A point object moves along the principal axis of a convex lens of focal length f such that its real image, also formed on the principal axis at a distance `(4f)/3` (at t=0) moves away from the lens with uniform velocity `alpha`. Find the velocity of the point object as a function of time t.

Options

A. `(f/(f+alphat))^(2)alpha`
B. `((alpha)/(f+alphat))^(2)`
C. `((f//3+alphat^(2))/f)^(2)alpha`
D. `(f/(f/3+alphat))^(2)alpha`

Answer

To solve the problem step by step, we will use the lens formula and the relationship between the object and image distances.

### Step 1: Understand the initial conditions
At time \( t = 0 \), the image is formed at a distance \( v = \frac{4f}{3} \) from the lens. We need to find the object distance \( u \) at this time.

### Step 2: Apply the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{f} = \frac{1}{\frac{4f}{3}} - \frac{1}{u} \]
This simplifies to:
\[ \frac{1}{f} = \frac{3}{4f} - \frac{1}{u} \]

### Step 3: Rearranging to find \( u \)
Rearranging gives:
\[ \frac{1}{u} = \frac{3}{4f} - \frac{1}{f} \]
Finding a common denominator:
\[ \frac{1}{u} = \frac{3 - 4}{4f} = \frac{-1}{4f} \]
Thus, we find:
\[ u = -4f \]
The negative sign indicates that the object is on the same side as the incoming light (the left side of the lens).

### Step 4: Determine the image position at time \( t \)
At time \( t \), the image moves away from the lens with a uniform velocity \( \alpha \). Therefore, the position of the image at time \( t \) is:
\[ v(t) = \frac{4f}{3} + \alpha t \]

### Step 5: Apply the lens formula again at time \( t \)
Using the lens formula again with the new image position:
\[ \frac{1}{f} = \frac{1}{v(t)} - \frac{1}{u(t)} \]
Substituting for \( v(t) \):
\[ \frac{1}{f} = \frac{1}{\left(\frac{4f}{3} + \alpha t\right)} - \frac{1}{u(t)} \]

### Step 6: Rearranging to find \( u(t) \)
Rearranging gives:
\[ \frac{1}{u(t)} = \frac{1}{\left(\frac{4f}{3} + \alpha t\right)} - \frac{1}{f} \]
Finding a common denominator:
\[ \frac{1}{u(t)} = \frac{f - \left(\frac{4f}{3} + \alpha t\right)}{f\left(\frac{4f}{3} + \alpha t\right)} \]
This simplifies to:
\[ \frac{1}{u(t)} = \frac{f - \frac{4f}{3} - \alpha t}{f\left(\frac{4f}{3} + \alpha t\right)} = \frac{\frac{-f}{3} - \alpha t}{f\left(\frac{4f}{3} + \alpha t\right)} \]
Thus:
\[ u(t) = \frac{f\left(\frac{4f}{3} + \alpha t\right)}{-f/3 - \alpha t} \]

### Step 7: Find the velocity of the object
The velocity \( v_o \) of the object is given by the change in position over time:
\[ v_o = \frac{u(t) - u(0)}{t} \]
Substituting \( u(0) = -4f \):
\[ v_o = \frac{u(t) + 4f}{t} \]
Substituting \( u(t) \) from the previous step:
\[ v_o = \frac{\frac{f\left(\frac{4f}{3} + \alpha t\right)}{-f/3 - \alpha t} + 4f}{t} \]
After simplifying, we find:
\[ v_o = \frac{3f\alpha t}{f\left(\frac{t}{3} + \alpha t^2\right)} \]

### Final Result
The velocity of the point object as a function of time \( t \) is:
\[ v_o = \frac{3f\alpha t}{f\left(\frac{t}{3} + \alpha t^2\right)} \]