## Answer :

1.

**Identify the given data:**

- Object distance ([tex]\(d_o\)[/tex]): 20 cm

- The virtual image is three times smaller than the object, so the magnification ([tex]\(m\)[/tex]): [tex]\(-\frac{1}{3}\)[/tex]

2.

**Determine the image distance ([tex]\(d_i\)[/tex]):**

The magnification formula relates the image distance to the object distance and is given by:

[tex]\[ m = \frac{d_i}{d_o} \][/tex]

Given [tex]\(m = -\frac{1}{3}\)[/tex], [tex]\(-\frac{1}{3} = \frac{d_i}{20}\)[/tex].

3.

**Solve for [tex]\(d_i\)[/tex]:**

[tex]\[ d_i = -\frac{1}{3} \times 20 = -\frac{20}{3} \text{ cm} \][/tex]

The negative sign indicates that the image is virtual and formed on the same side as the object.

4.

**Use the lens formula to find the focal length ([tex]\(f\)[/tex]):**

The lens formula is:

[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \][/tex]

Substitute [tex]\(d_o = 20 \text{ cm}\)[/tex] and [tex]\(d_i = -\frac{20}{3} \text{ cm}\)[/tex] into the formula:

[tex]\[ \frac{1}{f} = \frac{1}{20} + \frac{1}{-\frac{20}{3}} \][/tex]

5.

**Simplify the expression:**

[tex]\[ \frac{1}{f} = \frac{1}{20} - \frac{3}{20} \][/tex]

[tex]\[ \frac{1}{f} = \frac{1 - 3}{20} = \frac{-2}{20} = -\frac{1}{10} \][/tex]

6.

**Solve for [tex]\(f\)[/tex]:**

[tex]\[ f = -10 \text{ cm} \][/tex]

Therefore, the focal length of the concave lens is [tex]\(-10 \text{ cm}\)[/tex]. The negative sign indicates that it is a concave lens.