There is a small air bubble inside a glass sphere (μ = 1.5) of radius 10 cm. The bubble is 4.0 cm below the surface and is viewed normally from outside as shown. The apparent depth of the bubble will be

3.0 cm below the surface

The correct answer is Option (3) → 3.0 cm below the surface

Given a glass sphere with refractive index $ \mu = 1.5 $ and radius $R = 10\ \text{cm}$. The bubble lies $4.0\ \text{cm}$ below the inner surface, i.e. at object distance $u = 4\ \text{cm}$ measured from the surface along the normal.

Use the spherical refraction formula (Cartesian sign convention):

$\displaystyle \mu\!\left(\frac{1}{u}\right) + 1\!\left(\frac{1}{v}\right) = \frac{1- \mu}{R}$

Here the center of curvature is inside the glass (opposite the refracted side), so $R = -10\ \text{cm}$. Substitute $ \mu = 1.5 $, $u = 4 $, $R=-10 $:

$1.5\!\left(\frac{1}{4}\right) + \frac{1}{v} = \frac{1-1.5}{-10}$

$\frac{1}{v} = \frac{1-1.5}{-10} - \frac{1.5}{4} = 0.05 - 0.375 = -0.325$

$\Rightarrow v = -\frac{1}{0.325} \approx -3.08\ \text{cm}.$

The negative sign shows the image is virtual and lies inside the glass on the same side as the bubble, at a distance $|v|\approx 3.08\ \text{cm}$ below the surface.

Therefore the apparent depth of the bubble (as seen from outside) is approximately $3.08\ \text{cm}\approx 3.0\ \text{cm}$ below the surface.