An object is placed `1cm` above the principal axis of a convex lens of focal length of 40cm. Object distance is 60cm. If the object now starts moving perpendicularly away from optical axis with a speed of 10cm/s, then what is the speed of the image ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
- Focal length of the convex lens (F) = 40 cm
- Object distance (u) = -60 cm (the negative sign indicates that the object is placed on the same side as the incoming light)
- Height of the object above the principal axis = 1 cm
- Speed of the object moving perpendicularly away from the optical axis = 10 cm/s

### Step 2: Use the lens formula to find the image distance (v)
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging this gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]

### Step 3: Substitute the known values into the lens formula
Substituting \( f = 40 \) cm and \( u = -60 \) cm:
\[ \frac{1}{v} = \frac{1}{40} + \frac{1}{-60} \]

### Step 4: Calculate the right-hand side
Finding a common denominator (120):
\[ \frac{1}{40} = \frac{3}{120}, \quad \frac{1}{-60} = \frac{-2}{120} \]
Thus:
\[ \frac{1}{v} = \frac{3}{120} - \frac{2}{120} = \frac{1}{120} \]

### Step 5: Solve for v
Taking the reciprocal gives:
\[ v = 120 \text{ cm} \]

### Step 6: Calculate the magnification (m)
The magnification (m) is given by:
\[ m = \frac{v}{u} \]
Substituting the values:
\[ m = \frac{120}{-60} = -2 \]

### Step 7: Relate the speed of the image to the speed of the object
The speed of the image (v_image) is related to the speed of the object (v_object) by the magnification:
\[ v_{\text{image}} = m \cdot v_{\text{object}} \]
Substituting the values:
\[ v_{\text{image}} = -2 \cdot 10 \text{ cm/s} = -20 \text{ cm/s} \]

### Step 8: Interpret the result
The negative sign indicates that the image moves in the opposite direction to the object. Therefore, the speed of the image is 20 cm/s downward.

### Final Answer:
The speed of the image is 20 cm/s downward.

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