A transparent block A having refractive index mu=1.25$\mu=1.25$ is surrounded by another medium of refractive index mu=1.0$\mu=1.0$ as shown in figure. A light ray is incident on the flat face of the block with incident angle theta$\theta$ as shown in figure. What is the maximum value of theta$\theta$ for which light suffers total internal reflection at the top surface of the block?

[cite: 169, 170, 171]

### 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