Question

34.18 A transparent liquid fills a cylindrical tank to a depth of 3.60m. There is air above the liquid. You look at normal incidence at a small pebble at the bottom of the tank. The apparent depth of the pebble below the liquid’s surface is 2.45m. What is the refractive index of this liquid?

Short answer

The refractive index of the liquid is 1.4694.

Step-by-step solution

Object - image relationship for spherical reflecting surface:

$\frac{{\mathbf{n}}_{\mathbf{1}}}{\mathbf{s}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{2}}}{\mathbf{s}\mathbf{\text{'}}}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{2}}\mathbf{-}{\mathbf{n}}_{\mathbf{1}}}{\mathbf{R}}$

${\mathit{n}}_{\mathbf{1}}\mathbf{,}{\mathit{n}}_{\mathbf{2}}$⇒The refraction index of both of the surfaces,

s⇒object distance from the vertex of the spherical surface,

s’⇒The image distance from the vertex of the spherical surface,

R⇒The radius of the spherical surface.

The sign rules for the variables in the equation:

1. Sign rule for the object distance (s): when the object is on the same side of the refracting surface as the incoming light, object distance s is positive; otherwise, it is negative.
2. Sign rule for the image distance (s′ dash): When the image is on the same side of the refracting surface as the outgoing light (the refracted light), image distance s′ is positive; otherwise, it is negative.
3. Sign rule for the radius of curvature of a spherical surface (R): When the center of curvature C is on the same side as the outgoing light (the refracted light), the radius of curvature is positive; otherwise, it is negative.

Apply: the above equation gives us the position of an image formed by refraction of light between a spherical interface of two different mediums. To obtain this position, we need to get the radius of the spherical interface and the two refraction indexes of the mediums and the object position.

Note that one can use this formula for a plane interface when we put R equals to infinity as if the line of interface is considered a part of a sphere which has a radius of infinity.

Determine the refractive index of the liquid

$$\begin{gathered} n_2=1\to air \\ s=3.6m \end{gathered}$$

The image is the opposite side of the outgoing (refracted light). Therefore, $s\text{'}$→ Is negative.

$$\begin{gathered} s\text{'}=-2.45m \\ R=\infty \\ \frac{n_1}{s}+\frac{n_2}{s\text{'}} \\ =\frac{n_2-n_1}{R} \\ =\frac{n_2-n_1}{\infty }=0 \\ \frac{n_1}{s}=-\frac{n_2}{s\text{'}} \\ -\left(3.6m\right)\times \frac{1}{\left(-2.45m\right)} \\ =1.4694 \end{gathered}$$

Hence, the refractive index is 1.4694.