### Related Formula sin mathrmc = frac1mu$\sin \mathrm{c} = \frac{1}{\mu}$ ### Core Logic To see the coin from edge point mathrmE$\mathrm{E}$ at grazing emergency, the light ray travelling from mathrmO$mathrm{O}$ to mathrmE$mathrm{E}$ must strike the flat upper boundary at the critical angle.

Given the hemispherical layout geometry, the ray's angle of incidence at the center of the surface plane satisfies: theta = mathrmc = 45^circ$\theta = \mathrm{c} = 45^{\circ}$ Substituting this value into the critical value expression: mu = frac1sin 45^circ = sqrt2$\mu = \frac{1}{\sin 45^{\circ}} = \sqrt{2}$ ### Step 1: Final Conclusion The minimum refractive index required is sqrt2$\sqrt{2}$, matching option (3). ### Pattern Recognition Grazing boundary ray vision configurations dictate evaluating the specific systemic geometric configuration to compute the critical boundary angle. Here, the radius profile fixes theta = 45^circ$\theta = 45^{\circ}$ deterministically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula delta = (mu - 1)mathrmA$\delta = (\mu - 1)\mathrm{A}$ ### Core Logic To achieve dispersion without deviation, the net deviation produced by the prism combination must be zero: delta_textnet = 0 implies (mu_1 - 1)mathrmA_1 - (mu_2 - 1)mathrmA_2 = 0$\delta_{\text{net}} = 0 \implies (\mu_1 - 1)\mathrm{A}_1 - (\mu_2 - 1)\mathrm{A}_2 = 0$ Substituting the given parameters into the equation: (1.54 - 1) cdot 4^circ - (1.72 - 1)mathrmA_2 = 0$(1.54 - 1) \cdot 4^{\circ} - (1.72 - 1)\mathrm{A}_2 = 0$ 0.54 cdot 4 = 0.72 cdot mathrmA_2$0.54 \cdot 4 = 0.72 \cdot \mathrm{A}_2$ mathrmA_2 = frac2.160.72 = 3^circ$\mathrm{A}_2 = \frac{2.16}{0.72} = 3^{\circ}$ ### Step 1: Final Angle Value The required angle for the second thin prism is 3^circ$3^{\circ}$, which matches option (2). ### Pattern Recognition For zero deviation conditions using thin components, balance the deviation equations directly: (mu-1)mathrmA = (mu'-1)mathrmA'$(\mu-1)\mathrm{A} = (\mu'-1)\mathrm{A}'$. ### 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)$\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ where, f$f$ = focal length of the lens, mu$\mu$ = refractive index of the material, R_1, R_2$R_1, R_2$ = radii of curvature with standard Cartesian sign convention. ### Core Logic For a thin bi-convex lens, using standard coordinate conventions: - R_1 = +10mathrm~cm$R_1 = +10\mathrm{~cm}$ (positive since first surface centers to the right of light trajectory), - R_2 = -15mathrm~cm$R_2 = -15\mathrm{~cm}$ (negative since second surface centers to the left), - Focal length f = +12mathrm~cm$f = +12\mathrm{~cm}$. ### Step 1: Substituting in the Equation Substitute the values into Lens Maker's formula: frac112 = (mu - 1) left(frac110 - frac1-15right)$\frac{1}{12} = (\mu - 1) \left(\frac{1}{10} - \frac{1}{-15}\right)$ frac112 = (mu - 1) left(frac110 + frac115right)$\frac{1}{12} = (\mu - 1) \left(\frac{1}{10} + \frac{1}{15}\right)$ frac112 = (mu - 1) left(frac3 + 230right)$\frac{1}{12} = (\mu - 1) \left(\frac{3 + 2}{30}\right)$ frac112 = (mu - 1) left(frac530right) = (mu - 1) left(frac16right)$\frac{1}{12} = (\mu - 1) \left(\frac{5}{30}\right) = (\mu - 1) \left(\frac{1}{6}\right)$ mu - 1 = frac612 = 0.5 implies mu = 1.5$\mu - 1 = \frac{6}{12} = 0.5 \implies \mu = 1.5$ ### Pattern Recognition Convex lenses always have opposite signs for R_1$R_1$ and R_2$R_2$. The term left(frac1R_1 - frac1R_2right)$\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ is additive: left(frac1|R_1| + frac1|R_2|right)$\left(\frac{1}{|R_1|} + \frac{1}{|R_2|}\right)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula For a prism of angle A$A$: - Deviation: delta = i + e - A$\delta = i + e - A$ - Minimum deviation delta_textmin$\delta_{\text{min}}$ occurs when: i = e quad textand quad r_1 = r_2 = fracA2$i = e \quad \text{and} \quad r_1 = r_2 = \frac{A}{2}$ - Under minimum deviation, the ray inside an equilateral/isosceles prism travels symmetrically, making it parallel to the prism base. ### Core Logic Let's check the validity of each statement: - **Statement (A)**: The refracted ray inside the prism becomes parallel to the base at minimum deviation. (True for symmetric prisms) - **Statement (B)**: Larger angle prisms provide smaller minimum deviation. By minimum deviation equation: mu = fracsinleft(fracA+delta_textmin2right)sinleft(fracA2right)$\mu = \frac{\sin\left(\frac{A+\delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ As A$A$ increases, delta_textmin$\delta_{\text{min}}$ generally increases, not decreases. (False) - **Statement (C)**: Angle of incidence i$i$ and angle of emergence e$e$ become equal (i = e$i = e$) during the minimum deviation state. (True) - **Statement (D)**: The delta-i$\delta-i$ curve is asymmetric and parabolic-like; for any deviation delta > delta_textmin$\delta > \delta_{\text{min}}$, there are always exactly two different incident angles (i$i$ and e$e$) that yield the same deviation, except at the minimum deviation point (which has a single unique value). (True) - **Statement (E)**: Angle of refraction r = A/2$r = A/2$, which is half of the prism angle, not double. (False) ### Step 1: Conclusion Since statements A, C, and D are true, the correct option is (1). ### Pattern Recognition Review the classic parabolic shape of the deviation vs. angle of incidence (delta-i$\delta-i$) graph. Notice that any horizontal line above the minimum point intersects twice (representing i$i$ and e$e$ for that deviation). Minimum deviation is the unique local minimum, where i = e$i = e$ and r_1 = r_2 = A/2$r_1 = r_2 = A/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments: Refraction through Prism

### Related Formula Mirror Formula: frac1v + frac1u = frac1f$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ ### Core Logic For the convex mirror, u = -30text cm$u = -30\text{ cm}$ and f = +30text cm$f = +30\text{ cm}$. frac1v - frac130 = frac130 implies frac1v = frac230 implies v = +15text cm$\frac{1}{v} - \frac{1}{30} = \frac{1}{30} \implies \frac{1}{v} = \frac{2}{30} \implies v = +15\text{ cm}$ So, the convex mirror forms a virtual image 15 cm behind its surface. ### Step 1: Align Plane Mirror Image The total distance from the object to the image location is 30 + 15 = 45text cm$30 + 15 = 45\text{ cm}$. For a plane mirror to create an image at this same exact location, it must be placed precisely midway between the object and the image. Distance from object to plane mirror: d = frac452 = 22.5text cm$d = \frac{45}{2} = 22.5\text{ cm}$ Therefore, the clearance distance between the convex mirror and the plane mirror surface is: textDistance = 30 - 22.5 = 7.5text cm$\text{Distance} = 30 - 22.5 = 7.5\text{ cm}$ ### Pattern Recognition Coinciding images imply identical coordinate endpoints. Calculate the convex position explicitly, find the total path length from the real source object, and slice it in half for the plane mirror location. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments