/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 A small bulb is placed at the bo... [FREE SOLUTION] | 91Ó°ÊÓ

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A small bulb is placed at the bottom of a tank containing water to a depth of \(80 \mathrm{~cm}\). What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is \(1.33 .\) (Consider the bulb to be a point source.)

Short Answer

Expert verified
The area is approximately 26,500 cm².

Step by step solution

01

Understand the Concept of Critical Angle

For light traveling from a denser medium (water) to a less dense medium (air), there exists a specific angle known as the critical angle. Beyond this angle, light undergoes total internal reflection. The critical angle can be calculated using Snell's Law, which is given by \( n_1 \sin(\theta_c) = n_2 \sin(90^\circ) \), where \( n_1 \) is the refractive index of water (1.33), \( n_2 \) is the refractive index of air approximately (1.00), and \( \theta_c \) is the critical angle.
02

Calculate the Critical Angle

Applying Snell's Law, we have: \( \sin(\theta_c) = \frac{n_2}{n_1} = \frac{1}{1.33} \). Calculate \( \sin^{-1}(\frac{1}{1.33}) \) to find \( \theta_c \). This results in \( \theta_c \approx 48.75^\circ \).
03

Determine the Geometry of Emerging Light

From the point source at the bottom of the tank, light rays that meet the surface at angles less than the critical angle will emerge. The shape of the light emerging forms a circle directly above the bulb beneath the surface.
04

Calculate the Radius of the Circle

The light spreads outwards symmetrically, forming a cone with the vertex at the bulb. Using trigonometry, \( \tan(\theta_c) = \frac{r}{d} \), where \( r \) is the radius of the circle, and \( d \) is the depth of water (80 cm). Therefore, \( r = d \times \tan(\theta_c) \). Substituting the values, \( r = 80 \times \tan(48.75^\circ) \). Calculating this gives \( r \approx 91.82 \) cm.
05

Calculate the Area of the Circle

The area of the circle is given by the formula \( A = \pi r^2 \). Substituting the calculated radius, \( A = \pi \times (91.82)^2 \). This results in \( A \approx 26500 \) cm\(^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refractive Index
The refractive index is a crucial concept in understanding how light behaves when it moves between different mediums. It is a measure of how much a medium can bend (or refract) light. In simple terms, it tells you how ‘optically dense’ a material is. The refractive index is represented by the symbol \( n \). For example, the refractive index of water is \( 1.33 \), meaning light travels 1.33 times slower in water than in a vacuum. When light enters a medium with a different refractive index, its speed changes, which causes the light ray to bend. This bending effect is what makes underwater objects appear closer than they really are. Key points about refractive index:
  • Calculated as \( n = \frac{c}{v} \), where \( c \) is the speed of light in a vacuum and \( v \) is the speed of light in the medium.
  • Refractive indices are greater than 1 for any material with a density greater than vacuum.
  • The larger the refractive index, the more light bends when entering the material.
Critical Angle
The critical angle is the specific angle of incidence for which the angle of refraction is \( 90^\circ \). This occurs when light tries to move from a medium with a higher refractive index to a medium with a lower refractive index, for example, from water into air. Beyond this angle, light does not pass into the second medium; instead, it reflects entirely back into the first medium, a phenomenon known as total internal reflection. To find the critical angle \( \theta_c \), we can use the formula derived from Snell's Law:\[ \sin(\theta_c) = \frac{n_2}{n_1} \]where \( n_1 \) is the refractive index of the first medium (water, in this case, 1.33) and \( n_2 \) is that of the second medium (air, approximately 1.00).In our example:
  • Substituting the refractive indices gives: \( \sin(\theta_c) = \frac{1}{1.33} \).
  • The inverse sine function or \( \sin^{-1} \) results in \( \theta_c \approx 48.75^\circ \).
The calculated critical angle of \( 48.75^\circ \) means that light rays hitting the water surface at less than this angle will be refracted out into the air.
Snell's Law
Snell's Law is the fundamental law of refraction in optics, which explains how light is refracted—or bent—when it passes through boundaries of different optical materials. Named after Dutch mathematician Willebrord Snellius, this law is given by:\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]Where:
  • \( n_1 \) and \( n_2 \) are the refractive indices of the first and second mediums respectively.
  • \( \theta_1 \) is the angle of incidence, and \( \theta_2 \) is the angle of refraction.
This law helps us understand not just the critical angle, but also the general behavior of light as it traverses through various mediums. **Applications of Snell's Law:**
  • Used in calculating the critical angle for total internal reflection.
  • Helps design lenses and optical instruments.
  • Aids in explaining natural phenomena such as mirages.
In our example, Snell's Law allowed us to find the critical angle by setting the angle of refraction to \( 90^\circ \) (as at the critical angle), simplifying calculations between water and air. Without Snell's Law, we wouldn’t easily predict how light behaves at water's surface or solve problems involving optics.

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Most popular questions from this chapter

A man with normal near point \((25 \mathrm{~cm})\) reads a book with small print using a magnifying glass: a thin convex lens of focal length \(5 \mathrm{~cm}\). (a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass? (b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?

At what angle should a ray of light be incident on the face of a prism of refracting angle \(60^{\circ}\) so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is \(1.524 .\)

A tank is filled with water to a height of \(12.5 \mathrm{~cm}\). The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be \(9.4 \mathrm{~cm}\). What is the refractive index of water? If water is replaced by a liquid of refractive index \(1.63\) up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be \(40^{\circ}\). What is the refractive index of the material of the prism? The refracting angle of the prism is \(60^{\circ} .\) If the prism is placed in water (refractive index \(1.33)\), predict the new angle of minimum deviation of a parallel beam of light.

Use the mirror equation to deduce that: (a) an object placed between \(f\) and \(2 f\) of a concave mirror produces a real image beyond \(2 f\). (b) a convex mirror always produces a virtual image independent of the location of the object. (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

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