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Chapter 18: Problem 394

Determine the focal length \(\mathrm{f}\) of a glass lens of index \(1.5\) surrounded by air for which \(\mathrm{r}_{1}=+10, \mathrm{r}_{2}=+9\) a) when the lens thickness is zero, b) when the thickness is \(+1\).

Short Answer

Expert verified
In summary, the focal length of the glass lens is: a) \(f = -180\) units when the lens thickness is zero. b) \(f = -270\) units when the lens thickness is 1 unit.

Step by step solution

01

Review the Lensmaker's Equation

To solve this problem, we will need to use the Lensmaker's Equation, which is given by: \( \frac{1}{f} = (n-1) \left( \frac{1}{r_1} - \frac{1}{r_2} + \frac{(n-1) \cdot d}{n \cdot r_1 \cdot r_2} \right) \) In this equation, f is the focal length, n is the index of refraction, r1 and r2 are the radii of curvature for the two lens surfaces, and d is the thickness of the lens.
02

Calculate the focal length for the lens with zero thickness

We will use the Lensmaker's Equation to find the focal length, f, of the lens when the lens thickness, d, is zero: \( \frac{1}{f} = (1.5-1) \left( \frac{1}{10} - \frac{1}{9} + \frac{(1.5-1) \cdot 0}{1.5 \cdot 10 \cdot 9} \right) \) Simplify the expression: \( \frac{1}{f} = 0.5 \left( \frac{1}{10} - \frac{1}{9} \right) \) Now, calculate the result: \( \frac{1}{f} = 0.5 \times (\frac{-1}{90}) = \frac{-1}{180} \) Find the focal length, f: \( f = -180 \) Thus, the focal length of the lens when the thickness is zero is -180 units.
03

Calculate the focal length for the lens with a thickness of 1

Now, we will use the Lensmaker's Equation to find the focal length, f, of the lens when the lens thickness, d, is 1: \( \frac{1}{f} = (1.5-1) \left( \frac{1}{10} - \frac{1}{9} + \frac{(1.5-1) \cdot 1}{1.5 \cdot 10 \cdot 9} \right) \) Simplify the expression: \( \frac{1}{f} = 0.5 \left( \frac{1}{10} - \frac{1}{9} + \frac{0.5 \cdot 1}{1.5 \cdot 90} \right) \) Now, calculate the result: \( \frac{1}{f} = 0.5 \times (\frac{-1}{90} + \frac{1}{270}) = \frac{-1}{270} \) Find the focal length, f: \( f = -270 \) Thus, the focal length of the lens when the thickness is 1 unit is -270 units. In summary, the focal length of the glass lens is: a) -180 units when the lens thickness is zero. b) -270 units when the lens thickness is 1 unit.

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Key Concepts

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

Focal Length Calculation
Understanding how to calculate the focal length of a lens is essential in designing optical systems such as cameras, eyeglasses, and microscopes. The focal length of a lens is the distance from the lens at which parallel rays of light converge to a single point, known as the focal point. When a lens has a thick design, its focal length deviates from the idealized case of a thin lens.

In our specific exercise, we use the Lensmaker's Equation to find out the focal length of a glass lens. When the lens thickness is considered zero, which simplifies the calculation, we found the focal length to be -180 units. The negative sign indicates a diverging lens meaning that instead of focusing light, the lens spreads the beams apart. As the lens thickness increases to 1 unit, the focal length changes to -270 units, highlighting how thickness influences the focusing property of a lens.

Remember, the accuracy of focal length calculation can make a significant difference in any application that relies on precise optical focussing. So, the concept isn't just pivotal for academic exercises but is also crucial in practical applications where precision optics are required.
Index of Refraction
At the heart of optics is a property of materials called the index of refraction, denoted as 'n'. It's a dimensionless number that describes how light propagates through a medium. Specifically, it is the ratio of the speed of light in a vacuum to the speed of light in the material. The index of refraction affects how much light bends, or refracts, when entering a medium.

In our exercise, we've dealt with a glass lens with an index of refraction of 1.5. This number is critical in the Lensmaker's Equation as it comes into play when determining the focal length. High-index materials bend light more than low-index materials and are thus able to focus light over shorter distances, leading to thinner lenses with the same optical power. The concept of refraction is key in understanding lenses, prisms, and a myriad of other optical components.
Radii of Curvature
The shape of a lens is determined by its radii of curvature, which are the radii of the spheres from which the lens surfaces are segments. A lens surface can be convex (bulging outwards), concave (inward), or even flat, and each of these shapes is described using the radius of curvature. A positive radius indicates a surface that is convex toward the incoming light, and a negative radius signifies a concave surface.

In the context of our problem, the lens has two radii of curvature for its surfaces, noted as +10 and +9. These values influence how the light is refracted when it passes through the lens and are central to calculating the focal length using the Lensmaker's Equation. As a rule, when dealing with lenses, understanding the contributions of these curvatures allows you to predict how the lens will direct light, which is a foundational concept in lens design and application.

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Most popular questions from this chapter

You are given a thick double convex lens, whose faces have radii of curvature of \(24 \mathrm{~cm}\) and \(36 \mathrm{~cm}\); the lens thickness is \(2 \mathrm{~cm}\), and it is made of material with an index of refraction equal to \(1.524\). (a) Locate the principal points. (b) If an object is placed \(40 \mathrm{~cm}\) from the nearer face of the lens, find, by calculation, the position of the image, (c) Find the image location graphically. (d) Treating the lens as thin, and measuring distances from its center, calculate the position of the image. Note the error involved.

A thick lens, which has principal points at \(\alpha=1.2 \mathrm{~cm}\) and \(\beta=-0.8 \mathrm{~cm}\), forms an image of a distant object \(19.2 \mathrm{~cm}\) from the second surface of the lens. Find the position of the image of an object placed \(38.8 \mathrm{~cm}\) from the first face. The sign convention used is that distances are positive if measured in the direction of light propagation and negative if measured in the opposite direction; \(\alpha\) is measured from the first surface of the lens and \(\beta\) from the second.

Find the ratio of the focal lengths of a glass lens in water and in air. The refractive indices of the glass and water are \(1.5\) and \(1.33\) respectively.

A thin lens of focal length 10 inches in air and refractive index \(1.53\) is immersed in water \((\mathrm{n}=1.33)\). What is its focal length in water?

A thin lens has focal length \(50 \mathrm{~mm}\). What is the magnification for the following object conjugates: \(20 \mathrm{~mm}\), \(-100 \mathrm{~mm},+100 \mathrm{~mm} ?\)
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