### Related Formula 1. Mirror focal length: f = fracR2$f = \frac{R}{2}$ 2. Magnification (image size relative to object size): m = frach_ih_o = fracff - u$m = \frac{h_i}{h_o} = \frac{f}{f - u}$ where u$u$ is the object distance. ### Core Logic Given parameters: - Radius of curvature R = 12 \ mathrmcm implies |f| = 6 \ mathrmcm$R = 12 \ \mathrm{cm} \implies |f| = 6 \ \mathrm{cm}$ - Object distance u = -18 \ mathrmcm$u = -18 \ \mathrm{cm}$ Let's calculate magnification for both mirrors: 1. **For Convex Mirror:** - Focal length f_textconvex = +6 \ mathrmcm$f_{\text{convex}} = +6 \ \mathrm{cm}$ (using Cartesian sign convention) - Magnification: m_1 = fracf_textconvexf_textconvex - u = frac66 - (-18) = frac624 = frac14$m_1 = \frac{f_{\text{convex}}}{f_{\text{convex}} - u} = \frac{6}{6 - (-18)} = \frac{6}{24} = \frac{1}{4}$ Thus, image size is frac14 h_o$\frac{1}{4} h_o$. ### Step 1: Calculate magnification of concave mirror 2. **For Concave Mirror:** - Focal length f_textconcave = -6 \ mathrmcm$f_{\text{concave}} = -6 \ \mathrm{cm}$ - Magnification:

m_2 = fracf_textconcavef_textconcave - u = frac-6-6 - (-18) = frac-612 = -frac12$m_2 = \frac{f_{\text{concave}}}{f_{\text{concave}} - u} = \frac{-6}{-6 - (-18)} = \frac{-6}{12} = -\frac{1}{2}$ Thus, image size is frac12 h_o$\frac{1}{2} h_o$. ### Step 2: Calculate the ratio of image sizes Since the objects are identical (same height h_o$h_o$), the ratio of the sizes of the images is:

textRatio = frac|h_i1||h_i2| = frac|m_1||m_2| = frac1/41/2 = frac12$\text{Ratio} = \frac{|h_{i1}|}{|h_{i2}|} = \frac{|m_1|}{|m_2|} = \frac{1/4}{1/2} = \frac{1}{2}$ Thus, the ratio of sizes is 1/2$1/2$. ### Pattern Recognition Sees: Parallel convex vs concave mirror magnification. Trap: Reversing the signs of focal lengths (Convex focal length is +$+$, Concave is -$-$ in standard coordinate systems). Shortcut: Use direct magnification equation m = fracff-u$m = \frac{f}{f-u}$. For convex, m = frac624 = frac14$m = \frac{6}{24} = \frac{1}{4}$. For concave, m = frac-612 = -frac12$m = \frac{-6}{12} = -\frac{1}{2}$. Ratio of absolute values is frac1/41/2 = 1/2$\frac{1/4}{1/2} = 1/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula Lens Maker's Formula: frac1f = (mu - 1) left(frac1R_1 - frac1R_2right)$\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ For a bi-convex lens of equal radii (R_1 = +R$R_1 = +R$, R_2 = -R$R_2 = -R$): frac1f = (mu - 1) frac2R$\frac{1}{f} = (\mu - 1) \frac{2}{R}$ Power P propto frac1f$P \propto \frac{1}{f}$. ### Core Logic For the initial lens: - R = frac16 \ mathrmcm$R = \frac{1}{6} \ \mathrm{cm}$ - frac1f_1 = (mu - 1) frac21/6 = 12 (mu - 1)$\frac{1}{f_1} = (\mu - 1) \frac{2}{1/6} = 12 (\mu - 1)$ For the replacement lens (R_1 = +R_1$R_1 = +R_1$, R_2 = -R_2$R_2 = -R_2$): frac1f_2 = (mu - 1) left(frac1R_1 + frac1R_2right)$\frac{1}{f_2} = (\mu - 1) \left(\frac{1}{R_1} + \frac{1}{R_2}\right)$ Since power must be preserved (f_1 = f_2$f_1 = f_2$): frac1R_1 + frac1R_2 = frac2R = 12 \ mathrmcm^-1$\frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} = 12 \ \mathrm{cm}^{-1}$ We need to find a combination where the sum of the reciprocals of the radii equals 12$12$. ### Step 1: Verify the options Let's check each choice: - **Option (1):** R_1 = frac13$R_1 = \frac{1}{3}$, R_2 = frac13$R_2 = \frac{1}{3}$: frac11/3 + frac11/3 = 3 + 3 = 6 neq 12$\frac{1}{1/3} + \frac{1}{1/3} = 3 + 3 = 6 \neq 12$ - **Option (2):** R_1 = frac15$R_1 = \frac{1}{5}$, R_2 = frac17$R_2 = \frac{1}{7}$: frac11/5 + frac11/7 = 5 + 7 = 12 quad checkmark quad text(Correct Combination)$\frac{1}{1/5} + \frac{1}{1/7} = 5 + 7 = 12 \quad \checkmark \quad \text{(Correct Combination)}$ - **Option (3):** R_1 = frac13$R_1 = \frac{1}{3}$, R_2 = frac17$R_2 = \frac{1}{7}$: frac11/3 + frac11/7 = 3 + 7 = 10 neq 12$\frac{1}{1/3} + \frac{1}{1/7} = 3 + 7 = 10 \neq 12$ - **Option (4):** R_1 = frac16$R_1 = \frac{1}{6}$, R_2 = frac19$R_2 = \frac{1}{9}$: frac11/6 + frac11/9 = 6 + 9 = 15 neq 12$\frac{1}{1/6} + \frac{1}{1/9} = 6 + 9 = 15 \neq 12$ Thus, only Option (2) meets the physical conditions. ### Pattern Recognition Sees: Equivalent thin lens power with modified surfaces. Trap: Neglecting sign convention for the second spherical surface during substitution. Shortcut: If the radii are of the form 1/n$1/n$, then the sum of n_1 + n_2$n_1 + n_2$ must equal 2 times n_textinitial$2 \times n_{\text{initial}}$. Since initial n = 6$n = 6$, 2 times 6 = 12$2 \times 6 = 12$. The only pairing whose denominators add up to 12 is 5 + 7$5 + 7$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula 1. Prism Formula relating refractive index to minimum deviation: mu = fracsinleft(fracA + delta_m2right)sinleft(fracA2right)$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ 2. Under the condition of minimum deviation: i = fracA + delta_m2$i = \frac{A + \delta_m}{2}$ Thus, the formula simplifies to: mu = fracsin isin(A/2)$\mu = \frac{\sin i}{\sin(A/2)}$ ### Core Logic Given parameters: - Angle of prism A = 60^circ$A = 60^\circ$ - Refractive index of material mu = sqrt2$\mu = \sqrt{2}$ Substitute the parameters into the simplified formula: sqrt2 = fracsin isin(60^circ / 2)$\sqrt{2} = \frac{\sin i}{\sin(60^\circ / 2)}$ sqrt2 = fracsin isin(30^circ)$\sqrt{2} = \frac{\sin i}{\sin(30^\circ)}$ ### Step 1: Solve for angle of incidence Since sin(30^circ) = 0.5$\sin(30^\circ) = 0.5$: sin i = sqrt2 times frac12 = frac1sqrt2$\sin i = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$ Solving for i$i$: i = 45^circ$i = 45^\circ$ Thus, the angle of incidence is 45$45$ degrees. ### Pattern Recognition Sees: Prism minimum deviation and refracting index relations. Trap: Mistaking the computed angle of incidence (i$i$) for the angle of minimum deviation (delta_m$\delta_m$). Shortcut: Snell's law at the symmetrical boundary reduces to mu = sin i / sin(A/2)$\mu = \sin i / \sin(A/2)$. Because A = 60^circ$A = 60^\circ$, the denominator is sin 30^circ = 1/2$\sin 30^\circ = 1/2$. Therefore, sin i = 0.5 times sqrt2 = 1/sqrt2 implies i = 45^circ$\sin i = 0.5 \times \sqrt{2} = 1/\sqrt{2} \implies i = 45^\circ$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula The refraction equation at a single spherical interface is given by: fracmu_2v - fracmu_1u = fracmu_2 - mu_1R$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$ where: - mu_1$\mu_1$ is the refractive index of the initial medium (air, mu_1 = 1.0$\mu_1 = 1.0$) - mu_2$\mu_2$ is the refractive index of the second medium (glass, mu_2 = 1.5$\mu_2 = 1.5$) - u$u$ is the object distance - v$v$ is the image distance - R$R$ is the radius of curvature ### Core Logic Given parameters: - mu_1 = 1.0$\mu_1 = 1.0$, mu_2 = 1.5$\mu_2 = 1.5$ - Radius of curvature R = +50mathrm~cm$R = +50\mathrm{~cm}$ - Image distance v = +200mathrm~cm$v = +200\mathrm{~cm}$ (real image inside glass)

### Step 1: Substitute parameters into refraction formula $frac1.5200 - frac1u = frac1.5 - 1.050$$\frac{1.5}{200} - \frac{1}{u} = \frac{1.5 - 1.0}{50}$ frac3400 - frac1u = frac0.550 = frac1100$\frac{3}{400} - \frac{1}{u} = \frac{0.5}{50} = \frac{1}{100}$ ### Step 2: Solve for object distance ($ ### Step 2: Solve for object distance ($u) $) $-frac1u = frac1100 - frac3400$-\frac{1}{u} = \frac{1}{100} - \frac{3}{400}$ $ $-frac1u = frac4 - 3400 = frac1400$-\frac{1}{u} = \frac{4 - 3}{400} = \frac{1}{400}$ $ $u = -400mathrm~cm = -4mathrm~m$u = -400\mathrm{~cm} = -4\mathrm{~m}$ The magnitude of the distance of the light source is $ The magnitude of the distance of the light source is $4\mathrm{~m}. ### Pattern Recognition Remember standard sign convention: Light travels from object to refracting surface. Distances measured in the direction of incident light are positive. Here, $. ### Pattern Recognition Remember standard sign convention: Light travels from object to refracting surface. Distances measured in the direction of incident light are positive. Here, $v and $ and $R are positive, whereas $ are positive, whereas $u$ is negative. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula For light propagation, wave velocity, frequency, and wavelength are related by: v = f lambda Rightarrow lambda_textair = fraccf$v = f \lambda \Rightarrow \lambda_{\text{air}} = \frac{c}{f}$ When light passes into a medium of refractive index mu$\mu$, the frequency remains constant, but the wavelength scales down to: lambda_textmedium = fraclambda_textairmu$\lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{\mu}$ ### Core Logic Given parameters: - Frequency f = 5 times 10^14mathrm~Hz$f = 5 \times 10^{14}\mathrm{~Hz}$ - Speed of light in vacuum/air c approx 3 times 10^8mathrm~m/s$c \approx 3 \times 10^8\mathrm{~m/s}$ - Refractive index of medium mu = 2$\mu = 2$ ### Step 1: Calculate Wavelength in Air (Vacuum) lambda_textair = frac3 times 10^8mathrm~m/s5 times 10^14mathrm~Hz = 0.6 times 10^-6mathrm~m = 600mathrm~nm$\lambda_{\text{air}} = \frac{3 \times 10^8\mathrm{~m/s}}{5 \times 10^{14}\mathrm{~Hz}} = 0.6 \times 10^{-6}\mathrm{~m} = 600\mathrm{~nm}$ ### Step 2: Calculate Refracted Wavelength in Medium lambda_textmedium = fraclambda_textairmu = frac600mathrm~nm2 = 300mathrm~nm$\lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{\mu} = \frac{600\mathrm{~nm}}{2} = 300\mathrm{~nm}$ ### Pattern Recognition Remember: Frequency is a source characteristic and never changes during refraction. Speed and wavelength both decrease by a factor of mu$\mu$ inside the medium. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments