A woman cannot see objects closer than 40 cm from her left eye and...

For reading glasses, the object distance is taken as the standard near point, $u = -25 \text{ cm}$. The lenses form virtual images at the respective near points of the defective eyes.

For the left eye, the image distance is $v_L = -40 \text{ cm}$. Using the lens formula:

$\dfrac{1}{f_L} = \dfrac{1}{v_L} - \dfrac{1}{u}$

$\dfrac{1}{f_L} = \dfrac{1}{-40} - \dfrac{1}{-25} = \dfrac{-5 + 8}{200} = \dfrac{3}{200} \text{ cm}^{-1}$

$f_L = \dfrac{200}{3} \text{ cm}$

For the right eye, the image distance is $v_R = -100 \text{ cm}$. Using the lens formula:

$\dfrac{1}{f_R} = \dfrac{1}{v_R} - \dfrac{1}{u}$

$\dfrac{1}{f_R} = \dfrac{1}{-100} - \dfrac{1}{-25} = \dfrac{-1 + 4}{100} = \dfrac{3}{100} \text{ cm}^{-1}$

$f_R = \dfrac{100}{3} \text{ cm}$

An astronomical telescope requires an objective lens with a larger focal length and an eyepiece with a smaller focal length to achieve a magnification greater than $1$.

Thus, the objective focal length is $f_o = f_L = \dfrac{200}{3} \text{ cm}$ and the eyepiece focal length is $f_e = f_R = \dfrac{100}{3} \text{ cm}$.

The magnification of an astronomical telescope for a relaxed eye (normal adjustment) is given by:

$M = \dfrac{f_o}{f_e}$

$M = \dfrac{\dfrac{200}{3}}{\dfrac{100}{3}} = 2$