Question

An object is approaching a thin convex lens of focal length $0.3m$ with a speed of $0.01m/s$.The magnitude of the rate of change of position of image when the object is at a distance of $0.4m$ from the lens is

Answer 1

Step 1: Given data

1. The focal length of the convex lens $0.3m$
2. The object is approaching at a speed of $0.01m/s$
3. The object is at a distance of $0.4m$

Step 2: Formula used

1. Lens Formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$where $v$ is the image distance, $u$ is the object distance and $f$ is the focal length of the lens
2. Magnification of a lens: $m=-\frac{v}{u}$

Step 3: Find the image distance

Let the image distance be $v$

$u=0.4m$

$\frac{du}{dt}=0.01m/s$

$f=0.3m$

$$\begin{gathered} \therefore \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\ \Rightarrow v=\frac{fu}{u+f} \\ =\frac{\left(0.3\right)\left(0.4\right)}{0.3+0.4} \\ =1.2m \end{gathered}$$

Step 4: Find the magnitude of the rate of change of position of the image

Since we want to find the image velocity too, we will differentiate the lens formula with respect to time as

$$\begin{gathered} \Rightarrow \frac{d}{dt}\left(\frac{1}{v}\right)=\frac{d}{dt}\left(\frac{1}{f}\right)+\frac{d}{dt}\left(\frac{1}{u}\right) \\ \Rightarrow -\frac{dv}{dt}\left(\frac{1}{v^2}\right)=-\frac{du}{dt}\left(\frac{1}{u^2}\right)\left[\because f\mathrm{constant}\right] \\ \Rightarrow \frac{dv}{dt}={\left(\frac{v}{u}\right)}^2\frac{du}{dt} \\ \Rightarrow \frac{dv}{dt}={\left(\frac{1.2}{0.4}\right)}^2\left(0.01\right) \\ =0.09m/s \end{gathered}$$

Hence the magnitude of the rate of change in the position of the image is $0.09$