Solution to Question 19
1. Analyzing the Transverse Rod CD
For the image of the frame to be “exactly the same size” as the object, the transverse magnification $m_T$ for rod CD must have a magnitude of 1 ($|m_T| = 1$). Since the image is formed by a single lens, a real image with unit magnification implies the object is placed at $2f$. Therefore, the cross intersection (position of CD) is at: $$ u_{cross} = 2f = 2(10) = 20 \text{ cm} $$
2. Analyzing the Longitudinal Rod AB
From the grid diagram, point $B$ is 2 grid units to the left of the cross, and point $A$ is 1 grid unit to the right. Let the grid unit length be $x$. Using Newton’s form of the lens equation ($x_1 x_2 = f^2$), where distances are measured from the focus $F$ (located $f$ in front of the lens):
- Focus $F$ is at $f = 10$ cm from the lens.
- The Cross is at $2f = 20$ cm from the lens, which is distance $f$ from $F$.
- Point $A$ (closer to lens) is at distance $x$ from the cross towards the lens. Distance from $F$: $d_A = f – x$.
- Point $B$ (further from lens) is at distance $2x$ from the cross away from the lens. Distance from $F$: $d_B = f + 2x$.
3. Condition for Longitudinal Length Preservation
For the length of the image of AB to equal the length of object AB, the endpoints A and B must be geometric conjugates with respect to the condition $x_A x_B = f^2$. $$ d_A \cdot d_B = f^2 $$ Substituting the distances: $$ (f – x)(f + 2x) = f^2 $$ Expanding the equation: $$ f^2 + 2fx – fx – 2x^2 = f^2 $$ $$ fx – 2x^2 = 0 $$ Since $x \neq 0$: $$ f – 2x = 0 \implies x = \frac{f}{2} $$
4. Calculation of Length
Given $f = 10$ cm: $$ x = \frac{10}{2} = 5 \text{ cm} $$ The total length of rod AB is the sum of the segments from the cross: $$ L_{AB} = 2x + x = 3x $$ $$ L_{AB} = 3(5) = 15 \text{ cm} $$