In an experiment with a convex lens, the length of an image is 1 cm, and the object length is 5 cm. If the object is placed at a distance of 40 cm from the lens, then the focal length of the lens is

The correct option is 6.67 cm.

To find the focal length of the convex lens, we can use the formula for linear magnification ($m$) and the lens formula.

First, let us note the given information from the question:
- Height or length of the object, $h_o=5\text{cm}$
- Height or length of the image, $h_i=-1\text{cm}$ (since a real image formed by a convex lens is inverted, its height is taken as negative)
- Object distance, $u=-40\text{cm}$ (according to the sign convention, the object distance is negative)

The magnification $m$ of a lens is defined as the ratio of the height of the image to the height of the object, which is also equal to the ratio of the image distance ($v$) to the object distance ($u$):
$m=\frac{h_i}{h_o}$

Substituting the given values into the magnification formula:
$\frac{-1}{5}=\frac{v}{-40}$

Solving for the image distance $v$:
$v=\frac{-1\times (-40)}{5}=8\text{cm}$

Now, we use the lens formula to find the focal length ($f$):
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Substituting the values of $v=8\text{cm}$ and $u=-40\text{cm}$ into the lens formula:
$\frac{1}{f}=\frac{1}{8}-\frac{1}{-40}$

Simplifying the signs:
$\frac{1}{f}=\frac{1}{8}+\frac{1}{40}$

Finding a common denominator (which is 40) to add the fractions:
$\frac{1}{f}=\frac{5}{40}+\frac{1}{40}=\frac{6}{40}$

Taking the reciprocal to solve for $f$:
$f=\frac{40}{6}\approx 6.67\text{cm}$

Therefore, the focal length of the lens is 6.67 cm.