A cylindrical glass rod ( μ = 1.5 ) with a radius of 1.0 cm has one...

Using the formula for refraction at a single spherical surface:

$\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}$

Given:
Refractive index of the surrounding medium (water), $\mu_1 = \dfrac{4}{3}$
Refractive index of the rod (glass), $\mu_2 = 1.5 = \dfrac{3}{2}$
Object distance, $u = -8.0 \text{ cm}$
Radius of curvature, $R = +1.0 \text{ cm}$ (positive since the center of curvature lies in the direction of incident light)

Substituting the values into the formula:

$\dfrac{3/2}{v} - \dfrac{4/3}{-8.0} = \dfrac{3/2 - 4/3}{1.0}$

$\dfrac{3}{2v} + \dfrac{4}{24} = \dfrac{9/6 - 8/6}{1}$

$\dfrac{3}{2v} + \dfrac{1}{6} = \dfrac{1}{6}$

$\dfrac{3}{2v} = 0$

$v = \infty$

The image is formed at infinity.