Figure 34-34 shows a small light bulb suspended at distance ${\mathit{d}}_{\mathbf{1}}\mathbf{=}\mathbf{250}\mathbf{\text{cm}}$above the surface of the water in a swimming pool where the water depth ${\mathit{d}}_{\mathbf{2}}\mathbf{=}\mathbf{200}\mathbf{\text{cm}}$. The bottom of the pool is a large mirror. How far below the mirror surface is the image of the bulb? (Hint: Assume that the rays are close to a vertical axis through the bulb, and use the small-angle approximation in which $\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{\approx }\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathbf{\approx }\mathit{\theta }$)

• The small light bulb is suspended above the surface of the water at a distance,$d_1=250\text{cm}$
• Water depth of a swimming pool,$d_2=200\text{cm}$
• Small-angle approximation,$\sin \theta \approx \tan \theta \approx \theta$
• The refraction index of water,$n_w=1.33$
• The refractive index of air,$n_{air}=1$

The light ray coming from the object kept above the water surface in the air medium undergoes refraction when it travels through the water surface. As per the refraction law, the light coming from a medium of lower refractive index to a medium of higher refractive index slows down and bends towards the normal incidence line. Now, as it travels through the water surface it gets reflected from the mirror surface. Now, as per the concept of equal values of angle of incidence and reflection, we calculate the distance of the image that is formed below the mirror due to optical transmission.

Formulae:

The tangent angle of a triangle,

$\tan \theta =\frac{p}{b}$ …..(i)

Snell’s law of refraction,

$n_1\sin i=n_2\sin r$ …..(ii)

Here, $n_1$is the refractive index of incident medium, $n_2$is the refractive medium of refractive medium, $i$is the incident angle, and $r$is the refraction angle.

For the given problem, observe that the case involves refraction at air–water interface and reflection from a plane mirror at the bottom of the pool.

The above figure depicts the path of the light ray undergoing the process of refraction and then reflection from the mirror under water. Here, $\theta$is the angle of incidence and is $\theta \text{'}$the angle of refraction.

The object O is a vertical distance $d_1$above the water, and the water surface is a vertical distance $d_2$above the mirror.

You are looking for a distance $d$(treated as a positive number) below the mirror where the image I of the object is formed.

Now, considering triangle OAB, and using equation (i), you get that

$\left|AB\right|=d_1\tan \theta$

From the small angle approximation, you get

$\left|AB\right|\approx d_1\theta$ ….. (iii)

Now, using equation (ii) and the given small angle approximation, you get that

$\frac{\sin \theta }{\sin \theta \text{'}}=\frac{n_w}{n_{air}}$

Here, the refractive index of air is $n_{air}=1$. Therefore,

$\theta \text{'}\approx \frac{\theta }{n_w}$ ….. (iv)

Similarly, for the case of reflection from the mirror surface within the water surface, you get that the angle of refraction $\theta \text{'}$becomes the angle of incidence for this case.

Thus, applying equation (i) in triangle CBD, you get that

$\begin{array}{rcl}\left|\frac{BC}{2}\right|& =& d_2\tan \theta \text{'}\left(\because \text{Angle of incidence = angle of reflection}\right)\\ & \approx & d_2\theta \text{'}\end{array}$

From equation (iv), you get

$\left|\frac{BC}{2}\right|\approx \frac{d_2\theta }{n_w}$

$\left|BC\right|=\frac{2d_2\theta }{n_w}$ ….. (v)

Finally from triangle ACI, you obtain that

$\left|AI\right|=d+d_2$

Thus, using the above values, the distance$d$at which image is formed below the mirror surface can be calculated as follows:

$\begin{array}{rcl}d& =& \left|AI\right|-d_2\\ & =& \frac{\left|AC\right|}{\tan \theta }-d_2\\ & =& \frac{\left|AB\right|+\left|BC\right|}{\tan \theta }-d_2\\ & =& \left(d_1\theta +\frac{2d_2\theta }{n_w}\right)\frac{1}{\theta }-d_2\end{array}$

$\begin{array}{rcl}d& =& d_1+\frac{2d_2}{n_w}-d_2\\ & =& 250\text{cm}+\frac{2\left(200\text{cm}\right)}{1.33}-200\text{cm}\\ & =& 351\text{cm}\end{array}$

Hence, the required distance below the mirror’s surface is $351\text{cm}$.