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 textMagnification m = fracD_textoutD_textin = fracf_2f_1$$\text{Magnification } m = \frac{D_{\text{out}}}{D_{\text{in}}} = \frac{f_2}{f_1}$$ ### Core Logic For a beam expander using two convex lenses, the lenses are arranged such that their focal points coincide. This creates an afocal system where parallel input rays remain parallel upon output.

By similar triangles at the focal point: tan theta = fracD_1 / 2f_1 = fracD_2 / 2f_2$$\tan \theta = \frac{D_1 / 2}{f_1} = \frac{D_2 / 2}{f_2}$$ ### Step 1: Solving for f2 fracD_1f_1 = fracD_2f_2$$\frac{D_1}{f_1} = \frac{D_2}{f_2}$$ frac240 = frac14f_2$$\frac{2}{40} = \frac{14}{f_2}$$ f_2 = 14 times frac402 = 14 times 20 = 280text mm$$f_2 = 14 \times \frac{40}{2} = 14 \times 20 = 280\text{ mm}$$ ### Pattern Recognition A standard beam expander (Keplerian telescope design used in reverse) has f_2/f_1 = D_2/D_1$f_2/f_1 = D_2/D_1$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula m approx fracLf_0 times fracDf_e$$m \approx \frac{L}{f_0} \times \frac{D}{f_e}$$ ### Core Logic For a compound microscope in normal adjustment (image formed at infinity), the magnifying power is given by the standard approximation: m simeq fracl Df_0 f_e$$m \simeq \frac{l D}{f_{0} f_{e}}$$ where l$l$ is the tube length, D = 25text cm$D = 25\text{ cm}$ is the least distance of distinct vision. ### Step 1: Substitute Values Given values: l = 32text cm$l = 32\text{ cm}$ f_0 = 2text cm$f_0 = 2\text{ cm}$ f_e = 4text cm$f_e = 4\text{ cm}$ D = 25text cm$D = 25\text{ cm}$ (standard assumption when not given) m = frac322 times frac254$$m = \frac{32}{2} \times \frac{25}{4}$$ m = 16 times frac254 = 4 times 25 = 100$$m = 16 \times \frac{25}{4} = 4 \times 25 = 100$$

### Pattern Recognition Compound microscope formulas: Normal adjustment to$\to$ image at infty$\infty$, m = (L/f_0)(D/f_e)$m = (L/f_0)(D/f_e)$. Image at near point D$D$, m = (L/f_0)(1 + D/f_e)$m = (L/f_0)(1 + D/f_e)$. Default to standard approximation when given tube length. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### 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