Two transparent media A and B , having refractive indices 1.33 and...

For refraction at a single spherical surface, the formula is given by:

$\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}$

Given that the object is in medium $A$, $n_1 = 1.33$ and $n_2 = 1.48$.

Since medium $A$ lies on the convex side of the surface, the surface is convex towards the incident light. Therefore, the center of curvature lies in medium $B$, making the radius of curvature positive: $R = +30 \text{ cm}$.

The paraxial rays emerge parallel after refraction, which means the image is formed at infinity, so $v = \infty$.

Substituting the values into the formula:

$\dfrac{1.48}{\infty} - \dfrac{1.33}{u} = \dfrac{1.48 - 1.33}{30}$

$0 - \dfrac{1.33}{u} = \dfrac{0.15}{30}$

$-\dfrac{1.33}{u} = \dfrac{1}{200}$

$u = -1.33 \times 200 = -266 \text{ cm}$

The negative sign indicates that the object is placed in front of the refracting surface. The distance of the object from the surface is $266.0 \text{ cm}$.