A small object is embedded inside a glass sphere ( μ = 1.5 ) of...

The observer is to the right of the sphere, so refraction occurs at the right surface. Light travels from left to right, which is taken as the positive direction.

Let the pole $P$ be the rightmost point of the sphere.

Radius of curvature of the right surface is $R = -5.0 \text{ cm}$.

The object is $1.5 \text{ cm}$ to the left of the centre, so its distance from the pole $P$ is $u = -(5.0 + 1.5) = -6.5 \text{ cm}$.

Refractive index of the glass sphere is $\mu_1 = 1.5$ and for air is $\mu_2 = 1$.

Using the spherical refraction formula:

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

$\dfrac{1}{v} - \dfrac{1.5}{-6.5} = \dfrac{1 - 1.5}{-5.0}$

$\dfrac{1}{v} + \dfrac{3}{13} = \dfrac{-0.5}{-5.0}$

$\dfrac{1}{v} + \dfrac{3}{13} = \dfrac{1}{10}$

$\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{3}{13} = \dfrac{13 - 30}{130} = -\dfrac{17}{130}$

$v = -\dfrac{130}{17} \approx -7.65 \text{ cm}$

The negative sign indicates that the image is formed $7.65 \text{ cm}$ to the left of the right surface (pole $P$).

The distance of the image from the centre of the sphere is $7.65 - 5.0 = 2.65 \text{ cm}$.

Since the image is $7.65 \text{ cm}$ left of the pole and the centre is $5.0 \text{ cm}$ left of the pole, the image is $2.65 \text{ cm}$ left of the centre.