If $f$ is the focal length of a convex lens and an object is placed at a distance $v$ from the lens, then its image will be at a distance $u$ from the lens, where $f , v$, and $u$ are related by the lens equation $\frac { 1 } { f } = \frac { 1 } { v } + \frac { 1 } { u } .$
Find the rate of change of $v$ with respect to $u$.
A) $\frac { d v } { d u } = - \frac { f } { ( u - f ) ^ { 2 } }$
B) $\frac { d v } { d u } = - \frac { f ^ { 2 } } { u - f }$
C) $\frac { d v } { d u } = \frac { 2 f ^ { 2 } } { ( u - f ) ^ { 2 } }$
D)
$$\frac { d v } { d u } = \frac { f ^ { 2 } } { ( u - f ) ^ { 2 } }$$
E)
$$\frac { d v } { d u } = - \frac { f ^ { 2 } } { ( u - f ) ^ { 2 } }$$

Question

Quizplus · Accepted Answer

To find the rate of change of $v$ with respect to $u$, we differentiate the lens equation with respect to $u$. The lens equation is $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. Rearranging gives $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$. Differentiating both sides with respect to $u$, we get $-\frac{1}{v^2} \frac{dv}{du} = \frac{1}{u^2}$. Solving for $\frac{dv}{du}$ gives $\frac{dv}{du} = -\frac{v^2}{u^2}$. Substituting $v = \frac{uf}{u-f}$ from the original equation and simplifying, we get $\frac{dv}{du} = -\frac{f^2}{(u-f)^2}$.

If fff Is the Focal Length of a Convex Lens and an and an Object

If $f$ Is the Focal Length of a Convex Lens and an and an Object