Problem 6: Displacement Method for Same Size Images
1. Relationship between Object Distances
Let the focal length be $f$. The magnification produced by a lens is given by $m = \frac{f}{f-u}$.
We are given that the two images are equal in size. This implies the magnitudes of magnification are equal ($|m_1| = |m_2|$). Since the object positions are different, one image must be real ($m$ is negative) and the other virtual ($m$ is positive). $$ \frac{f}{f-u_1} = – \frac{f}{f-u_2} $$ $$ f – u_2 = -(f – u_1) \implies u_1 + u_2 = 2f $$
2. Using the Shift Information
The lens is moved vertically by a distance $f$. Assuming the lens moves away from the object: $$ u_2 – u_1 = f $$
We now have a system of two equations:
- $u_1 + u_2 = 2f$
- $u_2 – u_1 = f$
Subtracting the second from the first: $2u_1 = f \implies u_1 = \frac{f}{2}$.
3. Calculating Magnification
Substitute $u_1 = f/2$ into the magnification formula:
$$ m_1 = \frac{f}{f – u_1} = \frac{f}{f – f/2} = \frac{f}{f/2} = 2 $$The magnification factor is 2. This means the image is twice the size of the object.
$$ \text{Image Diameter} = |m| \times \text{Object Diameter} $$ $$ \text{Image Diameter} = 2 \times 2.0 \, \text{cm} = 4.0 \, \text{cm} $$