### Related Formula Magnification formula for spherical mirrors: m = -fracvu$m = -\frac{v}{u}$ Mirror formula: frac1f = frac1v + frac1u$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ where: * u$u$ is the object distance * v$v$ is the image distance * f$f$ is the focal length ### Core Logic Since the image is magnified (m = 2$m = 2$) and virtual, the mirror must be concave (f < 0$f < 0$). Let us use standard coordinate geometry signs: object is on the left (u$u$ is negative, say -u_0$-u_0$), virtual image is on the right (v$v$ is positive, say +v_0$+v_0$). Given magnification: m = 2 = -fracvu implies v = -2u$m = 2 = -\frac{v}{u} \implies v = -2u$ In terms of magnitudes: v_0 = 2u_0$v_0 = 2u_0$ ### Step 1: Use Distance Condition The distance between the object and virtual image is 15text cm$15\text{ cm}$. Since the object is in front of the mirror and the virtual image is behind it: textDistance = u_0 + v_0 = 15text cm$\text{Distance} = u_0 + v_0 = 15\text{ cm}$ Substitute v_0 = 2u_0$v_0 = 2u_0$: u_0 + 2u_0 = 15 implies 3u_0 = 15 implies u_0 = 5text cm$u_0 + 2u_0 = 15 \implies 3u_0 = 15 \implies u_0 = 5\text{ cm}$ Thus: * u_0 = 5text cm implies u = -5text cm$u_0 = 5\text{ cm} \implies u = -5\text{ cm}$ * v_0 = 10text cm implies v = +10text cm$v_0 = 10\text{ cm} \implies v = +10\text{ cm}$

### Step 2: Calculate Focal Length Using the mirror formula: frac1f = frac1v + frac1u$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ frac1f = frac110 + frac1-5$\frac{1}{f} = \frac{1}{10} + \frac{1}{-5}$ frac1f = frac1 - 210 = -frac110 implies f = -10text cm$\frac{1}{f} = \frac{1 - 2}{10} = -\frac{1}{10} \implies f = -10\text{ cm}$ Thus, the focal length is -10text cm$-10\text{ cm}$. ### Pattern Recognition Virtual and magnified image → always concave mirror. Distance D$D$ between object and virtual image is given by D = |u| + v$D = |u| + v$. Since v = m|u|$v = m|u|$, we can simplify to |u| = fracDm+1$|u| = \frac{D}{m+1}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula mu = fracsinleft(fracA + delta_textmin2right)sinleft(fracA2right)$\mu = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ ### Core Logic Given mu = cotleft(fracA2right) = fraccosleft(fracA2right)sinleft(fracA2right)$\mu = \cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. Equating this to the prism formula: fraccosleft(fracA2right)sinleft(fracA2right) = fracsinleft(fracA + delta_textmin2right)sinleft(fracA2right)$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ cosleft(fracA2right) = sinleft(fracA + delta_textmin2right)$\cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + \delta_{\text{min}}}{2}\right)$ ### Step 1: Trigonometric Substitution Convert the cosine term to sine: sinleft(fracpi2 - fracA2right) = sinleft(fracA + delta_textmin2right)$\sin\left(\frac{\pi}{2} - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_{\text{min}}}{2}\right)$ fracpi2 - fracA2 = fracA + delta_textmin2$\frac{\pi}{2} - \frac{A}{2} = \frac{A + \delta_{\text{min}}}{2}$ pi - A = A + delta_textmin$\pi - A = A + \delta_{\text{min}}$ delta_textmin = pi - 2A$\delta_{\text{min}} = \pi - 2A$ ### Pattern Recognition When mu$\mu$ equals a cotangent function of half-angle, the sine dynamic simplifications lead directly to linear functions involving complements of angle A$A$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula h_textapparent = sum frach_imu_i$h_{\text{apparent}} = \sum \frac{h_i}{\mu_i}$ ### Core Logic Sum the contributions of both shifting mediums: h_textapparent = frach_1mu_1 + frach_2mu_2$h_{\text{apparent}} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$ Given h_1 = h_2 = 6text cm$h_1 = h_2 = 6\text{ cm}$, mu_1 = frac32$\mu_1 = \frac{3}{2}$, mu_2 = frac85$\mu_2 = \frac{8}{5}$: ### Step 1: Compute fraction value h_textapparent = frac63/2 + frac68/5 = 4 + frac308 = 4 + frac154$h_{\text{apparent}} = \frac{6}{3/2} + \frac{6}{8/5} = 4 + \frac{30}{8} = 4 + \frac{15}{4}$ h_textapparent = frac16 + 154 = frac314text cm$h_{\text{apparent}} = \frac{16 + 15}{4} = \frac{31}{4}\text{ cm}$ ### Step 2: Match to target format Comparing with fracalpha4$\frac{\alpha}{4}$ directly yields: alpha = 31$\alpha = 31$ ### Pattern Recognition Apparent depth across multi-layered planar mediums expands additively via separate individual medium thickness-to-refractive-index ratios. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula The magnification (m$m$) of a spherical mirror is given by: m = fracff - u$m = \frac{f}{f - u}$ where, f$f$ = focal length (R/2$R/2$) u$u$ = object distance ### Core Logic Given radius of curvature, R = 30 mathrm~cm$R = 30 \mathrm{~cm}$. For a convex mirror: f = +fracR2 = +15 mathrm~cm$f = +\frac{R}{2} = +15 \mathrm{~cm}$ Since a convex mirror always forms a virtual, erect, and diminished image of a real object, the magnification m$m$ must be positive: m = +frac12$m = +\frac{1}{2}$

### Step 1: Solve for Object Distance Using the magnification formula: +frac12 = frac1515 - u$+\frac{1}{2} = \frac{15}{15 - u}$ 15 - u = 30 implies u = -15 mathrm~cm$15 - u = 30 \implies u = -15 \mathrm{~cm}$ Thus, the object distance is -15 mathrm~cm$-15 \mathrm{~cm}$. ### Pattern Recognition Remember: Convex mirrors produce *only* virtual images for real objects, which means m$m$ is always positive and less than 1$1$. If the question mentioned a concave mirror with a diminished image of half size, the image would be real, and m$m$ would be negative. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments