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A spherical interface, with radius of curvature \(10 \mathrm{cm},\) separates media of refractive index 1 and \(\frac{4}{3}\). The center of curvature is located on the side of the higher index. Find the focal lengths for light incident from each side. How do the results differ when the two refractive indices are interchanged?

Short Answer

Expert verified
For original indices: f = 30 cm, f' = -40 cm. Interchanged: f = -30 cm, f' = 40 cm.

Step by step solution

01

Understand the Refractive Indices and Medium Positions

The refractive index of the first medium is 1 (usually indicating air), and the refractive index of the second medium is \( \frac{4}{3} \). The spherical interface has a radius of curvature \( R = 10 \) cm, with its center of curvature on the side of the higher refractive index medium.
02

Apply the Lensmaker's Formula for Spherical Surfaces

Use the formula \( \frac{n_2}{f_2} - \frac{n_1}{f_1} = \frac{n_2-n_1}{R} \). Here, \( n_1 = 1 \) and \( n_2 = \frac{4}{3} \). When light enters from the first medium into the second medium, the equation simplifies to \( \frac{1}{f} = \frac{1}{3} \cdot \frac{1}{10} \).
03

Calculate the First Focal Length

Solving \( \frac{1}{f} = \frac{1}{30} \), we find \( f = 30 \) cm. Thus, the focal length for light entering from the first medium is 30 cm.
04

Consider Light Incident from the Second Medium

When light is incident from the second medium (refractive index \( = \frac{4}{3} \)), switch the roles of \( n_1 \) and \( n_2 \). The formula becomes \( \frac{1}{f'} = \left(1 - \frac{3}{4}\right) \cdot \frac{1}{10} = -\frac{1}{40} \).
05

Calculate the Second Focal Length

Solving \( \frac{1}{f'} = -\frac{1}{40} \), we find \( f' = -40 \) cm. Thus, the focal length for light entering from the second medium is -40 cm, indicating a virtual focus.
06

Interchange the Refractive Indices

Reevaluate the focal lengths by interchanging the refractive indices. For light entering from medium with \( n = \frac{4}{3} \) to air, use \( \frac{1}{f} = -\frac{1}{30} \), so \( f = -30 \) cm. For light entering from air to \( n = \frac{4}{3} \), use \( \frac{1}{f'} = \frac{1}{40} \), so \( f' = 40 \) cm.

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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 fundamental concept in optics. It measures how much light bends, or refracts, as it passes through a medium. In simplest terms, it is a ratio of the speed of light in a vacuum to the speed of light in the medium. The higher the refractive index, the more the light bends. Mathematically, it is represented as\[n = \frac{c}{v}\]where \(n\) is the refractive index, \(c\) is the speed of light in a vacuum, and \(v\) is the speed of light in the medium.
In our exercise, we have two media with different refractive indices: air (with a refractive index of 1) and another medium (with a refractive index of \(\frac{4}{3}\)). Light travels faster in air compared to the second medium. Understanding the refractive index is crucial for predicting how light will behave at the interface of these two materials. This determines how lenses and optical interfaces direct or focus light, playing a key role in focal length calculations.
Lensmaker's Formula
The Lensmaker's Formula is pivotal for designing lenses to achieve desired focal properties. However, it’s also applicable to spherical surfaces separating two media with different refractive indices. This scenario is exactly what we have in our exercise.
The formula used in the context of a spherical interface is given by:\[\frac{n_2}{f_2} - \frac{n_1}{f_1} = \frac{n_2-n_1}{R}\]where
  • \(n_1\) and \(n_2\) are the refractive indices of the two media,
  • \(f_1\) and \(f_2\) are the respective focal lengths when light enters the first and second medium,
  • \(R\) is the radius of curvature of the interface.
When the light travels from air (\(n_1 = 1\)) to the second medium (\(n_2 = \frac{4}{3}\)), or vice versa, this formula helps calculate the focal lengths to understand how the curved interface bends the light. The interplay between refractive indices, curvature, and focal points can significantly affect how imaging systems or optical devices are designed and used.
Focal Length Calculation
Focal length is the distance over which initially collimated rays are brought to a focus. For optical systems such as lenses, mirrors, or interfaces, understanding focal length is key to determining how a system will focus light.
In the context of the exercise, we apply the modified Lensmaker’s Formula to calculate the focal lengths when light approaches from either side of the spherical interface.
  • When light enters from air into the medium with a refractive index \(\frac{4}{3}\), the formula simplifies and results in a focal length \(f = 30\) cm. This focal point is real since the radius curvature is centered on the side of the medium with the higher index.
  • Conversely, when light enters the medium from the other side, with the roles of refractive indices switched, it yields a focal length \(f' = -40\) cm, implying a virtual focal point. A virtual focus means the rays appear to diverge from a common point.
If we evaluate the scenario with the refractive indices interchanged, the signs of the focal lengths reverse, providing a fascinating insight into how simple changes can alter light behavior significantly. These calculations highlight the predictable nature of light based on principles of geometric optics.

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Most popular questions from this chapter

A plano-cylindrical lens in air has a radius of curvature of \(10 \mathrm{cm},\) a refractive index of \(1.50,\) and an axial length of \(5 \mathrm{cm}\) Light from a point object is incident on the concave, cylindrical surface from a distance of \(25 \mathrm{cm}\) to the left of the lens. Find the position and length of the image formed by the lens

A diverging thin lens and a concave mirror have focal lengths of equal magnitude. An object is placed (3/2) f from the diverging lens, and the mirror is placed a distance \(3 f\) on the other side of the lens. Using Gaussian optics, determine the final image of the system, after two refractions (a) by a three-ray diagram and (b) by calculation.

A ray of light makes an angle of incidence of \(45^{\circ}\) at the center of the top surface of a transparent cube of index 1.414 Trace the ray through the cube.

Derive an expression for the transit time of a ray of light that travels a distance \(x_{1}\) through a medium of index \(n_{1},\) a distance \(x_{2}\) through a medium of index \(n_{2}, \ldots,\) and a distance \(x_{m}\) through a medium of index \(n_{m} .\) Use a summation to express your result.

A double convex lens has a diameter of \(5 \mathrm{cm}\) and zero thickness at its edges. A point object on an axis through the center of the lens produces a real image on the opposite side. Both object and image distances are \(30 \mathrm{cm},\) measured from a plane bisecting the lens. The lens has a refractive index of 1.52. Using the equivalence of optical paths through the center and edge of the lens, determine the thickness of the lens at its center.

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