A spherical glass bottle having negligible wall thickness is placed in air. When the bottle is completely filled with water, its focal length is $f$. If the water is replaced by another transparent liquid of higher refractive index, then the focal length changes to $f'$. Then the correct option is

The spherical glass bottle with negligible wall thickness filled with liquid acts as a solid liquid sphere of radius $R$.

The system can be considered as a thick lens with two spherical refracting surfaces of radius $R$ and $-R$, separated by a distance $t = 2R$.

The power of the first surface is $P_1 = \dfrac{\mu - 1}{R}$ and the second surface is $P_2 = \dfrac{1 - \mu}{-R} = \dfrac{\mu - 1}{R}$.

The equivalent power of the spherical lens is given by:

$P = P_1 + P_2 - \dfrac{t}{\mu} P_1 P_2$

$P = \dfrac{2(\mu - 1)}{R} - \dfrac{2R}{\mu} \left( \dfrac{\mu - 1}{R} \right)^2$

$P = \dfrac{2(\mu - 1)}{R} \left( 1 - \dfrac{\mu - 1}{\mu} \right) = \dfrac{2(\mu - 1)}{\mu R}$

The focal length $f$ is the reciprocal of the equivalent power:

$f = \dfrac{1}{P} = \dfrac{\mu R}{2(\mu - 1)} = \dfrac{R}{2} \left( 1 + \dfrac{1}{\mu - 1} \right)$

When the water is replaced by a liquid of higher refractive index, $\mu$ increases. As $\mu$ increases, the term $\dfrac{1}{\mu - 1}$ decreases, which implies that the focal length $f$ decreases.

Therefore, $f' < f$.

Answer: $f' < f$