What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D from 2.5 D ? ['D' stands for dioptre]

### 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