A rod is placed along the principal axis as shown. Find the length of the image. The focal length of the mirror is \(f = 10\,\text{cm}\).
Show Hint
If an object is placed along the principal axis, find the image of both ends separately using the mirror formula.
The difference of their image positions gives the \textbf{image length}.
Concept:
When an object is placed along the principal axis of a spherical mirror, the image of each end of the object forms at different positions.
The image positions can be found using the mirror formula:
\[
\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
\]
where
\(f\) = focal length,
\(u\) = object distance,
\(v\) = image distance.
The length of the image is the difference between the image positions of the two ends.
Step 1: Identify object distances from the figure.}
Point \(A\) is \(20\,\text{cm}\) from the mirror.
\[
u_A = -20\,\text{cm}
\]
Point \(B\) is \(10\,\text{cm}\) further left.
\[
u_B = -30\,\text{cm}
\]
Step 2: Find the image position of point \(A\).}
\[
\frac{1}{f}=\frac{1}{v_A}+\frac{1}{u_A}
\]
\[
\frac{1}{10}=\frac{1}{v_A}-\frac{1}{20}
\]
\[
\frac{1}{v_A}=\frac{1}{10}+\frac{1}{20}
\]
\[
\frac{1}{v_A}=\frac{3}{20}
\]
\[
v_A=\frac{20}{3}\,\text{cm}
\]
Step 3: Find the image position of point \(B\).}
\[
\frac{1}{10}=\frac{1}{v_B}-\frac{1}{30}
\]
\[
\frac{1}{v_B}=\frac{1}{10}+\frac{1}{30}
\]
\[
\frac{1}{v_B}=\frac{4}{30}
\]
\[
v_B=7.5\,\text{cm}
\]
Step 4: Calculate the image length.}
\[
\text{Image length} = |v_B - v_A|
\]
\[
= \left|7.5 - \frac{20}{3}\right|
\]
\[
= \frac{45 - 40}{6}
\]
\[
= \frac{5}{6}\times12
\]
\[
= 10\,\text{cm}
\]
