### 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 mu = fracsinleft(fracA + delta_m2right)sinleft(fracA2right)$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ ### Core Logic Given that the refractive index mu = cotleft(fracA2right)$\mu = \cot\left(\frac{A}{2}\right)$. Substituting this into the prism formula: cotleft(fracA2right) = fracsinleft(fracA + delta_m2right)sinleft(fracA2right)$\cot\left(\frac{A}{2}\right) = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ fraccosleft(fracA2right)sinleft(fracA2right) = fracsinleft(fracA + delta_m2right)sinleft(fracA2right)$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ Equating the numerators: cosleft(fracA2right) = sinleft(fracA + delta_m2right)$\cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right)$ We can rewrite cosine in terms of sine: sinleft(fracpi2 - fracA2right) = sinleft(fracA + delta_m2right)$\sin\left(\frac{\pi}{2} - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right)$ ### Step 2: Solve for Deviation Comparing the angles inside the sine functions: fracpi2 - fracA2 = fracA2 + fracdelta_m2$\frac{\pi}{2} - \frac{A}{2} = \frac{A}{2} + \frac{\delta_m}{2}$ Multiply the entire equation by 2: pi - A = A + delta_m$\pi - A = A + \delta_m$ delta_m = pi - 2A$\delta_m = \pi - 2A$ Converting radians to degrees: delta_m = 180^circ - 2A$\delta_m = 180^{\circ} - 2A$ ### Pattern Recognition Whenever refractive index mu = cot(A/2)$\mu = \cot(A/2)$, the relation sin(90^circ - A/2)$\sin(90^\circ - A/2)$ strictly matches the prism sine equation, meaning minimum deviation delta_m$\delta_m$ is always 180^circ - 2A$180^\circ - 2A$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics And Optical Instruments