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 n_1 sin i = n_2 sin r$$n_1 \sin i = n_2 \sin r$$ ### Core Logic

By Snell\'s law matching standard boundary normal configurations : n_1 sin(90^circ - theta_1) = n_2 sin theta_3 implies n_1 cos theta_1 = n_2 sin theta_3$$n_1 \sin(90^{\circ} - \theta_1) = n_2 \sin \theta_3 \implies n_1 \cos \theta_1 = n_2 \sin \theta_3$$ [cite: 711, 712] Substituting the angle identity macro given [cite: 2, 713]: n_1 left(fracn_22n_1right) = n_2 sin theta_3 implies sin theta_3 = frac12 implies theta_3 = 30^circ$$n_1 \left(\frac{n_2}{2n_1}\right) = n_2 \sin \theta_3 \implies \sin \theta_3 = \frac{1}{2} \implies \theta_3 = 30^{\circ}$$ ### Step 1: Geometrical Thickness Resolution From the block triangles geometry : tan 30^circ = fracd/2t implies frac1sqrt3 = fracd2t$$\tan 30^{\circ} = \frac{d/2}{t} \implies \frac{1}{\sqrt{3}} = \frac{d}{2t}$$ t = fracdsqrt32 = frac4sqrt3 cdot sqrt32 = 6text cm$$t = \frac{d\sqrt{3}}{2} = \frac{4\sqrt{3} \cdot \sqrt{3}}{2} = 6\text{ cm}$$ ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula Lens magnification formula: m = fracvu$$m = \frac{v}{u}$$ Thin lens equation: frac1v - frac1u = frac1f$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ ### Core Logic For a virtual image formed by a convex lens, both the object and the image lie on the same side of the lens. The image distance is three times the object distance: v = 3u$v = 3u$ The distance between the object and its virtual image is given as 20mathrm~cm$20\mathrm{~cm}$: v - u = 20mathrm~cm implies 3u - u = 20mathrm~cm$$v - u = 20\mathrm{~cm} \implies 3u - u = 20\mathrm{~cm}$$ 2u = 20mathrm~cm implies u = 10mathrm~cm$$2u = 20\mathrm{~cm} \implies u = 10\mathrm{~cm}$$ Applying Cartesian sign conventions: object distance u = -10mathrm~cm$u = -10\mathrm{~cm}$ and image distance v = -30mathrm~cm$v = -30\mathrm{~cm}$. ### Step 1: Calculate Focal Length Substitute these values into the lens formula: frac1-30 - frac1-10 = frac1f$$\frac{1}{-30} - \frac{1}{-10} = \frac{1}{f}$$ -frac130 + frac110 = frac1f$$-\frac{1}{30} + \frac{1}{10} = \frac{1}{f}$$ frac-1 + 330 = frac230 = frac115 = frac1f implies f = 15mathrm~cm$$\frac{-1 + 3}{30} = \frac{2}{30} = \frac{1}{15} = \frac{1}{f} \implies f = 15\mathrm{~cm}$$ ### Pattern Recognition A virtual image from a convex lens means the object is placed inside the focal point (u < f$u < f$). This serves as a quick sanity check for your final value (10mathrm~cm < 15mathrm~cm$10\mathrm{~cm} < 15\mathrm{~cm}$). ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula Magnification formula for spherical mirrors: m = -fracvu$$m = -\frac{v}{u}$$ Mirror formula: frac1f = frac1v + frac1u$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ where: * u$u$ is the object distance * v$v$ is the image distance * f$f$ is the focal length ### Core Logic Since the image is magnified (m = 2$m = 2$) and virtual, the mirror must be concave (f < 0$f < 0$). Let us use standard coordinate geometry signs: object is on the left (u$u$ is negative, say -u_0$-u_0$), virtual image is on the right (v$v$ is positive, say +v_0$+v_0$). Given magnification: m = 2 = -fracvu implies v = -2u$$m = 2 = -\frac{v}{u} \implies v = -2u$$ In terms of magnitudes: v_0 = 2u_0$v_0 = 2u_0$ ### Step 1: Use Distance Condition The distance between the object and virtual image is 15text cm$15\text{ cm}$. Since the object is in front of the mirror and the virtual image is behind it: textDistance = u_0 + v_0 = 15text cm$$\text{Distance} = u_0 + v_0 = 15\text{ cm}$$ Substitute v_0 = 2u_0$v_0 = 2u_0$: u_0 + 2u_0 = 15 implies 3u_0 = 15 implies u_0 = 5text cm$$u_0 + 2u_0 = 15 \implies 3u_0 = 15 \implies u_0 = 5\text{ cm}$$ Thus: * u_0 = 5text cm implies u = -5text cm$u_0 = 5\text{ cm} \implies u = -5\text{ cm}$ * v_0 = 10text cm implies v = +10text cm$v_0 = 10\text{ cm} \implies v = +10\text{ cm}$

### Step 2: Calculate Focal Length Using the mirror formula: frac1f = frac1v + frac1u$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ frac1f = frac110 + frac1-5$$\frac{1}{f} = \frac{1}{10} + \frac{1}{-5}$$ frac1f = frac1 - 210 = -frac110 implies f = -10text cm$$\frac{1}{f} = \frac{1 - 2}{10} = -\frac{1}{10} \implies f = -10\text{ cm}$$ Thus, the focal length is -10text cm$-10\text{ cm}$. ### Pattern Recognition Virtual and magnified image → always concave mirror. Distance D$D$ between object and virtual image is given by D = |u| + v$D = |u| + v$. Since v = m|u|$v = m|u|$, we can simplify to |u| = fracDm+1$|u| = \frac{D}{m+1}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula mu = fracsinleft(fracA + delta_textmin2right)sinleft(fracA2right)$$\mu = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$ ### Core Logic Given mu = cotleft(fracA2right) = fraccosleft(fracA2right)sinleft(fracA2right)$\mu = \cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. Equating this to the prism formula: fraccosleft(fracA2right)sinleft(fracA2right) = fracsinleft(fracA + delta_textmin2right)sinleft(fracA2right)$$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$ cosleft(fracA2right) = sinleft(fracA + delta_textmin2right)$$\cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + \delta_{\text{min}}}{2}\right)$$ ### Step 1: Trigonometric Substitution Convert the cosine term to sine: sinleft(fracpi2 - fracA2right) = sinleft(fracA + delta_textmin2right)$$\sin\left(\frac{\pi}{2} - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_{\text{min}}}{2}\right)$$ fracpi2 - fracA2 = fracA + delta_textmin2$$\frac{\pi}{2} - \frac{A}{2} = \frac{A + \delta_{\text{min}}}{2}$$ pi - A = A + delta_textmin$$\pi - A = A + \delta_{\text{min}}$$ delta_textmin = pi - 2A$$\delta_{\text{min}} = \pi - 2A$$ ### Pattern Recognition When mu$\mu$ equals a cotangent function of half-angle, the sine dynamic simplifications lead directly to linear functions involving complements of angle A$A$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula h_textapparent = sum frach_imu_i$$h_{\text{apparent}} = \sum \frac{h_i}{\mu_i}$$ ### Core Logic Sum the contributions of both shifting mediums: h_textapparent = frach_1mu_1 + frach_2mu_2$$h_{\text{apparent}} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$$ Given h_1 = h_2 = 6text cm$h_1 = h_2 = 6\text{ cm}$, mu_1 = frac32$\mu_1 = \frac{3}{2}$, mu_2 = frac85$\mu_2 = \frac{8}{5}$: ### Step 1: Compute fraction value h_textapparent = frac63/2 + frac68/5 = 4 + frac308 = 4 + frac154$$h_{\text{apparent}} = \frac{6}{3/2} + \frac{6}{8/5} = 4 + \frac{30}{8} = 4 + \frac{15}{4}$$ h_textapparent = frac16 + 154 = frac314text cm$$h_{\text{apparent}} = \frac{16 + 15}{4} = \frac{31}{4}\text{ cm}$$ ### Step 2: Match to target format Comparing with fracalpha4$\frac{\alpha}{4}$ directly yields: alpha = 31$\alpha = 31$ ### Pattern Recognition Apparent depth across multi-layered planar mediums expands additively via separate individual medium thickness-to-refractive-index ratios. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments