There is a concave mirror of radius of curvature 20cm. A point object is placed at distance 15cm from the mirror. If point object start moving with speed `2mm//s` perpendicular to principal axis then at this instant what will be the speed of image in `mm//s` ?

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Explanation

To solve the problem step by step, we will use the mirror formula and the concept of magnification.

### Step 1: Determine the focal length of the concave mirror.
The radius of curvature (R) is given as 20 cm. The focal length (f) is given by the formula:
\[ f = \frac{R}{2} \]
Substituting the value of R:
\[ f = \frac{20 \text{ cm}}{2} = 10 \text{ cm} \]
Since it is a concave mirror, the focal length is negative:
\[ f = -10 \text{ cm} \]

**Hint:** Remember that for concave mirrors, the focal length is negative.

### Step 2: Use the mirror formula to find the image distance (v).
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Where:
- \( f \) is the focal length,
- \( v \) is the image distance,
- \( u \) is the object distance (which is negative for real objects).

Given that the object is placed at a distance of 15 cm from the mirror, we have:
\[ u = -15 \text{ cm} \]

Substituting the values into the mirror formula:
\[ \frac{1}{-10} = \frac{1}{v} + \frac{1}{-15} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{-10} + \frac{1}{15} \]
Finding a common denominator (30):
\[ \frac{1}{v} = \frac{-3}{30} + \frac{2}{30} = \frac{-1}{30} \]
Thus:
\[ v = -30 \text{ cm} \]

**Hint:** Make sure to keep track of the signs when using the mirror formula.

### Step 3: Calculate the magnification (m).
The magnification (m) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values of \( v \) and \( u \):
\[ m = -\frac{-30}{-15} = -2 \]

**Hint:** Magnification tells you the ratio of the height of the image to the height of the object and its sign indicates the orientation of the image.

### Step 4: Relate the velocities of the object and image.
The velocity of the image (\( v_i \)) is related to the velocity of the object (\( v_o \)) by the formula:
\[ v_i = m \cdot v_o \]
Given that the speed of the object is \( v_o = 2 \text{ mm/s} \):
\[ v_i = -2 \cdot 2 = -4 \text{ mm/s} \]

**Hint:** The negative sign indicates the direction of the image's movement relative to the object.

### Step 5: Conclusion
The speed of the image is \( 4 \text{ mm/s} \) in the opposite direction to the movement of the object.

**Final Answer:** The speed of the image is \( 4 \text{ mm/s} \).