Solution

Linear magnification $k$ is the ratio of the image size to the object size. For a thin lens, linear magnification can be expressed as the ratio of the image distance $b$ (from the lens to the image) to the object distance $a$ (from the object to the lens):

$$k = \frac{b}{a} \tag{1}$$

The lens equation is:

$$\frac{1}{a} + \frac{1}{b} = \frac{1}{f} \tag{2}$$

where $f$ is the focal length of the lens. From equation (1), we find $a = \frac{b}{k}$. Substituting this into equation (2):

$$\frac{k}{b} + \frac{1}{b} = \frac{1}{f} \Rightarrow \frac{k + 1}{b} = \frac{1}{f} \Rightarrow b = f(k + 1)$$

The illumination on the film is inversely proportional to the square of the distance from the lens to the film (according to the inverse square law):

$$E \sim \frac{1}{b^2} = \frac{1}{f^2(k + 1)^2}$$

If the focal length $f$ is fixed, then illumination is inversely proportional to $(k + 1)^2$:

$$E \sim \frac{1}{(k + 1)^2}$$

To maintain the same exposure when illumination decreases, the exposure time must be increased proportionally:

$$t \sim (k + 1)^2$$

So, for two magnifications:

$$\frac{t_1}{t_2} = \frac{(k_1 + 1)^2}{(k_2 + 1)^2}$$

Hence:

$$t_1 = t_2 \left( \frac{k_1 + 1}{k_2 + 1} \right)^2$$