A small bulb is placed at the bottom of a tank containing water to a depth of $80 ext{ cm}$. What is the area of the surface of water through which light from the bulb can emerge out? Given, the refractive index of water is $\frac{4}{3}$. (Consider the bulb to be a point source).

Q: A small bulb is placed at the bottom of a tank containing water to a depth of $80\text{ cm}$. What is the area of the surface of water through which light from the bulb can emerge out? Given, the refractive index of water is $\frac{4}{3}$. (Consider the bulb to be a point source).

Given Data: Depth of water in the tank ($h$) = $80\text{ cm} = 0.8\text{ m}$ Refractive index of water ($n$) = $\frac{4}{3}$ Formula and Concept:\nWhen a point source of light is placed at the bottom of a tank of water, the light rays strike the water-air interface at various angles of incidence. Rays incident at an angle greater than the critical angle ($C$) undergo total internal reflection and remain inside the water. Rays incident at angles less than or equal to the critical angle emerge into the air. \nThus, the area of the water surface through which light emerges is a circle of radius $r$ (often called the radius of the circle of illumination). \nThe relation for critical angle $C$ is given by: $$\sin C = \frac{1}{n}$| \nFrom the geometry of the cone of light, the radius $r$ of the circular patch is related to the depth $h$ and critical angle $C$ by: $$r = h \tan C$$ Step-by-Step Calculation: Find $\sin C$: $$\sin C = \frac{1}{4/3} = \frac{3}{4} = 0.75$$ Find $\cos C$ using trigonometric identity: $$\cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$ Find $\tan C$: $$\tan C = \frac{\sin C}{\cos C} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}$$ Calculate the radius $r$: $$r = h \tan C = 80 \times \frac{3}{\sqrt{7}} = \frac{240}{\sqrt{7}}\text{ cm}$$ Given $\sqrt{7} \approx 2.6457$: $$r = \frac{240}{2.6457} \approx 90.71\text{ cm} = 0.907\text{ m}$$ Calculate the area ($A$) of the surface: $$A = \pi r^2 = \pi \left(\frac{240}{\sqrt{7}}\right)^2 = \pi \times \frac{57600}{7}$| Using $\pi \approx 3.1416$: $$A = 3.1416 \times \frac{57600}{7} = \frac{180956.16}{7} \approx 25850.88\text{ cm}^2 = 2.585\text{ m}^2$$ Answer:\nThe area of the surface of water through which light from the bulb can emerge is approximately $25850.88\text{ cm}^2$ (or $2.585\text{ m}^2$).

Question

A small bulb is placed at the bottom of a tank containing water to a depth of $80	ext{ cm}$. What is the area of the surface of water through which light from the bulb can emerge out? Given, the refractive index of water is $\frac{4}{3}$. (Consider the bulb to be a point source).

Accepted Answer

Given Data: Depth of water in the tank ($h$) = $80	ext{ cm} = 0.8	ext{ m}$ Refractive index of water ($n$) = $\frac{4}{3}$ Formula and Concept:
When a point source of light is placed at the bottom of a tank of water, the light rays strike the water-air interface at various angles of incidence. Rays incident at an angle greater than the critical angle ($C$) undergo total internal reflection and remain inside the water. Rays incident at angles less than or equal to the critical angle emerge into the air. 
Thus, the area of the water surface through which light emerges is a circle of radius $r$ (often called the radius of the circle of illumination). 
The relation for critical angle $C$ is given by: $$\sin C = \frac{1}{n}$| 
From the geometry of the cone of light, the radius $r$ of the circular patch is related to the depth $h$ and critical angle $C$ by: $$r = h 	an C$$ Step-by-Step Calculation: Find $\sin C$: $$\sin C = \frac{1}{4/3} = \frac{3}{4} = 0.75$$ Find $\cos C$ using trigonometric identity: $$\cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - \left(\frac{3}{4}ight)^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$ Find $	an C$: $$	an C = \frac{\sin C}{\cos C} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}$$ Calculate the radius $r$: $$r = h 	an C = 80 	imes \frac{3}{\sqrt{7}} = \frac{240}{\sqrt{7}}	ext{ cm}$$ Given $\sqrt{7} \approx 2.6457$: $$r = \frac{240}{2.6457} \approx 90.71	ext{ cm} = 0.907	ext{ m}$$ Calculate the area ($A$) of the surface: $$A = \pi r^2 = \pi \left(\frac{240}{\sqrt{7}}ight)^2 = \pi 	imes \frac{57600}{7}$| Using $\pi \approx 3.1416$: $$A = 3.1416 	imes \frac{57600}{7} = \frac{180956.16}{7} \approx 25850.88	ext{ cm}^2 = 2.585	ext{ m}^2$$ Answer:
The area of the surface of water through which light from the bulb can emerge is approximately $25850.88	ext{ cm}^2$ (or $2.585	ext{ m}^2$).