91Ó°ÊÓ

The object image relationship for a spherical reflecting surface is

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

Where${\mathit{n}}_{\mathbf{2}}$ and${\mathit{n}}_{\mathbf{1}}$ are the refraction index of the surfaces

sis th object distance from the vertex of the spherical surface and

$\mathit{s}\mathbf{\text{'}}$is the image distance from the vertex of the spherical surface andRis the radius of curvature of the spherical surface

Here we have

$n_2=1$ Since it is air.

$n_1=1.33$

The image is located at

role="math" localid="1663910899695" $$\begin{gathered} s\text{'}=-\left(7-3\right)m \\ s\text{'}=-4m \end{gathered}$$

It is negative since the image is on the opposite side of the outgoing light.

We know since the surface of water-air interaction is planar hence $R=∞$

Input this information in the formula

$$\begin{gathered} \frac{n_1}{s}+\frac{n_2}{s\text{'}}=\frac{n_2-n_1}{∞} \\ \frac{n_1}{s}+\frac{n_2}{s\text{'}}=0 \\ \frac{n_1}{s}=-\frac{n_2}{s\text{'}} \end{gathered}$$

Rearranging we have

$$\begin{gathered} s=\frac{n_{1}s\text{'}}{n_2} \\ =\frac{1.33×-4m}{-1} \\ s=5.332m \end{gathered}$$

The depth of water from this point is 5.332m