A small bulb is placed at the bottom of a tank containing water to a depth of \sqrt7 m. The refractive index of water is 4 \over 3 . The area of the surface of water through which light from the bulb can emerge out is x \pi m2. The value of x is \underline\hspace2cm .

Geometrical Optics PYQ — JEE Main 2022 · 26 Jun · Shift 2 · Q70 with Solution

View written solutionFree

Correct answer: 9

Use the condition for emergence of light from water to air

Light from the bulb will emerge out of the water only if its angle of incidence at the water surface is less than or equal to the critical angle $C$ .

For water to air,

\sin C = \frac{n_{\text{air}}}{n_{\text{water}}} = \frac{1}{4/3} = \frac{3}{4}

So,

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

Then,

\cos C = \sqrt{1-\sin^2 C} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt7}{4}

Hence,

\tan C = \frac{\sin C}{\cos C} = \frac{3/4}{\sqrt7/4} = \frac{3}{\sqrt7}

Find the radius of the circular patch on the surface

The bulb is at depth

h = \sqrt7\ \text{m}

The limiting ray that can emerge strikes the surface at angle $C$ . Thus, if $r$ is the radius of the circular area on the surface through which light can emerge,

\tan C = \frac{r}{h}

So,

r = h \tan C = \sqrt7 \cdot \frac{3}{\sqrt7} = 3\ \text{m}

Find the area of the emergent patch

The area is a circle of radius $3$ m:

A = \pi r^2 = \pi (3)^2 = 9\pi\ \text{m}^2

Given that the area is $x\pi\ \text{m}^2$ , we get

x = 9

Compare with stored answer

Derived answer: $9$

Stored correct answer: $9$

They match.

Geometrical Optics question

Geometrical Optics question

More from Geometrical Optics

Geometrical Optics question

Geometrical Optics question

More from Geometrical Optics