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Geometrical Optics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsGeometrical Optics5 Marks
Question
A particle is moving at a constant speed V from a large distance towards a concave mirror of radius R along its principal axis. Find the speed of the image formed by the mirror as a function of the distance x of the particle from the mirror.
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Answer

Given that, u = distance of the object = -x f = focal length $=-\frac{\text{R}}{2}$ and, V = velocity of object $=\frac{\text{dx}}{\text{dt}}$ From mirror equation, $\frac{1}{-\text{x}}+\frac{1}{\text{v}}=-\frac{2}{\text{R}}$$\frac{1}{\text{v}}=-\frac{2}{\text{R}}+\frac{1}{\text{x}}=\frac{\text{R}-2\text{x}}{\text{R}\text{x}}\Rightarrow\text{v}=\frac{\text{Rx}}{\text{R}-2\text{x}}=$ Image distance
So, velocity of the image is given by,$\text{V}_1=\frac{\text{dv}}{\text{dt}}=\frac{\Big[\frac{\text{d}}{\text{dt}}(\text{xR})(\text{R}-2\text{x})\Big]-\Big[\frac{\text{d}}{\text{dt}}(\text{R}-2\text{x})(\text{xR})\Big]}{(\text{R}-2\text{x})^2}$
$=\frac{\text{R}\Big[\frac{\text{dx}}{\text{dt}}(\text{R}-2\text{x})\Big]-\Big[-2\frac{\text{d}}{\text{dt}}\text{x}\Big]}{(\text{R}-2\text{x})^2}=\frac{\text{R}[\text{v}(\text{R}-2\text{x})+2\text{vx}0}{(\text{R}-2\text{x})^2}$
$=\frac{\text{VR}^2}{(2\text{x}-\text{R})^2}=\frac{\text{R}\big[\text{VR}-2\text{xV}\big)+2\text{xV}}{(\text{R}-2\text{x})^2}$

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