Question

The figure shows a meter stick lying on the bottom of a 100-cm-long tank with its zero mark against the left edge. You look into the tank at a 30 angle, with your line of sight just grazing the upper left edge of the tank.

a. What mark do you see on the meter stick if the tank is empty?

b. What mark do you see on the meter stick if the tank is half full of water?

c. What mark do you see on the meter stick if the tank is completely full of water?

Answer 1

Concepts and reason

The problem deals with the concept of the Snell’s law as the mark on the meter stick is to be determined which might get changed because of the refraction.

Fundamentals

The Snell’s law tells the degree of refraction and relation between the angle of incidence, the angle of refraction and refractive indices of given pair of media.

The Snell’s law can be represented as:

${n_a}\sin {\theta _i} = {n_b}\sin {\theta _r}$

Here ${n_a}$ is the index of refraction in material a, ${n_b}$ is the index of refraction in material b, ${\theta _i}$ is the angle of incidence and ${\theta _r}$ is the angle of refraction.

And in a right angle triangle as:

$\sin \theta = \frac{{{\rm{Perpendicular}}}}{{{\rm{Hypotenuse}}}}$ , $\cos \theta = \frac{{{\rm{Base}}}}{{{\rm{Hypotenuse}}}}$ and $\tan \theta = \frac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}}$

The depth of the tank is $h = 50cm$

The width of the tank is $w = 100cm$

The angle at which the tank can be looked inside is ${30^0}$ , then for the empty tank we can get the below diagram as:

Suppose the mark on the stick is at $x$ cm distance from left end, then in the right angle triangle:

$\begin{array}{l}\\\tan \theta = \frac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}}\\\\\tan {60^0} = \frac{x}{{{\rm{50cm}}}}\\\end{array}$

Rearrange the terms as:

$\begin{array}{c}\\x = \left( {50cm} \right)\tan {60^o}\\\\ = 86.6cm\\\end{array}$

(b)

When the tank is half full of water then the angle of incidence and refraction will as shown in the below diagram:

Suppose the mark at which incident ray hit the water is $x$ .

From right angle formed in the air with angle ${60^o}$ as:

$\begin{array}{l}\\\tan \theta = \frac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}}\\\\\tan {60^0} = \frac{x}{{\frac{h}{2}}}\\\\\tan {60^0} = \frac{x}{{\frac{{50cm}}{2}}}\\\\\tan {60^0} = \frac{x}{{25cm}}\\\end{array}$

Rearrange the terms as:

$\begin{array}{c}\\x = \left( {25cm} \right)\tan {60^o}\\\\ = 43.3cm\\\end{array}$

The angle of incidence is ${\theta _i} = {60^o}$

The refractive index of the air is ${n_a} = 1$ as air is the medium a.

The refractive index of the water is ${n_b} = 1.33$ as water is the medium b.

Then using Snell’s law detriment the refraction angle ${\theta _r}$ as:

$\begin{array}{l}\\{n_a}\sin {\theta _i} = {n_b}\sin {\theta _r}\\\\1\left( {\sin {{60}^o}} \right) = 1.33\left( {\sin {\theta _r}} \right)\\\\\sin {\theta _r} = \frac{{\sin {{60}^o}}}{{1.33}}\\\\\sin {\theta _r} = 0.6511\\\end{array}$

Then,

${\theta _r} = {40.62^0}$

Then from the right angle triangle with ${40.62^0}$ refraction angle formed in the water is:

$\begin{array}{l}\\\tan \theta = \frac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}}\\\\\tan {40.62^0} = \frac{y}{{\frac{h}{2}}}\\\\\tan {40.62^0} = \frac{y}{{\frac{{50cm}}{2}}}\\\\\tan {40.62^0} = \frac{y}{{25cm}}\\\end{array}$

Rearrange the terms as:

$\begin{array}{c}\\y = \left( {25cm} \right)\tan {40.62^0}\\\\ = 21.44cm\\\end{array}$

Then the mark which can be seen if the tank is half full is

$x + y = 43.3cm + 21.44cm = 64.74cm$

(c)

When the tank is full of water then the angle of incidence and refraction will as shown in the below diagram:

When the tank is full of water then the incident ray will hit the water at mark of $0cm$ as shown in the above diagram.

The angle of incidence is ${\theta _i} = {60^o}$

The refractive index of the air is ${n_a} = 1$ as air is the medium a.

The refractive index of the water is ${n_b} = 1.33$ as water is the medium b.

Then using Snell’s law detriment the refraction angle ${\theta _r}$ as:

$\begin{array}{l}\\{n_a}\sin {\theta _i} = {n_b}\sin {\theta _r}\\\\1\left( {\sin {{60}^o}} \right) = 1.33\left( {\sin {\theta _r}} \right)\\\\\sin {\theta _r} = \frac{{\sin {{60}^o}}}{{1.33}}\\\\\sin {\theta _r} = 0.6511\\\end{array}$

Then,

${\theta _r} = {40.62^0}$

Then from the right angle triangle with ${40.62^0}$ refraction angle formed in the water is:

$\begin{array}{l}\\\tan \theta = \frac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}}\\\\\tan {40.62^0} = \frac{y}{h}\\\\\tan {40.62^0} = \frac{y}{{50cm}}\\\end{array}$

Rearrange the terms as:

$\begin{array}{c}\\y = \left( {50cm} \right)\tan {40.62^0}\\\\ = 42.85cm\\\end{array}$

Ans: Part a

If the tank is empty then the $86.6cm$ mark can be seen.

Part b

If the tank is half full then the $64.74cm$ mark can be seen.

Part b

If the tank is full then the $42.85cm$ mark can be seen.