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Determine the focal length of a planar-concave lens \(\left(n_{l}=1.5\right)\) having a radius of curvature of \(10 \mathrm{cm} .\) What is its power in diopters?

Short Answer

Expert verified
The focal length is \(20\, \text{cm}\) and the lens power is \(5 \, \text{D}\).

Step by step solution

01

Understand the Lens Maker's Formula

The Lens Maker's formula for a lens in the air is given by \( \frac{1}{f} = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \) where \(f\) is the focal length, \(n_l\) is the refractive index of the lens, \(R_1\) is the radius of curvature of the lens's first surface, and \(R_2\) is the radius of the second surface. For a planar-concave lens, the first surface is flat \((R_1 = \infty)\) and the second one is concave \((R_2 = -10\, \text{cm})\).
02

Plug in the Known Values

Using the formula, substitute \(n_l = 1.5\), \(R_1 = \infty\), and \(R_2 = -10 \, \text{cm}\). The equation becomes:\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{\infty} - \left(-\frac{1}{10}\right) \right) \]This simplifies to:\[ \frac{1}{f} = 0.5 \times \frac{1}{10} \]
03

Calculate the Focal Length

Simplify the equation:\[ \frac{1}{f} = 0.5 \times 0.1 = 0.05 \]Thus, the focal length \(f\) is:\[ f = \frac{1}{0.05} = 20 \, \text{cm} \]
04

Convert the Focal Length to Meters

Since the focal length needs to be in meters for the calculation of diopters, convert \(f = 20 \, \text{cm}\) to meters:\[ f = 0.20 \, \text{m} \]
05

Calculate the Power of the Lens

The optical power \(P\) in diopters (D) is given by \(P = \frac{1}{f}\) where \(f\) is in meters. Therefore:\[ P = \frac{1}{0.20} = 5 \, \text{D} \]

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Key Concepts

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

Planar-Concave Lens
A planar-concave lens is characterized by one flat surface and one inward-curving surface. This specific design causes light rays passing through the lens to diverge or spread out. The diverging effect is due to the concave portion bending the light outwards, making the lens useful in applications where light spreading is advantageous, such as in beam expanders or to correct certain vision imperfections.

In terms of optical engineering, this lens shape has a unique curvature configuration where one side is perfectly flat (also known as planar or infinite curvature), while the other side possesses a negative curvature radius. This negative curvature is crucial when applying the lens maker's formula, as it influences the behavior of light passing through.

Understanding planar-concave lenses is essential in optics, particularly when studying how different shapes of lenses affect light propagation. These lenses are integral in crafting devices that depend on the controlled divergence of light.
Focal Length Calculation
To calculate the focal length of a planar-concave lens, the lens maker's formula is very useful. This formula ties together the characteristics of the lens with its refractive index to determine how the lens bends light. The formula is expressed as:
  • \( \frac{1}{f} = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \) where \(n_l\) is the refractive index of the lens, and \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces.

For a planar-concave lens, the flat surface means \(R_1 = \infty\), simplifying the formula. Here,
  • \( \frac{1}{f} = (1.5 - 1) \left( \frac{1}{\infty} - \frac{1}{-10} \right) \)
  • Which reduces to \( \frac{1}{f} = 0.05 \)

The inverse of this value gives the focal length:
  • \(f = 20\ cm\)
Consequently, a negative focal length implies that the lens decreases convergence of light rays, consistent with its diverging nature.
Optical Power in Diopters
Optical power is a measure of a lens's ability to converge or diverge light, expressed in diopters (D). It is calculated using the focal length of the lens in meters. For lenses in diopter measurements, the formula used is:
  • \(P = \frac{1}{f}\)

where \(f\) is the focal length in meters. After converting the focal length from centimeters to meters (for example, 20 cm becomes 0.20 m), the calculation of optical power is straightforward.
In this case:
  • \(P = \frac{1}{0.20} = 5\ D\)

This value indicates that the lens has a power of -5 diopters, demonstrating its role as a diverging lens. The negative sign is typical for concave lenses, signifying their function in spreading light rather than focusing it.
Refractive Index
The refractive index (ulapped lies below 1. This ratio reflects how much light slows down as it passes through a material compared to its speed in a vacuum.

The refractive index is central to optical calculations because it affects how much the lens will refract (bend) light. Its value is essential when using the lens maker's formula to determine the focal length of a lens:
  • Refractive index \(n_l = 1.5\) implies light travels 1.5 times slower in the lens material than in a vacuum.

This property implies that light is bending significantly as it enters and exits the lens. A higher refractive index results in stronger bending, which is why glass lenses can typically focus light more effectively than other materials. Understanding the refractive index helps in designing sophisticated optical systems as it lawfully defines how light travels through varied media.

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Most popular questions from this chapter

Imagine that you are standing 5 feet from, and looking directly toward, a brass ball I foot in diameter hanging in front of a pawn shop. Describe the image you would see in the ball.

We would like to place an object \(45 \mathrm{cm}\) in front of a lens and have its image appear on a screen \(90 \mathrm{cm}\) behind the lens. What must be the focal length of the appropriate positive lens?

In an amusement park a large upright convex spherical mirror is facing a plane mirror \(10.0 \mathrm{m}\) away. A girl \(1.0 \mathrm{m}\) tall standing midway between the two sees herself twice as tall in the plane mirror as in the spherical one. In other words, the angle subtended at the observer by the image in the plane mirror is twice the angle subtended by the image in the spherical mirror. What is the focal length of the latter?

A jeweler is examining a diamond 5.0 mm in diameter with a loupe having a focal length of \(25.4 \mathrm{mm}\) (a) Determine the maximum angular magnification of the loupe. (b) How big does the stone appear through the magnifier? (c) What is the angle subtended by the diamond at the unaided eye when held at the near point? (d) What angle does it subtend at the aided eye?

A convenient way to measure the focal length of a positive lens makes use of the following fact. If a pair of conjugate object and (real) image points \((S \text { and } P \text { ) are separated by a distance } L>4 f\), there will be two locations of the lens, a distance \(d\) apart, for which the same pair of conjugates obtain. Show that $$f=\frac{L^{2}-d^{2}}{4 L}$$ Note that this avoids measurements made specifically from the vertex, which are generally not easy to do.

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