Q17E - University Physics with Modern Physics

Question

34.17 A speck of dirt is embedded 3.50 cm below the surface of a sheet of ice (n=1.309). What is its apparent depth when viewed at normal incidence?

Step-by-Step Solution

Answer

The apparent depth when viewed at normal incidence is 2.67 cm

Step 1: Formula for Spherical refracting surface with plane surface between two optical materials

$\frac{{\mathbf{n}}_{\mathbf{a}}}{\mathbf{s}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{b}}}{\mathbf{s}\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$

Step 2: Determine the apparent depth of dirt

There is a special case for the spherical refracting surface which is the plane surface between two materials where $R=\infty$. In this case, use the below equation for a plane refracting surface in the form

$\frac{n_a}{s}+\frac{n_b}{s\text{'}}=0$

where $n_a$ is the refractive index of ice and equals 1.309, $n_b$ is the refractive index of air and equals 1.0. The distance is the distance from the air to the dirt and equals 3.5 cm. $s\text{'}$ is the apparent depth. So, solve equation for $s\text{'}$

$s\text{'}=-\left(\frac{n_b}{n_a}\right)s$

Put the values in above equation,

$s\text{'}=-\left(\frac{n_b}{n_a}\right)s=-\left(\frac{1.0}{1.309}\right)\left(3.5cm\right)=-2.67cm$

Hence, the apparent depth of dirt is 2.67cm.