/2}{t} \implies \frac{1}{\sqrt{3}} = \frac{d}{2t}$ t = fracdsqrt32 = frac4sqrt3 cdot sqrt32 = 6text cm$t = \frac{d\sqrt{3}}{2} = \frac{4\sqrt{3} \cdot \sqrt{3}}{2} = 6\text{ cm}$ ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula Lens magnification formula: m = fracvu$m = \frac{v}{u}$ Thin lens equation: frac1v - frac1u = frac1f$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ ### Core Logic For a virtual image formed by a convex lens, both the object and the image lie on the same side of the lens. The image distance is three times the object distance: v = 3u$v = 3u$ The distance between the object and its virtual image is given as 20mathrm~cm$20\mathrm{~cm}$: v - u = 20mathrm~cm implies 3u - u = 20mathrm~cm$v - u = 20\mathrm{~cm} \implies 3u - u = 20\mathrm{~cm}$ 2u = 20mathrm~cm implies u = 10mathrm~cm$2u = 20\mathrm{~cm} \implies u = 10\mathrm{~cm}$ Applying Cartesian sign conventions: object distance u = -10mathrm~cm$u = -10\mathrm{~cm}$ and image distance v = -30mathrm~cm$v = -30\mathrm{~cm}$. ### Step 1: Calculate Focal Length Substitute these values into the lens formula: frac1-30 - frac1-10 = frac1f$\frac{1}{-30} - \frac{1}{-10} = \frac{1}{f}$ -frac130 + frac110 = frac1f$-\frac{1}{30} + \frac{1}{10} = \frac{1}{f}$ frac-1 + 330 = frac230 = frac115 = frac1f implies f = 15mathrm~cm$\frac{-1 + 3}{30} = \frac{2}{30} = \frac{1}{15} = \frac{1}{f} \implies f = 15\mathrm{~cm}$ ### Pattern Recognition A virtual image from a convex lens means the object is placed inside the focal point (u < f$u < f$). This serves as a quick sanity check for your final value (10mathrm~cm < 15mathrm~cm$10\mathrm{~cm} < 15\mathrm{~cm}$). ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments