Question

14. All lights around a swimming pool are switched off,
except for a bright point-light source kept at the
bottom of a swimming pool filled with clear water
of refractive index 4/3. As a result, only a circular
patch of 6 m diameter of the water surface is visible
to spectators standing around the swimming pool
Which of the following gives the nearest value of
the depth of the pool?
A 1.6 m
B. 5.0 m
C. 2.6 m
D. 4.0 m​

Answer 1

Answer:

2.6 is the correct answer

Answer 2

The nearest value of the depth is C. 2.6 m.

Given,

The swimming pool is filled with clear water of refractive index 4/3. A circular patch of 6 m diameter of the water surface is visible.

To Find,

The nearest value of the depth of the pool =?

Solution,

We can solve the given as follows:

It is given that the refractive index of water is 4/3. The diameter of the circular patch is 6 m.

$Refractive\: index\: of\: water, u_{w} = \frac{4}{3}$

$Diameter = 6\: m$

The given situation is observed when there is a total internal reflection of light. Total internal reflection of light is a phenomenon in which light travels from a denser medium to a rarer medium.

Now, the formula for finding the critical angle is:

$sin\: C = \frac{1}{u_{w} }$

Substituting the value,

$sin\: C = \frac{3}{4}$

Also, sin is given as the opposite divided by the hypetenuse.

Therefore,

$sin\: C = \frac{r}{\sqrt{r^{2} + h^{2} } }$

Since the diameter is 6 m, the radius will be 3 m.

Substituting the values,

$\frac{3}{4} = \frac{3}{\sqrt{3^{2} + h^{2} } }$

$\frac{3}{4} = \frac{3}{\sqrt{9 + h^{2} } }$

$4 = \sqrt{9 + h^{2}$

Squaring both sides,

$16 = 9 + h^{2}$

$h^{2} = 16 - 9 = 7$

$h = \sqrt{7} = 2.6\: m$

Hence, the nearest value of the depth is 2.6 m.

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