An object is kept on the principal axis of a convex mirror of focal length `10 cm` at a distance of `10 cm` from the pole. The object starts moving at a velocity `20 mm//sce` towards the mirror at angle `30^(@)` with the principal axis. What will be the speed of its image and direction with the principal axis at that instant.

Options

A. speed `=5 (sqrt7)/(4)mm//sec`
B. speed `=(5 sqrt7)/(2)mm//sec`
C. `tan^(-1) ((2)/(sqrt3))` with the principal axis
D. none of these

Explanation

To solve the problem step-by-step, we will follow the outlined procedure to find the speed of the image and its direction with respect to the principal axis.

### Step 1: Identify Given Data
- Focal length of the convex mirror, \( f = +10 \, \text{cm} \)
- Object distance, \( u = -10 \, \text{cm} \) (negative as per the sign convention for mirrors)
- Velocity of the object, \( v_0 = 20 \, \text{mm/s} = 2 \, \text{cm/s} \)
- Angle of movement with respect to the principal axis, \( \theta = 30^\circ \)

### Step 2: Calculate Image Distance Using the Mirror Formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{10} = \frac{1}{v} + \frac{1}{-10} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5} \]
Thus, the image distance \( v \) is:
\[ v = 5 \, \text{cm} \]

### Step 3: Resolve the Velocity of the Object into Components
The object is moving at an angle of \( 30^\circ \) with the principal axis. We resolve its velocity into horizontal (x-axis) and vertical (y-axis) components:
- Horizontal component \( v_{0x} = v_0 \cos(30^\circ) = 20 \cos(30^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{mm/s} \)
- Vertical component \( v_{0y} = v_0 \sin(30^\circ) = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10 \, \text{mm/s} \)

### Step 4: Calculate the Velocity of the Image Along the X-axis
Using the formula for the velocity of the image along the x-axis:
\[ V_i - V_m = -\frac{v^2}{u^2}(V_0 - V_m) \]
Since the mirror is stationary, \( V_m = 0 \):
\[ V_i = -\frac{v^2}{u^2} V_0 \]
Where \( v = 5 \, \text{cm} \) and \( u = 10 \, \text{cm} \):
\[ V_i = -\left(\frac{5^2}{10^2}\right)(10\sqrt{3}) = -\left(\frac{25}{100}\right)(10\sqrt{3}) = -\frac{5\sqrt{3}}{2} \, \text{mm/s} \]

### Step 5: Calculate the Velocity of the Image Along the Y-axis
Using the magnification relation:
\[ \frac{h_i}{h_o} = -\frac{v}{u} \]
Differentiating with respect to time gives:
\[ u \frac{dh_i}{dt} + h_i \frac{du}{dt} = -v \frac{dh_o}{dt} - h_o \frac{dv}{dt} \]
At the instant, \( h_i = 0 \) and \( h_o = 0 \), so:
\[ u \frac{dh_i}{dt} = -v \frac{dh_o}{dt} \]
Substituting \( u = -10 \, \text{cm} \), \( v = 5 \, \text{cm} \), and \( \frac{dh_o}{dt} = 10 \, \text{mm/s} \):
\[ -10 \frac{dh_i}{dt} = -5 \times 10 \]
Thus:
\[ \frac{dh_i}{dt} = \frac{5 \times 10}{10} = 5 \, \text{mm/s} \]

### Step 6: Calculate the Resultant Velocity of the Image
The resultant velocity \( V_i \) can be expressed as:
\[ V_i = V_{ix} \hat{i} + V_{iy} \hat{j} \]
Substituting the values:
\[ V_i = -\frac{5\sqrt{3}}{2} \hat{i} + 5 \hat{j} \]

### Step 7: Calculate the Magnitude of the Resultant Velocity
The magnitude of the resultant velocity is:
\[ |V_i| = \sqrt{\left(-\frac{5\sqrt{3}}{2}\right)^2 + (5)^2} = \sqrt{\frac{75}{4} + 25} = \sqrt{\frac{75 + 100}{4}} = \sqrt{\frac{175}{4}} = \frac{5\sqrt{7}}{2} \, \text{mm/s} \]

### Step 8: Calculate the Direction of the Resultant Velocity
The direction angle \( \theta \) with respect to the principal axis is given by:
\[ \tan \theta = \frac{V_{iy}}{V_{ix}} = \frac{5}{-\frac{5\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} \]
Thus:
\[ \theta = \tan^{-1}\left(-\frac{2}{\sqrt{3}}\right) \]

### Final Answer
The speed of the image is \( \frac{5\sqrt{7}}{2} \, \text{mm/s} \) and the direction is \( \tan^{-1}\left(-\frac{2}{\sqrt{3}}\right) \) with respect to the principal axis.