Questions: If (f) is the focal length of a convex lens and an object is placed at a distance (p) from the lens, then its image will be at a distance (q) from the lens, where (f, p), and (q) are related by the lens equation [ frac1f=frac1p+frac1q ] What is the rate of change of (p) with respect to (q) if (q=3) and (f=8) ? (Make sure you have the correct sign for the rate.)

Solution Steps

The lens equation is given by:

\[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \]

We need to find the rate of change of \( p \) with respect to \( q \), denoted as \( \frac{dp}{dq} \), when \( q = 3 \) and \( f = 8 \).

Differentiate both sides of the lens equation with respect to \( q \):

\[ 0 = -\frac{1}{p^2} \frac{dp}{dq} + \frac{1}{q^2} \]

Rearrange the differentiated equation to solve for \( \frac{dp}{dq} \):

\[ \frac{1}{p^2} \frac{dp}{dq} = \frac{1}{q^2} \]

\[ \frac{dp}{dq} = \frac{p^2}{q^2} \]

Substitute \( f = 8 \) and \( q = 3 \) into the original lens equation to find \( p \):

\[ \frac{1}{8} = \frac{1}{p} + \frac{1}{3} \]

\[ \frac{1}{p} = \frac{1}{8} - \frac{1}{3} = \frac{3 - 8}{24} = -\frac{5}{24} \]

\[ p = -\frac{24}{5} \]

Substitute \( p = -\frac{24}{5} \) and \( q = 3 \) into the expression for \( \frac{dp}{dq} \):

\[ \frac{dp}{dq} = \left(-\frac{24}{5}\right)^2 \div 3^2 = \frac{576}{25} \div 9 = \frac{576}{225} \]

Simplify:

\[ \frac{dp}{dq} = \frac{64}{25} \]

Final Answer