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The critical angle for total internal reflection at a liquid-air interface is \(42.5^{\circ} .\) (a) If a ray of light traveling in the liquid has an angle of incidence at the interface of \(35.0^{\circ},\) what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence at the interface of \(35.0^{\circ}\) , what angle does the refracted ray in the liquid make with the normal?

Short Answer

Expert verified
(a) \(\theta_2 \approx 26.1^{\circ}\) in air, (b) \(\theta_2 \approx 42.9^{\circ}\) in the liquid.

Step by step solution

01

Understanding the Critical Angle

The critical angle, given in this problem as \(42.5^{\circ}\), is the angle of incidence beyond which total internal reflection occurs when light transitions from a medium with higher index of refraction to one with a lower index.
02

Using Snell's Law for Part (a)

Since the angle of incidence \(35.0^{\circ}\) is less than the critical angle, we can use Snell's Law, \(n_1\sin(\theta_1) = n_2\sin(\theta_2)\). Here, \(n_1\) is the refractive index of the liquid, \(\theta_1 = 35.0^{\circ}\), and \(n_2 = 1\) for air. We need to find \(\theta_2\), the angle in air.
03

Calculate the Refracted Angle in Air

Reorganize Snell's Law to solve for \(\theta_2\): \(\sin(\theta_2) = \frac{n_1}{n_2} \sin(\theta_1)\). Given that \(n_2 = 1\), it simplifies to \(\sin(\theta_2) = n_1 \sin(35.0^{\circ})\). Assuming \(n_1\) can be approximated using the critical angle as reference, rearrange the relation: \(n_1 = \frac{1}{\sin(42.5^{\circ})}\), yielding \(\theta_2 = \arcsin(n_1 \cdot \sin(35.0^{\circ}))\). Calculate \(\theta_2\) to find the angle.
04

Using Snell's Law for Part (b)

Since the light travels from air to the liquid, Snell's Law \(n_2\sin(\theta_2) = n_1\sin(\theta_1)\) is used. Here, \(n_2 = 1\), and you need to calculate \(\theta_2\), the angle in the liquid.
05

Calculate the Refracted Angle in the Liquid

Rearrange Snell's Law to solve for \(\theta_2\): \(\sin(\theta_2) = \frac{n_2}{n_1} \sin(35.0^{\circ})\). Here, \(n_1 = \frac{1}{\sin(42.5^{\circ})}\) as earlier calculated from the critical angle. Solve for \(\theta_2 = \arcsin(\frac{\sin(35.0^{\circ})}{\frac{1}{\sin(42.5^{\circ})}})\) to find the angle in the liquid.

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Key Concepts

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

Total Internal Reflection
Total internal reflection is a fascinating phenomenon that occurs when a light ray travels through a medium with a higher refractive index and reaches an interface with a second medium of lower refractive index. Here's an interesting part: if the incoming angle, known as the angle of incidence, is greater than a certain specific angle called the critical angle, the light will not pass through. Instead, it will reflect back into the medium it came from.
This means for total internal reflection to occur, two critical conditions must be met:
  • The light must be moving from a denser to a rarer medium (e.g., from water to air).
  • The angle of incidence must be greater than the critical angle.
When these conditions are satisfied, total internal reflection happens, causing stunning effects like the sparkling appearance of diamonds.
Critical Angle
The critical angle is key to unlocking the magic of total internal reflection. Its value depends on the refractive indices of the two media involved. It is defined as the minimum angle of incidence where light is not refracted out of the denser medium but is instead reflected back.
For example, in the exercise provided, the critical angle is given as \(42.5^{\circ}\). This means that light traveling in the liquid, approaching the boundary at \(42.5^{\circ}\) or more, won't pass into the air; it will totally reflect back. To calculate the critical angle between two media, you can use:
  • \( \theta_c = \arcsin\left(\frac{n_2}{n_1}\right) \)
where \(\theta_c\) is the critical angle, \(n_1\) is the refractive index of the denser medium, and \(n_2\) is lighter.
Refractive Index
The refractive index is a measure that describes how much light bends, or refracts, as it passes from one medium to another. Every material has its distinct refractive index, indicating how dense it is compared to a vacuum.
For light traveling from one medium to another, the refractive index determines the change in speed and direction. It is denoted by the symbol \(n\). Mathematically, Snell's Law describes the relationship:
  • \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \)
where \(n_1\) and \(n_2\) are the refractive indices of the two media, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, respectively. Knowing the refractive index is crucial for predicting how light will behave when moving between different materials.
Angle of Incidence
The angle of incidence is the angle that an incoming ray, like a light beam, makes with the perpendicular or "normal" at the point of contact on a surface or interface of two media. Understanding this concept is crucial when studying light behavior such as reflection or refraction.
When compared to the critical angle, we can predict if light will continue into the next medium or reflect back. In the original exercise:
  • If the incident angle \(35.0^{\circ}\) is less than \(42.5^{\circ}\) critical angle, light refracts into air.
  • If the incident angle exceeds \(42.5^{\circ}\), total internal reflection would occur.
This showcases how knowing the angle of incidence affects light paths and optical phenomena.
Angle of Refraction
The angle of refraction is the angle formed between the refracted light ray and the normal. It represents the direction in which light travels after entering a new medium.
Snell’s Law helps calculate this using the equation:
  • \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \)
where \(\theta_2\) is the angle of refraction. Understanding how light bends when entering a new medium allows you to determine the direction of light travel, crucial when designing lenses and optical devices.
In our exercise, calculating the angle of refraction shows how light transitions between air and the liquid, illustrating real-world applications of Snell's Law.

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Most popular questions from this chapter

A beam of light strikes a sheet of glass at an angle of \(57.0^{\circ}\) with the normal in air. You observe that red light makes an angle of \(38.1^{\circ}\) with the normal in the glass, while violet light makes a \(36.7^{\circ}\) angle. (a) What are the indexes of refraction of this glass for these colors of light? (b) What are the speeds of red and violet light in the glass?

A certain birefringent material has indexes of refraction \(n_{1}\) and \(n_{2}\) for the two per- pendicular components of linearly polarized light passing through it. The corresponding wavelengths are \(\lambda_{1}=\lambda_{0} / n_{1}\) and \(\lambda_{0} / n_{2},\) where \(\lambda_{0}\) is the wavelength in vacuum. (a) If the crystal is to function as a quarter-wave plate, the number of wavelengths of each component within the material must differ by \(\frac{1}{4}\) . Show that the minimum thickness for a quarter-wave plate is $$d=\frac{\lambda_{0}}{4\left(n_{1}-n_{2}\right)}$$ (b) Find the minimum thickness of a quarter-wave plate made of siderite \(\left(\mathrm{FeO} \cdot \mathrm{CO}_{2}\right)\) if the indexes of refraction are \(n_{1}=1.875\) and \(n_{2}=1.635\) and the wavelength in vacuum is \(\lambda_{0}=589 \mathrm{nm} .\)

The refractive index of a certain glass is \(1.66 .\) For what incident angle is light reflected from the surface of this glass completely polarized if the glass is immersed in (a) air and (b) water?

(a) At what angle above the horizontal is the sun if sunlight reflected from the surface of a calm lake is completely polarized? (b) What is the plane of the electric-field vector in the reflected light?

At the very end of Wagner's series of operas Ring of the Nibelung, Brunnhilde takes the golden ring from the finger of the dead Siegfried and throws it into the Rhine, where it sinks to the bottom of the river. Assuming that the ring is small enough compared to the depth of the river to be treated as a point and that the Rhine is 10.0 \(\mathrm{m}\) deep where the ring goes in, what is the area of the largest circle at the surface of the water over which light from the ring could escape from the water?

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