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 $\frac{1}{f}=\frac{1}{v}+\frac{8}{u}$ . Find the rate of change of v with respect to u.
A) $\frac{d v}{d u}=\frac{f^{2}}{(u-f)^{2}}$
B) $\frac{d v}{d u}=-\frac{8 f^{2}}{(u-8 f)^{2}}$
C) $\frac{d v}{d u}=\frac{2 f^{2}}{(u-f)^{2}}$
D) $\frac{d v}{d u}=-\frac{f}{(u-f)^{2}}$
E) $\frac{d v}{d u}=-\frac{f^{2}}{u-f}$

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The Answer of 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 $\frac{1}{f}=\frac{1}{v}+\frac{8}{u}$ . Find the rate of change of v with respect to u.
A) $\frac{d v}{d u}=\frac{f^{2}}{(u-f)^{2}}$
B) $\frac{d v}{d u}=-\frac{8 f^{2}}{(u-8 f)^{2}}$
C) $\frac{d v}{d u}=\frac{2 f^{2}}{(u-f)^{2}}$
D) $\frac{d v}{d u}=-\frac{f}{(u-f)^{2}}$
E) $\frac{d v}{d u}=-\frac{f^{2}}{u-f}$

If F Is the Focal Length of a Convex Lens 1f=1v+8u\frac{1}{f}=\frac{1}{v}+\frac{8}{u}f1​=v1​+u8​

If F Is the Focal Length of a Convex Lens $\frac{1}{f}=\frac{1}{v}+\frac{8}{u}$