### 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

### Related Formula frac1f_texteq = frac1f_textliquid + frac1f_textglass$\frac{1}{f_{\text{eq}}} = \frac{1}{f_{\text{liquid}}} + \frac{1}{f_{\text{glass}}}$ frac1f = (mu - 1)left(frac1R_1 - frac1R_2right)$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ ### Core Logic Let's find the focal length of each individual lens in the combination: 1. **Liquid Lens**: - Refractive index, mu_l = 1.3$\mu_l = 1.3$ - The upper surface is flat (exposed to air): R_1 = infty$R_1 = \infty$ - The lower surface matches the upward concave surface of the glass lens: R_2 = -30mathrm~cm$R_2 = -30\mathrm{~cm}$ frac1f_textliquid = (1.3 - 1) left(frac1infty - frac1-30right) = 0.3 times frac130 = frac1100mathrm~cm^-1$\frac{1}{f_{\text{liquid}}} = (1.3 - 1) \left(\frac{1}{\infty} - \frac{1}{-30}\right) = 0.3 \times \frac{1}{30} = \frac{1}{100}\mathrm{~cm}^{-1}$ 2. **Glass Lens**: - Refractive index, mu_g = 1.5$\mu_g = 1.5$ - First surface radius (concave upward), R_1 = -30mathrm~cm$R_1 = -30\mathrm{~cm}$ - Second surface radius (convex downward), R_2 = -20mathrm~cm$R_2 = -20\mathrm{~cm}$ (following light path downward) frac1f_textglass = (1.5 - 1) left(frac1-30 - frac1-20right) = 0.5 left(-frac130 + frac120right) = 0.5 left(frac160right) = frac1120mathrm~cm^-1$\frac{1}{f_{\text{glass}}} = (1.5 - 1) \left(\frac{1}{-30} - \frac{1}{-20}\right) = 0.5 \left(-\frac{1}{30} + \frac{1}{20}\right) = 0.5 \left(\frac{1}{60}\right) = \frac{1}{120}\mathrm{~cm}^{-1}$ ### Step 1: Combination Focal Length Add the powers of both lenses: frac1f_texteq = frac1100 + frac1120 = frac6 + 5600 = frac11600$\frac{1}{f_{\text{eq}}} = \frac{1}{100} + \frac{1}{120} = \frac{6 + 5}{600} = \frac{11}{600}$ f_texteq = frac60011mathrm~cm$f_{\text{eq}} = \frac{600}{11}\mathrm{~cm}$ ### Pattern Recognition Sees: Glass lens with liquid poured on top → Think of it as a double lens system (liquid lens + glass lens). Trap: Be extremely careful with sign conventions for radii of curvature of the boundaries! Assume light travels from air through the liquid and then through the glass. This defines a consistent spatial propagation direction. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics

### Related Formula $frac1f_texteq = frac1f_1 + frac1f_2$$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2}$ $ $frac1v - frac1u = frac1f$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ where, $ where, $f_1 = focal length of the convex lens $ = focal length of the convex lens $f_2 = focal length of the concave lens $ = focal length of the concave lens $u = object distance $ = object distance $v = image distance ### Core Logic Given parameters: - Convex lens: $ = image distance ### Core Logic Given parameters: - Convex lens: $f_1 = +30\mathrm{~cm} - Concave lens: $ - Concave lens: $f_2 = -20\mathrm{~cm} - Object distance: $ - Object distance: $u = -20\mathrm{~cm} (placed to the left of the lens system) First, find the equivalent focal length of the lens combination in contact: $ (placed to the left of the lens system) First, find the equivalent focal length of the lens combination in contact: $frac1f_texteq = frac130 + frac1-20 = frac2 - 360 = -frac160 implies f_texteq = -60mathrm~cm$\frac{1}{f_{\text{eq}}} = \frac{1}{30} + \frac{1}{-20} = \frac{2 - 3}{60} = -\frac{1}{60} \implies f_{\text{eq}} = -60\mathrm{~cm}$ ### Step 1: Image Distance Calculation Use the thin lens formula: $ ### Step 1: Image Distance Calculation Use the thin lens formula: $frac1v - frac1u = frac1f_texteq$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{\text{eq}}}$ $ $frac1v - frac1-20 = frac1-60 implies frac1v + frac120 = -frac160$\frac{1}{v} - \frac{1}{-20} = \frac{1}{-60} \implies \frac{1}{v} + \frac{1}{20} = -\frac{1}{60}$ $ $frac1v = -frac160 - frac120 = frac-1 - 360 = -frac460 = -frac115$\frac{1}{v} = -\frac{1}{60} - \frac{1}{20} = \frac{-1 - 3}{60} = -\frac{4}{60} = -\frac{1}{15}$ $ $v = -15mathrm~cm$v = -15\mathrm{~cm}$ ### Pattern Recognition Sees: Lenses in contact + object distance → Combination focal length first, then thin lens formula. Trap: Keep proper sign conventions. An object to the left implies $ ### Pattern Recognition Sees: Lenses in contact + object distance → Combination focal length first, then thin lens formula. Trap: Keep proper sign conventions. An object to the left implies $u = -20\mathrm{~cm}. A negative image distance $. A negative image distance $v = -15\mathrm{~cm} means a virtual image formed on the same side as the object. The question asks for "distance", which is the magnitude: $ means a virtual image formed on the same side as the object. The question asks for "distance", which is the magnitude: $|-15| = 15\mathrm{~cm}$. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics