Question

An object is approaching a thin convex lens of focal length $0.3m$ with a speed of $0.01\,m/s$.The magnitude of the rate of change of lateral magnification of image when the object is at a distance of $0.4{\text{ }}m$ from the lens is
A) $0.3$
B) $0.6$
C) $0.15$
D) $- 0.3$

Answer 1

In this solution, we will first use the lens formula relation to determine the image position when the object is a certain distance away from the lens. Then we will use the formula for magnification and differentiate it with respect to time to find the magnitude of its rate of change.
Formula used:
Lens formula:$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ where $v$ is the image distance, $u$ is the object distance, and $f$ is the focal length of the lens
Magnification of a lens: $m = - \dfrac{v}{u}$

Complete step by step answer:
We’ve been given that an object is approaching a thin convex lens of focal length $0.3m$ with a speed of $0.01\,m/s$. When the object is $0.4{\text{ }}m$ away from the lens, the image position will be determined from the lens formula as
$\Rightarrow$ $\dfrac{1}{v} - \dfrac{1}{{ - 0.4}} = \dfrac{1}{{0.3}}$
Which gives us the image distance as
$\Rightarrow$ $v = 1.2\,m$
Since we want to find the image velocity too, we will differentiate the lens formula with respect to time as
$\Rightarrow$ $- \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} + \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}} = 0$ (The right-hand side will be zero since the focal length of the lens will be constant with respect to time)
Which can be simplified to
$\dfrac{{dv}}{{dt}} = {\left( {\dfrac{v}{u}} \right)^2}\dfrac{{du}}{{dt}}$
Substituting $v = 1.2\,m$ and $u = - 0.4\,m$ and $\dfrac{{du}}{{dt}} = 0.01\,m/s$, we get
$\dfrac{{dv}}{{dt}} = 0.09\,m/s$
So, the image is moving at a velocity of $0.09\,m/s.$
So now differentiating the formula for magnification $m = - \dfrac{v}{u}$, we get
$\Rightarrow$ $\dfrac{{dm}}{{dt}} = \dfrac{{u\dfrac{{dv}}{{dt}} - v\dfrac{{du}}{{dt}}}}{{{u^2}}}$
Substituting $\dfrac{{dv}}{{dt}} = 0.09\,m/s$, $v = 1.2\,m$ and $u = - 0.4\,m$ and $\Rightarrow$ $\dfrac{{du}}{{dt}} = 0.01\,m/s$, we get
$\Rightarrow$ $\dfrac{{dm}}{{dt}} = - \dfrac{{0.048}}{{0.16}}$
$\Rightarrow$ $\dfrac{{dm}}{{dt}} = - 0.3$

**Hence the rate of change of lateral magnification of the image will be $\dfrac{{dm}}{{dt}} = - 0.3$ which corresponds to option (D). **

Note: While using the lens formula and the formula for magnification, we need to take into account the sign convention of the lens in question and the image and object distance. While differentiating the magnification formula, we must differentiate both image and object position as both will be changing with respect to time.