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

The formula for refraction at a single spherical surface is:

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

The observer is to the left of the sphere, so the refraction occurs at the left surface. Let the pole be at the left surface. Taking the direction of incident light (from right to left) as positive, distances to the right of the pole are taken as negative.

Refractive index of the medium of the object (glass), $\mu_1 = 1.5$

Refractive index of the medium of the observer (air), $\mu_2 = 1$

The radius of the sphere is $5.0 \text{ cm}$. The centre of curvature is to the right of the pole, so $R = -5.0 \text{ cm}$.

The object is $1.5 \text{ cm}$ to the left of the centre. The position of the object from the pole is $u = -5.0 + 1.5 = -3.5 \text{ cm}$.

Substituting the values into the refraction formula:

$\dfrac{1}{v} - \dfrac{1.5}{-3.5} = \dfrac{1 - 1.5}{-5.0}$

$\dfrac{1}{v} + \dfrac{3}{7} = \dfrac{-0.5}{-5.0}$

$\dfrac{1}{v} + \dfrac{3}{7} = \dfrac{1}{10}$

$\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{3}{7} = \dfrac{7 - 30}{70} = -\dfrac{23}{70}$

$v = -\dfrac{70}{23} \approx -3.04 \text{ cm}$

The negative sign indicates that the image is formed $3.04 \text{ cm}$ to the right of the left surface.

The distance of the image from the centre of the sphere is $5.0 - 3.04 = 1.96 \text{ cm}$.

Since the image is between the left surface and the centre, its position is $1.96 \text{ cm} \approx 2 \text{ cm}$ to the left of the centre.