A buring candle is placed in front of a concave spherical mirror on its principal optical axis at a distance of `(4//3)F` form the pole of the mirror (here `F` is the focal length of the mirror). The candle is arranged at right angle to the axis. The image of the candle in the concave mirror impinges upon a convex mirror of focal length `2F` . The distance between the mirrors is `3F` and their axes coincide. The image of the candle in the first mirror plays the part of a virtual object with respect to the second mirror and gives a real image arranged between the two mirrors. Find the total linear magnification (magnitude only) of the system.

Explanation

To solve the problem step by step, we will analyze the situation involving the concave mirror and the convex mirror.

### Step 1: Understand the setup
We have a burning candle placed at a distance of \( \frac{4}{3}F \) from the pole of a concave mirror. The focal length \( F \) is considered negative for concave mirrors.

### Step 2: Apply the mirror formula for the concave mirror
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
For the concave mirror:
- Focal length \( f = -F \)
- Object distance \( u = -\frac{4}{3}F \) (negative because the object is in front of the mirror)

Substituting these values into the mirror formula:
\[ -\frac{1}{F} = \frac{1}{v} - \frac{3}{4F} \]
Rearranging gives:
\[ \frac{1}{v} = -\frac{1}{F} + \frac{3}{4F} = -\frac{4}{4F} + \frac{3}{4F} = -\frac{1}{4F} \]
Thus, we find:
\[ v = -4F \]

### Step 3: Calculate the magnification for the concave mirror
The magnification \( m_1 \) for the concave mirror is given by:
\[ m_1 = -\frac{v}{u} \]
Substituting the values we found:
\[ m_1 = -\frac{-4F}{-\frac{4}{3}F} = -\frac{4F}{\frac{4}{3}F} = -3 \]

### Step 4: Determine the virtual object for the convex mirror
The distance between the two mirrors is \( 3F \). The image formed by the concave mirror acts as a virtual object for the convex mirror. The object distance \( u \) for the convex mirror is calculated as:
\[ u = 3F - |v| = 3F - 4F = -F \]
(Note: The negative sign indicates that the object is virtual and located on the same side as the object for the convex mirror.)

### Step 5: Apply the mirror formula for the convex mirror
For the convex mirror:
- Focal length \( f = +2F \)
- Object distance \( u = -F \)

Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{2F} = \frac{1}{v} - \frac{1}{F} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{2F} + \frac{1}{F} = \frac{1}{2F} + \frac{2}{2F} = \frac{3}{2F} \]
Thus, we find:
\[ v = \frac{2F}{3} \]

### Step 6: Calculate the magnification for the convex mirror
The magnification \( m_2 \) for the convex mirror is given by:
\[ m_2 = -\frac{v}{u} \]
Substituting the values:
\[ m_2 = -\frac{\frac{2F}{3}}{-F} = \frac{2}{3} \]

### Step 7: Calculate the total magnification
The total magnification \( m \) of the system is the product of the magnifications of the two mirrors:
\[ m = m_1 \times m_2 = (-3) \times \left(\frac{2}{3}\right) = -2 \]
In magnitude, the total linear magnification is:
\[ |m| = 2 \]

### Final Answer
The total linear magnification (magnitude only) of the system is:
\[ \boxed{6} \]