To solve the problem step by step, we will analyze the situation involving a double convex lens, the object, and the image formed on a screen.
### Step 1: Understand the Setup
We have a double convex lens forming a real image of an object on a fixed screen. The lens is moving away from the screen with a constant velocity \( v_0 = 1 \, \text{m/s} \). We need to find the velocity of the object when its size is doubled compared to the size of the image.
### Step 2: Use the Lens Formula
The lens formula is given by:
\[
\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
\]
where:
- \( f \) is the focal length of the lens,
- \( v \) is the image distance,
- \( u \) is the object distance.
### Step 3: Differentiate the Lens Formula
To find the relationship between the velocities of the object and the image, we differentiate the lens formula with respect to time \( t \):
\[
0 = -\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt}
\]
This can be rearranged to:
\[
\frac{1}{v^2} \frac{dv}{dt} = \frac{1}{u^2} \frac{du}{dt}
\]
### Step 4: Substitute Velocities
Let:
- \( \frac{dv}{dt} = v' \) (the velocity of the image),
- \( \frac{du}{dt} = u' \) (the velocity of the object).
Thus, we can write:
\[
\frac{v'}{v^2} = \frac{u'}{u^2}
\]
### Step 5: Express Object Velocity
From the above equation, we can express the object velocity \( u' \) in terms of the image velocity \( v' \):
\[
u' = \frac{u^2}{v^2} v'
\]
### Step 6: Determine the Size Relationship
Since the size of the object is double that of the image, we denote the size of the image as \( h \) and the size of the object as \( 2h \). The magnification \( m \) is given by:
\[
m = \frac{\text{size of object}}{\text{size of image}} = 2
\]
Thus, we can say:
\[
m = \frac{h_o}{h_i} = 2 \implies \frac{1}{m} = \frac{1}{2}
\]
### Step 7: Substitute Magnification into Velocity Equation
Now substituting \( m = 2 \) into the velocity equation:
\[
u' = \frac{u^2}{v^2} v' = \frac{u^2}{(m \cdot u)^2} v' = \frac{u^2}{(2u)^2} v' = \frac{1}{4} v'
\]
### Step 8: Calculate the Image Velocity
Since the lens is moving away from the screen with a velocity \( v_0 = 1 \, \text{m/s} \), the image velocity \( v' \) can be considered as:
\[
v' = v_0 = 1 \, \text{m/s}
\]
### Step 9: Find Object Velocity
Substituting \( v' \) into the equation for \( u' \):
\[
u' = \frac{1}{4} \cdot 1 = \frac{1}{4} \, \text{m/s}
\]
### Step 10: Calculate the Velocity of the Object Relative to the Screen
The velocity of the object with respect to the screen is given by:
\[
V_{os} = v_0 - u' = 1 - \frac{1}{4} = 1 - 0.25 = 0.75 \, \text{m/s}
\]
### Conclusion
Thus, the velocity of the object at the instant its size is doubled compared to the size of the image is:
\[
\boxed{0.75 \, \text{m/s}}
\]
