Q112P You look down at a coin that lie... [FREE SOLUTION]

Chapter 34: Q112P (page 1044)

You look down at a coin that lies at the bottom of a pool of liquid of depthand index of refraction(Fig. 34-57). Because you view with two eyes, which intercept different rays of light from the coin, you perceive the coin to bewhere extensions of the intercepted rays cross, at depth $d_{a}$ instead of $d$ . Assuming that the intercepted rays in Fig. 34-57 are close to a vertical axis through the coin, show that $d_{a} = \frac{d}{n}$ .

Short Answer

Expert verified

For the given figure assuming that the intercepted rays are close to a vertical axis through the coin is $d_{a} = \frac{d}{n}$ .

Step by step solution

The given data

Refraction index of water is $n$ .
Refraction index air is $1$
Pool depth is $d$ .
Intercepted rays at depth are. $d_{a}$

Understanding the concept of refraction

The laws of refraction state that the light coming from higher to lower refractive index medium moves away from the normal line while for the ray from lower to higher, the ray moves towards the normal line. Thus, considering the laws of refraction for the given media, the ray diagram of the given situation is traced.

Now, using the concept of trigonometric calculations for the drawn ray diagram, we can get the required relation for the case of small angle approximations.

Formula:

From the equation for small angle approximations, $\frac{(t a n 胃_{2})}{(t a n 胃_{1})} 鈮� \frac{(\sin 胃_{2})}{(\sin 胃_{1})}$ (i)

The Snell鈥檚 law of refraction, $\frac{\sin 胃_{2}}{\sin 胃_{1}} = \frac{n_{1}}{n_{2}}$ (ii)

The tangent equation of an angle of a right angled-triangle,

$\tan 胃 = \frac{P e r p e n d i c u l a r}{B a s e}$ (iii)

Calculation of the depth equation

Medium 1 is water and medium 2 is air. The light rays strike the water surface at point A and B.

Let, the midpoint between A and B be the point C. The penny $P$ is directly below point C. The location of apparent or virtual penny is V when the rays are traced back to the figure to get the position of the image in real.

Now, the angles $鈭� C V B$ are taken as $胃_{2}$ and angle $鈭� C P B$ as $胃_{1}$ . The triangles $C V B$ and $C P B$ share common horizontal side from $C$ to $B$ is $x .$

Now, using the given data in triangle CVB and equation (i), we get the tangent $鈭� C V B$ as follows:

localid="1663046799610" $t a n 胃_{2} = \frac{x}{d_{a}}$

Now, using the given data in triangle CPB and equation (i), we get the tangent $鈭� C P B$ as follows:

localid="1663046327890" $t a n 胃_{1} = \frac{x}{d}$

Using the above values in the condition of equation (i) for small angle approximation and the equation (ii) value, we get the required relation as follows:

$\frac{x}{\frac{d_{a}}{\frac{x}{d}} 鈮� \frac{n_{1}}{n_{2}}} \frac{d}{d_{a}} 鈮� n d = \frac{d}{n}$