Solution
For a pinhole camera, the magnification $m$ is defined by the geometry of similar triangles formed by the object and the image relative to the pinhole: $$ m = \frac{h_i}{h_o} = \frac{v}{u} $$ Here, $v$ is the length of the camera (distance from pinhole to screen) and $u$ is the distance from the object to the pinhole.
1. Initial Condition:
$$ h_i = h_o \left(\frac{l}{10}\right) \quad \dots(1) $$ Where $l$ is the initial length and $u=10$ m.
$$ h_i = h_o \left(\frac{l}{10}\right) \quad \dots(1) $$ Where $l$ is the initial length and $u=10$ m.
2. New Condition:
The camera is shifted 1.0 m farther, so $u’ = 10 + 1 = 11$ m.
To keep the image size ($h_i$) constant, the length must change to $l’$. $$ h_i = h_o \left(\frac{l’}{11}\right) \quad \dots(2) $$
The camera is shifted 1.0 m farther, so $u’ = 10 + 1 = 11$ m.
To keep the image size ($h_i$) constant, the length must change to $l’$. $$ h_i = h_o \left(\frac{l’}{11}\right) \quad \dots(2) $$
3. Calculation:
Equating (1) and (2): $$ \frac{l}{10} = \frac{l’}{11} \implies l’ = 1.1 l $$
Equating (1) and (2): $$ \frac{l}{10} = \frac{l’}{11} \implies l’ = 1.1 l $$
4. Percentage Change:
$$ \% \Delta l = \frac{l’ – l}{l} \times 100 = \frac{1.1l – l}{l} \times 100 = 10\% $$
$$ \% \Delta l = \frac{l’ – l}{l} \times 100 = \frac{1.1l – l}{l} \times 100 = 10\% $$
Answer: 10% Increase