### 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 frac1v - frac1u = frac1f$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ m = fracvu$m = \frac{v}{u}$ m_L = fracdvdu = m^2$m_L = \frac{dv}{du} = m^2$ ### Core Logic Let the pole of the lens be the origin. Point A$A$ of the slanted object lies on the principal axis at u = -30mathrm~cm$u = -30\mathrm{~cm}$ from the convex lens (f = +20mathrm~cm$f = +20\mathrm{~cm}$). Let's locate the image of A$A$: frac1v - frac1-30 = frac120 implies frac1v = frac120 - frac130 = frac160 implies v = +60mathrm~cm$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \implies v = +60\mathrm{~cm}$ Thus, the transverse magnification m$m$ at point A$A$ is: m = fracvu = frac60-30 = -2$m = \frac{v}{u} = \frac{60}{-30} = -2$ Since the longitudinal extension of the object is small (du = 1mathrm~cm$du = 1\mathrm{~cm}$ along the axis): dv = m^2 du = (-2)^2 times 1 = 4mathrm~cm$dv = m^2 du = (-2)^2 \times 1 = 4\mathrm{~cm}$ The height of the object at point B$B$ is h_o = 2mathrm~cm$h_o = 2\mathrm{~cm}$. Its image height is: h_i = m cdot h_o = (-2) times 2 = -4mathrm~cm$h_i = m \cdot h_o = (-2) \times 2 = -4\mathrm{~cm}$ Now, compute the angle beta$\beta$ made by the image with the principal axis: tanbeta = frach_idv = frac-4mathrm~cm4mathrm~cm = -1$\tan\beta = \frac{h_i}{dv} = \frac{-4\mathrm{~cm}}{4\mathrm{~cm}} = -1$ beta = -45^circ$\beta = -45^{\circ}$ ### Step 1: Final Conclusion The angle made by the image with the principal axis is $ ### Step 1: Final Conclusion The angle made by the image with the principal axis is $-45^{\circ}. ### Pattern Recognition For small objects tilted with respect to the principal axis: 1. Axial displacement scales by $. ### Pattern Recognition For small objects tilted with respect to the principal axis: 1. Axial displacement scales by $m^2. 2. Transverse height scales by $. 2. Transverse height scales by $m. 3. Slope scales by $. 3. Slope scales by $\frac{m}{m^2} = \frac{1}{m}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics

### Related Formula $fracmu_2v - fracmu_1u = fracmu_2 - mu_1R$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$ ### Core Logic From the given diagram, using the standard Cartesian sign convention with the pole of the surface as origin: - Refractive index of first medium, $ ### Core Logic From the given diagram, using the standard Cartesian sign convention with the pole of the surface as origin: - Refractive index of first medium, $\mu_1 = 1.0 - Refractive index of second medium, $ - Refractive index of second medium, $\mu_2 = 1.5 - Object distance, $ - Object distance, $u = -0.2\mathrm{~m} (left of surface) - Radius of curvature, $ (left of surface) - Radius of curvature, $R = +0.4\mathrm{~m} (convex surface towards first medium, center $ (convex surface towards first medium, center $C lies in second medium) Applying the formula for refraction at a spherical interface: $ lies in second medium) Applying the formula for refraction at a spherical interface: $frac1.5v - frac1-0.2 = frac1.5 - 10.4$\frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{0.4}$ $ $frac1.5v + 5.0 = frac0.50.4 = 1.25$\frac{1.5}{v} + 5.0 = \frac{0.5}{0.4} = 1.25$ $ $frac1.5v = 1.25 - 5.0 = -3.75$\frac{1.5}{v} = 1.25 - 5.0 = -3.75$ $ $v = frac1.5-3.75 = -0.4mathrm~m$v = \frac{1.5}{-3.75} = -0.4\mathrm{~m}$ The negative sign indicates that the image is formed to the left of the spherical refracting surface. ### Step 1: Final Conclusion The image of object 'O' is formed $ The negative sign indicates that the image is formed to the left of the spherical refracting surface. ### Step 1: Final Conclusion The image of object 'O' is formed $0.4\mathrm{~m}$ left to the spherical surface. ### Pattern Recognition Ensure to strictly implement the coordinate sign conventions: the direction of incident light is positive. Since light goes from left to right, left-side points have a negative coordinate, and right-side points have a positive coordinate. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics