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Chapter 5: Problem 44

In the process of constructing a doublet, an equiconvex thin lens \(L_{1}\) is positioned in intimate contact with a thin negative lens, \(L_{2}\) such that the combination has a focal length of \(50 \mathrm{cm}\) in air. If their indices are 1.50 and \(1.55,\) respectively, and if the focal length of \(L_{2}\) is \(-50 \mathrm{cm},\) determine all the radii of curvature.

Short Answer

Expert verified
The radii of curvature for the equiconvex lens are both 25 cm.

Step by step solution

01

Understand the Problem

We need to find the radii of curvature for a doublet lens system composed of two lenses in contact. Given the combined focal length, indices of refraction, and focal length of the negative lens, we will use lensmaker's equations.
02

Lensmaker's Equation for Thin Lenses

The lensmaker's equation for an equiconvex lens is given by: \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \), where \( n \) is the index of refraction, \( f \) is the focal length, \( R_1 \) and \( R_2 \) are the radii of curvature. For an equiconvex lens, \( R_1 = -R_2 = R \).
03

Calculate Combined Focal Length

For two thin lenses in contact, the combined power \( P \) is the sum of their individual powers: \( P = P_1 + P_2 \), where \( P = \frac{1}{f} \). Substituting, \( \frac{1}{50} = \frac{1}{f_1} + \frac{1}{-50} \).
04

Solve for Focal Length of Lens 1

Rearranging the equation from Step 3, we have \( \frac{1}{f_1} = \frac{1}{50} + \frac{1}{50} = \frac{2}{50} \), so \( f_1 = 25 \) cm.
05

Use Lensmaker's Equation for Lens 1

Using the lensmaker's equation: \( \frac{1}{25} = (1.50 - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) \). Simplifying gives: \( \frac{1}{25} = 0.5 \times \frac{2}{R} \), \( \frac{1}{25} = \frac{1}{R} \). Thus, \( R = 25 \) cm.

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Key Concepts

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

Lensmaker's Equation
The lensmaker's equation is an essential formula in optics that helps us calculate the focal length of a lens based on its physical attributes.
This equation takes into account the lens's refractive index and the radii of curvature of its surfaces.
  • The formula is given by: \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \) Where: \( f \) is the focal length of the lens, \( n \) is the refractive index, \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces. This formula provides a way to connect the physical construction of a lens with its optical qualities.
    When dealing with complex lens systems, it becomes a fundamental part of the calculation process to determine the lenses' behaviors in syncronization.
Focal Length
Focal length is the distance from the lens at which parallel rays of light converge to a single point after passing through the lens.
It is a crucial parameter that defines the imaging properties of a lens.
  • For a converging lens, the focal length is positive.
  • For a diverging lens, the focal length is negative. In an optical system involving multiple lenses, the combined focal length is influenced by the interaction between lenses. The focal length is instrumental in defining magnification and the field of view in optical devices.
    Shorter focal lengths mean stronger convergence or divergence, while longer focal lengths imply a weaker effect.
Radii of Curvature
The radii of curvature refer to the radii of the spherical surfaces that form the lens. For an equiconvex lens (like lens \( L_1 \) in the exercise), this means the front and back surfaces are symmetrical.
  • For such symmetrical lenses, we have: \( R_1 = -R_2 = R \).
  • A positive radius of curvature indicates a surface that bends outward, while a negative one implies an inward-curving surface. Determining these values accurately is crucial because they dictate how the lens refracts light, and ultimately, its focal length. In physical lens construction, these dimensions are precisely manipulated to achieve desired optical characteristics.
Refractive Index
The refractive index of a lens material is a measure of how much it reduces the speed of light passing through it.
It is a dimensionless number that indicates the bending effect a material has on light.
  • Materials with higher refractive indices slow down light more and bend light rays more aggressively.
  • The refractive index depends on the wavelength of light being used; however, typical calculations assume it to be consistent across common visible wavelengths. For example, in our exercise, lens \( L_1 \) has a refractive index of 1.50. This affects how it focuses light and combines with other lenses to affect the overall optical system.
    So, a meticulous choice of materials with the appropriate refractive index is imperative when designing and constructing lenses and lens systems.

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Most popular questions from this chapter

A thin convex lens \(L\) is positioned midway between two diaphragms: \(D_{1}, 4.0 \mathrm{cm}\) to its left, and \(D_{2}, 4.0 \mathrm{cm}\) to its right. The lens has a diameter of \(12 \mathrm{cm}\) and a focal length of \(12 \mathrm{cm} .\) The holes in \(D_{1}\) and \(D_{2}\) have diameters of \(12 \mathrm{cm}\) and \(8.0 \mathrm{cm},\) respectively. An axial object point is \(20 \mathrm{cm}\) to the left of \(D_{1} .\) (a) What is the image of \(D_{1}\) in the object space (i.e., as imaged by any lens to its left with light traveling left \() ?(\mathrm{b})\) What is the image of \(L\) in the object space? (c) What is the image of \(D_{2}\) in the object space? Give the size and location of that aperture's image. (d) Locate the entrance pupil and the aperture stop.

A nearsighted person with the same vision in both eyes has a far point at \(100 \mathrm{cm}\) and a near point at \(18 \mathrm{cm},\) each measured from her cornea. (a) Determine the focal length of the needed corrective contact lenses. (b) Find her new near point. Here you want to find the location of an object in front of the lens that will now be imaged at \(18 \mathrm{cm}\) in front of the lens.

A blade of grass standing \(10.0 \mathrm{mm}\) tall is \(150 \mathrm{mm}\) in front of a thin positive lens having a \(100 \mathrm{mm}\) focal length; \(250 \mathrm{mm}\) behind that first lens is a thin negative lens with a focal length of \(-75.0 \mathrm{mm}\). (a) Show that the first lens forms an image \(300 \mathrm{mm}\) behind it. (b) Describe that image. (c) What's its magnification? (d) Prove that the final image formed by both lenses is located \(150 \mathrm{mm}\) behind the negative lens. (e) What is the total magnification of the combination?

A jeweler is examining a diamond \(5.0 \mathrm{mm}\) 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 small planar mirror is attached to a thin vertical wire so that the mirror is parallel to a wall \(1.0 \mathrm{m}\) away. A horizontal scale is mounted flat on the wall opposite the mirror, whose center is directly opposite the zero mark on the scale. A horizontal laserbeam reflects off the mirror and hits the scale at \(5.0 \mathrm{cm}\) left of zero. The mirror is then rotated through an angle \(\alpha\) and the beam-scale spot of light moves left an additional \(15.0 \mathrm{cm} .\) Find \(\alpha\).
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