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A ray of light traveling in a block of glass \((n=1.52)\) is incident on the top surface at an angle of \(57.2^{\circ}\) with respect to the normal in the glass. If a layer of oil is placed on the top surface of the glass, the ray is totally reflected. What is the maximum possible index of refraction of the oil?

Short Answer

Expert verified
The maximum possible index of refraction of the oil should be equal to \(1.52 \sin(57.2^{\circ})\). Get the sine of \(57.2^{\circ}\), multiply it by 1.52, and that should be the answer.

Step by step solution

01

Apply Snell's Law

According to Snell's Law, the ratio of the sine of the angle of incidence \(i\) to the sine of the angle of refraction \(r\) is equal to the ratio of the indices of refraction of the two media. \[ n_{1} \sin(i) = n_{2} \sin(r) \]Here, \(n_{1}\) is the index of refraction of the glass, \(n_{2}\) is what we're solving for (the maximum index of refraction for the oil), \(i\) is the angle of incidence, and \(r\) is the angle of refraction.
02

Substitute given values

The angle of incidence is given as \(57.2^{\circ}\) and the total reflection implies that the angle of refraction is \(90^{\circ}\). The index of refraction of the glass \(n_{1}\) is given as 1.52. Substituting these into the equation, we have:\[1.52 \sin(57.2^{\circ}) = n_{2} \sin(90^{\circ})\]The sine of \(90^{\circ}\) is 1, so the equation becomes:\[1.52 \sin(57.2^{\circ}) = n_{2} \]
03

Solve the equation

Solving for \(n_{2}\) will get the maximum index of refraction of the oil:\[n_{2} = 1.52 \sin(57.2^{\circ})\]The value of \(\sin(57.2^{\circ})\) can be obtained using a calculator and multiply that by 1.52.

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Key Concepts

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

Snell's Law
Snell's Law is a fundamental principle in optics that describes how light bends when passing from one transparent medium to another with a different index of refraction. The law states that the product of the index of refraction of the first medium and the sine of the angle of incidence is equal to the product of the index of refraction of the second medium and the sine of the angle of refraction.

This concept can be mathematically expressed as \[ n_1 \sin(i) = n_2 \sin(r) \] where \(n_1\) and \(n_2\) are the indices of refraction for the first and second medium respectively, \(i\) is the angle of incidence, and \(r\) is the angle of refraction. Using this law, we can solve for unknown variables, such as the angle of refraction or the index of refraction of the second medium, given the other known values. In our original exercise, the goal is to find the maximum possible index of refraction of the oil to prevent light from refracting out of the glass.
Angle of Incidence
The angle of incidence is the angle formed between the incoming light ray and the normal (an imaginary line perpendicular to the surface) at the point of incidence. In simpler terms, it's how steeply a ray of light strikes a surface compared to its straight-up position.\
\
For the given exercise, the light travels within the glass and hits the surface atop at an angle of \(57.2^\circ\), making it the angle of incidence. This angle is vital for calculating how light will behave upon trying to move into a new medium, which is critical in determining whether total internal reflection will occur or not. Understanding the angle of incidence helps predict the path of a light ray and is an integral part of applying Snell's Law.
Angle of Refraction
Similar to the angle of incidence, the angle of refraction is the angle between the refracted ray and the normal at the point of refraction. It indicates the direction a light ray will take upon entering a new medium. Generally, the light will bend towards the normal if it's slower in the new medium and away from it if it's faster.\
\
According to Snell's Law, changes in speed caused by the varying indices of refraction in different media bend the light ray, resulting in a change in its angle. For total internal reflection, like in our textbook problem, the angle of refraction reaches \(90^\circ\), meaning that the light doesn't pass into the second medium (oil) but instead reflects back entirely into the glass.
Total Internal Reflection
Total internal reflection is an optical phenomenon that occurs when a light ray travels from a medium with a higher index of refraction towards a medium with a lower index of refraction and hits the interface at an angle greater than a particular critical angle. At this point, no refraction occurs; instead, the light is completely reflected back into the denser medium.\
\
This is precisely what happens in the textbook exercise with the glass and oil layer. If the index of refraction of oil were lower than that of glass, it would have allowed the light ray to pass through but with the conditions given (angle of incidence at \(57.2^\circ\) and total internal reflection), the index of refraction of oil must be at its maximum yet still lower than glass to satisfy the conditions for total internal reflection. In layman's terms, the light 'gives up' on passing through and decides to bounce back entirely, similar to how a ball would bounce off a wall if thrown at a certain angle.

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Most popular questions from this chapter

Unpolarized light with intensity \(I_{0}\) is incident on two polarizing filters. The axis of the first filter makes an angle of \(60.0^{\circ}\) with the vertical, and the axis of the second filter is horizontal. What is the intensity of the light after it has passed through the second filter?

Light is incident normally on the short face of a \(30^{\circ}-60^{\circ}-90^{\circ}\) prism (Fig. \(\mathbf{P 3 3 . 5 0}\) ). A drop of liquid is placed on the hypotenuse of the prism. If the index of refraction of the prism is \(1.56,\) find the maximum index that the liquid may have for the light to be totally reflected.

Optical fibers are constructed with a cylindrical core surrounded by a sheath of cladding material. Common materials used are pure silica \(\left(n_{2}=1.450\right)\) for the cladding and silica doped with germanium \(\left(n_{1}=1.465\right)\) for the core. (a) What is the critical angle \(\theta_{\text {crit }}\) for light traveling in the core and reflecting at the interface with the cladding material? (b) The numerical aperture (NA) is defined as the angle of incidence \(\theta_{i}\) at the flat end of the cable for which light is incident on the core-cladding interface at angle \(\theta_{\text {crit }}\) (Fig. \(\mathbf{P 3 3 . 4 6}\) ). Show that \(\sin \theta_{\mathrm{i}}=\sqrt{n_{1}^{2}-n_{2}^{2}}\). (c) What is the value of \(\theta_{\mathrm{i}}\) for \(n_{1}=1.465\) and \(n_{2}=1.450 ?\)

A horizontal, parallel-sided plate of glass having a refractive index of 1.52 is in contact with the surface of water in a tank. A ray coming from above in air makes an angle of incidence of \(35.0^{\circ}\) with the normal to the top surface of the glass. (a) What angle does the ray refracted into the water make with the normal to the surface? (b) What is the dependence of this angle on the refractive index of the glass?

Light is incident in air at an angle \(\theta_{a}\) (Fig. \(\mathbf{P 3 3 . 5 2}\) ) on the upper surface of a transparent plate, the surfaces of the plate being plane and parallel to each other. (a) Prove that \(\theta_{a}=\theta_{a}^{\prime}\). (b) Show that this is true for any number of different parallel plates. (c) Prove that the lateral displacement \(d\) of the emergent beam is given by the relationship $$ d=t \frac{\sin \left(\theta_{a}-\theta_{b}^{\prime}\right)}{\cos \theta^{\prime}} $$ where \(t\) is the thickness of the plate. (d) A ray of light is incident at an angle of \(66.0^{\circ}\) on one surface of a glass plate \(2.40 \mathrm{~cm}\) thick with an index of refraction of \(1.80 .\) The medium on either side of the plate is air. Find the lateral displacement between the incident and emergent rays.

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