A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is:

### Related Formula Lens Maker's Formula in a surrounding medium with refractive index mu_m$\mu_m$ is: frac1f = left( fracmu_Lmu_m - 1 right) left( frac1R_1 - frac1R_2 right)$$\frac{1}{f} = \left( \frac{\mu_L}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ ### Core Logic In air (mu_m = 1$\mu_m = 1$): frac112 = (1.6 - 1) left( frac1R_1 - frac1R_2 right)$$\frac{1}{12} = (1.6 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ frac112 = 0.6 left( frac1R_1 - frac1R_2 right) implies left( frac1R_1 - frac1R_2 right) = frac112 times 0.6 = frac1072$$\frac{1}{12} = 0.6 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{12 \times 0.6} = \frac{10}{72}$$ ### Step 1: Calculate Focal Length in Water In water (mu_m = 1.28$\mu_m = 1.28$): frac1f_w = left( frac1.61.28 - 1 right) left( frac1072 right)$$\frac{1}{f_w} = \left( \frac{1.6}{1.28} - 1 \right) \left( \frac{10}{72} \right)$$ Simplify the relative index factor: frac1.61.28 = frac160128 = 1.25$$\frac{1.6}{1.28} = \frac{160}{128} = 1.25$$ frac1f_w = (1.25 - 1) left( frac1072 right) = 0.25 times frac1072 = frac14 times frac1072 = frac10288$$\frac{1}{f_w} = (1.25 - 1) \left( \frac{10}{72} \right) = 0.25 \times \frac{10}{72} = \frac{1}{4} \times \frac{10}{72} = \frac{10}{288}$$ f_w = 28.8 mathrm~cm = 288 mathrm~mm$$f_w = 28.8 \mathrm{~cm} = 288 \mathrm{~mm}$$ ### Pattern Recognition Sees: Lens index $ ### Pattern Recognition Sees: Lens index $\mu_L = 1.6, focal length in air $, focal length in air $f_a, and focal length in medium $, and focal length in medium $f_m. Shortcut: Use the ratio of focal lengths directly: $. Shortcut: Use the ratio of focal lengths directly: $fracf_mf_a = fracmu_L - 1fracmu_Lmu_m - 1 = frac0.6frac1.61.28 - 1 = frac0.60.25 = 2.4$\frac{f_m}{f_a} = \frac{\mu_L - 1}{\frac{\mu_L}{\mu_m} - 1} = \frac{0.6}{\frac{1.6}{1.28} - 1} = \frac{0.6}{0.25} = 2.4$ $ $f_m = 2.4 times 12 = 28.8 mathrm~cm = 288 mathrm~mm$f_m = 2.4 \times 12 = 28.8 \mathrm{~cm} = 288 \mathrm{~mm}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula The equivalent focal length f_texteq$f_{\text{eq}}$ of two thin coaxially aligned lenses separated by distance d$d$ is given by: frac1f_texteq = frac1f_1 + frac1f_2 - fracdf_1 f_2$$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$$ The equivalent power in diopters (D$D$) when focal lengths are in meters is: P = frac1f_texteq$$P = \frac{1}{f_{\text{eq}}}$$ ### Core Logic Given parameters: - f_1 = 30 mathrm~cm = 0.3 mathrm~m$f_1 = 30 \mathrm{~cm} = 0.3 \mathrm{~m}$ - f_2 = 10 mathrm~cm = 0.1 mathrm~m$f_2 = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$ - d = 10 mathrm~cm = 0.1 mathrm~m$d = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$ ### Step 1: Calculate Power Substitute parameters into the equivalent focal length equation: frac1f_texteq = frac10.3 + frac10.1 - frac0.10.3 times 0.1$$\frac{1}{f_{\text{eq}}} = \frac{1}{0.3} + \frac{1}{0.1} - \frac{0.1}{0.3 \times 0.1}$$ frac1f_texteq = frac10.3 + 10 - frac10.3$$\frac{1}{f_{\text{eq}}} = \frac{1}{0.3} + 10 - \frac{1}{0.3}$$ frac1f_texteq = 10 mathrm~m^-1$$\frac{1}{f_{\text{eq}}} = 10 \mathrm{~m}^{-1}$$ P = 10 mathrm~D$$P = 10 \mathrm{~D}$$ ### Pattern Recognition Sees: Lenses separated by distance d$d$ where d = f_2$d = f_2$. Shortcut: Notice that d = f_2 = 10 mathrm~cm$d = f_2 = 10 \mathrm{~cm}$. When the separation distance between two thin lenses equals the focal length of the second lens, the equivalent power simplifies directly to P = P_2 = 1/f_2 = 10 mathrm~D$P = P_2 = 1/f_2 = 10 \mathrm{~D}$ since the terms 1/f_1$1/f_1$ and d/(f_1 f_2)$d/(f_1 f_2)$ cancel out. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### 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

### Related Formula 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 1. **Cutting along horizontal plane AB**: When a lens is cut along its principal axis, the radius of curvature of neither surface changes. Thus, the focal length of the split parts L_1$L_1$ and L_2$L_2$ remains exactly equal to the initial focal length: f_L_1 = f$f_{L_1} = f$ 2. **Cutting along vertical plane XY**: When a lens is cut perpendicular to the principal axis into two symmetric plano-convex lenses, one surface becomes flat (R_2 = infty$R_2 = \infty$). According to Lens Maker's Formula, the focal length of parts L_3$L_3$ and L_4$L_4$ doubles: f_L_3 = 2f$f_{L_3} = 2f$ 3. **Ratio Determination**: fracf_L_1f_L_3 = fracf2f = frac12$$\frac{f_{L_1}}{f_{L_3}} = \frac{f}{2f} = \frac{1}{2}$$ Hence, the ratio is 1:2$1:2$. ### Pattern Recognition Shortcut rule for lens cutting: - Horizontal cut (along axis) rightarrow$\rightarrow$ Focal length stays f$f$. - Vertical cut (perp to axis) rightarrow$\rightarrow$ Focal length doubles to 2f$2f$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments