Given:
- Length of the rod, L = 10 cm
- Focal length of the concave lens, f = -10 cm (negative sign indicates concave lens)
- Distance of the middle point of the rod from the lens, u = 20 cm

To find:
- Magnification of the length of the rod

Solution:

Step 1: Calculating the image distance (v):
Using the lens formula,

\f[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\f]

Substituting the given values, we can find the image distance (v):

\f[\frac{1}{-10} = \frac{1}{v} - \frac{1}{20}\f]

Simplifying,

\f[-\frac{1}{10} = \frac{1}{v} - \frac{1}{20}\f]

Combining the fractions,

\f[-\frac{1}{10} = \frac{2-v}{20v}\f]

Cross-multiplying,

\f[-2v + v = 10\f]

\f[-v = 10\f]

\f[v = -10 cm\f]

Since the image distance (v) is negative, it means the image formed is virtual and located on the same side as the object.

Step 2: Calculating the magnification (m):
The magnification (m) can be calculated using the formula,

\f[m = \frac{v}{u}\f]

Substituting the values,

\f[m = \frac{-10}{20}\f]

Simplifying,

\f[m = -0.5\f]

The negative sign indicates that the image formed is inverted.

Step 3: Calculating the magnification of the length of the rod:
The magnification of the length of the rod can be calculated using the formula,

\f[m_{\text{rod}} = \frac{L'}{L}\f]

where L' is the length of the image of the rod.

Since the image formed is inverted, the length of the image can be calculated as,

\f[L' = mL\f]

Substituting the values,

\f[L' = -0.5 \times 10\f]

Simplifying,

\f[L' = -5 cm\f]

The negative sign indicates that the image of the rod is inverted.

Finally, substituting the values in the formula for magnification of the length of the rod,

\f[m_{\text{rod}} = \frac{-5}{10}\f]

Simplifying,

\f[m_{\text{rod}} = -0.5\f]

The magnification of the length of the rod is -0.5.