A double convex lens has a focal length of 25 cm . The radius of...

Given $f = 25 \text{ cm}$ and $\mu = 1.5$.

Let the radii of curvature of the double convex lens be $R_1 = R$ and $R_2 = -2R$.

Using the lens maker's formula:

$\dfrac{1}{f} = (\mu - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)<br><br>Substituting the given values:<br><br>\dfrac{1}{25} = ($1.5 - 1) \left( \dfrac{1}{R} - \dfrac{1}{-2R} \right)$<br><br>$\dfrac{1}{25} = 0.5 \left( \dfrac{1}{R} + \dfrac{1}{2R} \right)<br><br>\dfrac{1}{25} = \dfrac{1}{2}\left( \dfrac{3}{2R} \right)

$\dfrac{1}{25} = \dfrac{3}{4R}$

$4R = 75 \Rightarrow R = 18.75$\text{ cm}

The other radius is $2R = 2 \times18.75 = 37.5$\text{ cm}.

The radii of curvature are $18.75$\text{ cm}and $37.5$\text{ cm}.