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 The magnification (m$m$) of a spherical mirror is given by: m = fracff - u$$m = \frac{f}{f - u}$$ where, f$f$ = focal length (R/2$R/2$) u$u$ = object distance ### Core Logic Given radius of curvature, R = 30 mathrm~cm$R = 30 \mathrm{~cm}$. For a convex mirror: f = +fracR2 = +15 mathrm~cm$$f = +\frac{R}{2} = +15 \mathrm{~cm}$$ Since a convex mirror always forms a virtual, erect, and diminished image of a real object, the magnification m$m$ must be positive: m = +frac12$$m = +\frac{1}{2}$$

### Step 1: Solve for Object Distance Using the magnification formula: +frac12 = frac1515 - u$$+\frac{1}{2} = \frac{15}{15 - u}$$ 15 - u = 30 implies u = -15 mathrm~cm$$15 - u = 30 \implies u = -15 \mathrm{~cm}$$ Thus, the object distance is -15 mathrm~cm$-15 \mathrm{~cm}$. ### Pattern Recognition Remember: Convex mirrors produce *only* virtual images for real objects, which means m$m$ is always positive and less than 1$1$. If the question mentioned a concave mirror with a diminished image of half size, the image would be real, and m$m$ would be negative. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula From the Lens Maker's Formula: frac1f = left( fracmu_textlensmu_textmedium - 1 right) left( frac1R_1 - frac1R_2 right)$$\frac{1}{f} = \left( \frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ Taking the ratio of focal length in liquid medium (f_m$f_m$) to focal length in air (f_a$f_a$): fracf_mf_a = frac(mu_1 - 1) mu_mmu_1 - mu_m$$\frac{f_m}{f_a} = \frac{(\mu_1 - 1) \mu_m}{\mu_1 - \mu_m}$$ where, mu_1$\mu_1$ = refractive index of the lens material = 1.5 mu_m$\mu_m$ = refractive index of the liquid medium = 1.6 f_a$f_a$ = focal length in air = 20 mathrm~cm$20 \mathrm{~cm}$ ### Core Logic Substitute the parameters into the relative ratio template equation: fracf_m20 = frac(1.5 - 1) times 1.61.5 - 1.6$$\frac{f_m}{20} = \frac{(1.5 - 1) \times 1.6}{1.5 - 1.6}$$ ### Step 1: Simplify and Compute $fracf_m20 = frac0.5 times 1.6-0.1$$$\frac{f_m}{20} = \frac{0.5 \times 1.6}{-0.1}$$ fracf_m20 = frac0.8-0.1 = -8$$\frac{f_m}{20} = \frac{0.8}{-0.1} = -8$$ f_m = -8 times 20 = -160 mathrm~cm$$f_m = -8 \times 20 = -160 \mathrm{~cm}$$ Therefore, the focal length in the liquid is -160 mathrm~cm$-160 \mathrm{~cm}$. ### Pattern Recognition Notice that since the surrounding liquid medium has a higher refractive index than the lens material itself (mu_m gt mu_1$\mu_m \gt \mu_1$), the sign of the focal length flips from positive to negative. The convex lens behaves as a diverging lens inside this specific liquid. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula textNewton's Lens Formula: x_1 x_2 = f^2$$\text{Newton's Lens Formula: } x_1 x_2 = f^2$$ where x_1$x_1$ and x_2$x_2$ are object and image distances from the focal points. ### Core Logic Since distances x$x$ and y$y$ are measured from the focal point (not the optical center), we use Newton's formula: x y = f^2$x y = f^2$. From the graph, a prominent point on the curve is (20, 20)$(20, 20)$. ### Step 1: Calculate Focal Length 20 times 20 = f^2$$20 \times 20 = f^2$$ f^2 = 400 implies f = 20 mathrm~cm$$f^2 = 400 \implies f = 20 \mathrm{~cm}$$ Alternatively using standard lens formula: Object distance from optical center u = -(f + x)$u = -(f + x)$ Image distance v = +(f + y)$v = +(f + y)$ frac1v - frac1u = frac1f$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$ frac1f+y - frac1-(f+x) = frac1f$$\frac{1}{f+y} - \frac{1}{-(f+x)} = \frac{1}{f}$$ If x = y = 20 mathrm~cm$x = y = 20 \mathrm{~cm}$: frac1f+20 + frac1f+20 = frac1f$$\frac{1}{f+20} + \frac{1}{f+20} = \frac{1}{f}$$ frac2f+20 = frac1f$$\frac{2}{f+20} = \frac{1}{f}$$ 2f = f + 20 implies f = 20 mathrm~cm$$2f = f + 20 \implies f = 20 \mathrm{~cm}$$ ### Pattern Recognition Whenever distances are specified relative to the focal point, Newton's formula (xy = f^2$xy = f^2$) instantly solves the problem. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula m = fracvu$$m = \frac{v}{u}$$ frac1f = frac1v - frac1u$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ ### Core Logic For a real image produced by a convex lens, the magnification m$m$ is negative. The distance between the object and the real image is the absolute sum of their distances from the lens: |v| + |u| = v - u = 45 mathrm~cm$|v| + |u| = v - u = 45 \mathrm{~cm}$ (since u$u$ is negative and v$v$ is positive). ### Step 1: Set Up Magnification and Distance fracvu = -2$$\frac{v}{u} = -2$$ v = -2u quad dots (i)$$v = -2u \quad \dots (i)$$ Distance between object and image: v - u = 45 quad dots (ii)$$v - u = 45 \quad \dots (ii)$$ ### Step 2: Solve for Object and Image Distances Substitute (i) into (ii): (-2u) - u = 45$(-2u) - u = 45$ -3u = 45 Rightarrow u = -15 mathrm~cm$$-3u = 45 \Rightarrow u = -15 \mathrm{~cm}$$ Then, v = -2(-15) = +30 mathrm~cm$v = -2(-15) = +30 \mathrm{~cm}$. ### Step 3: Calculate Focal Length frac1f = frac1v - frac1u$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ frac1f = frac130 - frac1-15$$\frac{1}{f} = \frac{1}{30} - \frac{1}{-15}$$ frac1f = frac130 + frac230 = frac330 = frac110$$\frac{1}{f} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10}$$ f = +10 mathrm~cm$$f = +10 \mathrm{~cm}$$ ### Pattern Recognition For a real image, total object-to-image distance D = v + |u|$D = v + |u|$. Applying sign convention naturally resolves this to D = v - u$D = v - u$. Two conditions (magnification + total distance) reliably solve for both u, v$u, v$ before hitting the lens equation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula fracmu_2v - fracmu_1u = fracmu_2 - mu_1R$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$ ### Core Logic For a convex refracting surface, the radius of curvature R$R$ is positive if the center of curvature lies in the denser medium. mu_1 = 1 text (air) $ \mu_1 = 1 \text{ (air)} $ mu_2 = 1.5 text (glass/medium) $ \mu_2 = 1.5 \text{ (glass/medium)} $ u = -100 text cm $ u = -100 \text{ cm} $ R = +20 text cm $ R = +20 \text{ cm} $

### Step 1: Calculate Image Position frac1.5v - frac1-100 = frac1.5 - 120$$\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}$$ frac1.5v + frac1100 = frac0.520$$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}$$ frac1.5v = frac140 - frac1100$$\frac{1.5}{v} = \frac{1}{40} - \frac{1}{100}$$ frac1.5v = frac5 - 2200 = frac3200$$\frac{1.5}{v} = \frac{5 - 2}{200} = \frac{3}{200}$$ v = frac1.5 times 2003 = 100 text cm$$v = \frac{1.5 \times 200}{3} = 100 \text{ cm}$$ ### Step 2: Distance from Object The image is formed at v = +100 text cm$v = +100 \text{ cm}$ from the pole, which is on the other side of the surface. The object is at |u| = 100 text cm$|u| = 100 \text{ cm}$ from the pole. Distance between object and image = |u| + v = 100 + 100 = 200 text cm$= |u| + v = 100 + 100 = 200 \text{ cm}$. ### Pattern Recognition Always read the final line of optics questions carefully. The question asks for distance "from the object" not "from the surface/pole". This is a classic trap where students answer 100 instead of 200. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments