Magnification (m):

The magnification is given by the relationship:

$ m = \frac{1}{4} = \frac{v}{u} $

Rearrange to find the object distance (u) in terms of the image distance (v):

$ u = 4v $

Distance between object and image:

It is given that the sum of the object distance and the image distance is 40 cm:

$ v + u = 40 $

Substituting $ u = 4v $ into the equation gives:

$ v + 4v = 40 $

$ 5v = 40 $

Solving for $ v $ gives:

$ v = 8 \, \text{cm} $

Find the object distance (u):

Substitute $ v = 8 \, \text{cm} $ back into $ u = 4v $:

$ u = 4 \times 8 = 32 \, \text{cm} $

Calculate the focal length (f):

Use the mirror formula:

$ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $

Substituting the values of $ v $ and $ u $:

$ \frac{1}{8} + \frac{1}{32} = \frac{1}{f} $

Simplify and solve for $ \frac{1}{f} $:

$ \frac{1}{8} - \frac{1}{32} = \frac{1}{f} $

$ \frac{4 - 1}{32} = \frac{1}{f} $

$ \frac{3}{32} = \frac{1}{f} $

Therefore, the focal length is:

$ f = \frac{32}{3} = 10.7 \, \text{cm} $