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Chapter 34: Problem 26

The left end of a long glass rod \(8.00 \mathrm{~cm}\) in diameter, with an index of refraction of 1.60 , is ground and polished to a convex hemispherical surface with a radius of \(4.00 \mathrm{~cm}\). An object in the form of an arrow \(1.50 \mathrm{~mm}\) tall, at right angles to the axis of the rod, is located on the axis \(24.0 \mathrm{~cm}\) to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

Short Answer

Expert verified
The position of the image formed by paraxial rays incident on the convex surface is 52.2 cm on the other side of the glass and the height is 3.27 mm. The image is erect.

Step by step solution

01

Calculate the Reciprocal of the Focal Length

We must first find the reciprocal of the focal length (1/f) using the lens maker's formula: 1/f = (n-1)(1/R1 - 1/R2). Since the left side of the lens is flat, R1 approaches infinity, thus 1/R1 equals 0. Therefore, the formula can be simplified to: 1/f = (n-1)(-1/R2) = (1.60 - 1)(-1/-0.04), resulting in 1/f = 0.015 m^-1.
02

Find the Image Distance Using the Lens Formula

Next, use the lens formula, which is 1/f = 1/do + 1/di. But in our case we take the direction from which the light comes as positive which means we take do as negative and we get 1/di = 1/f + 1/do = 0.015 - 1/(-0.24) = 0.015 +0.0041667 = 0.01916667, so di (image distance) is 1/0.0191667 = 52.2 cm.
03

Calculate the Magnification

To find the image height, we have to calculate the magnification. The magnification equation is: m = - di / do = - (52.2 cm) / -24.0 cm. Thus, the magnification factor is 2.18.
04

Find the Image Height

Next, we multiply the object height by the magnification factor to find the image height. Hence, the image height is: h = m * ho = 2.18 * 1.5 mm = 3.27 mm.
05

Determine the Orientation of the Image

The image is erect if it's orientation is the same as the object. Since the magnification is positive, this means that the image must also be in the same orientation as the object. Hence, the image is erect.

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Key Concepts

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

Index of Refraction
The index of refraction is a measure of how much light bends, or refracts, when it enters a material. This property is crucial in optics, as it helps determine how lenses and other optical devices focus light. An index of refraction is represented by the letter \( n \).
- If \( n = 1 \), the medium doesn't affect light speed, like in a vacuum. - If \( n > 1 \), light slows down, bending towards the normal, typical for lenses or glass.In our context, the glass rod has an index of refraction of 1.60. This means light entering the rod slows down significantly, resulting in a noticeable bend. The higher the index of refraction, the greater the ability of the material to bend light and focus it. This value directly influences calculations involving focal lengths and image positions, as it's a part of the lens maker鈥檚 formula.
Lens Maker's Formula
The lens maker's formula is vital for understanding how lenses shape light to form images. This formula calculates the focal length of a lens based on its shape and the index of refraction. It鈥檚 shown as:\[\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]- \( f \) is the focal length.- \( n \) is the index of refraction.- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.For a convex hemispherical surface, and considering a flat opposite side, the formula simplifies greatly, with \( 1/R_1 \) being zero. In our exercise involving the glass rod, applying this formula helps compute the reciprocal of the focal length necessary for accurate optical calculations. This simplified formula played a pivotal role in earlier steps of solving the problem by removing the variables associated with a non-curved side of the glass.
Paraxial Rays
Paraxial rays are beams of light that make small angles with an optical axis and stay close to it throughout their travel. These rays are essential in simplifying optical calculations because they help maintain focus and limit distortions. When discussing the position and height of an image formed by a lens or optical surface, paraxial ray approximation is often used: - Because paraxial rays are assumed to be close to the optical axis, their paths can be easily predicted. - They lead to more accurate calculations and conclusions regarding image formation. In practical applications, only rays very close to the axis are considered paraxial, but this idea aids in evaluating lenses like the one used in the exercise, effectively facilitating finding the image using geometric optics principles.
Magnification in Optics
Magnification in optics refers to how much larger or smaller an image appears, as compared to the original object. This concept is critical for understanding the outcomes of using lenses or mirrors in imaging.The formula to calculate magnification is:\[m = -\frac{d_i}{d_o}\]- \( d_i \) is the image distance from the lens, - \( d_o \) is the object distance. In our example, magnification helps determine both the orientation and size of the resulting image.
- A positive magnification signifies an erect image.- A negative magnification indicates an inverted image. By using the problem鈥檚 given values, calculating the magnification confirms the image is not only larger by a factor of about 2.18 but is also erect, since the magnification is positive. This deeper understanding allows students to relate mathematical outcomes to physical orientations and appearances of images formed by lenses.

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Most popular questions from this chapter

A \(1.20-\mathrm{cm}\) -tall object is \(50.0 \mathrm{~cm}\) to the left of a converging lens of focal length \(40.0 \mathrm{~cm}\). A second converging lens, this one having a focal length of \(60.0 \mathrm{~cm},\) is located \(300.0 \mathrm{~cm}\) to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it \(I_{1}\) ) formed by the lens with a focal length of \(40.0 \mathrm{~cm}\). (b) \(I_{1}\) is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

A small tropical fish is at the center of a water-filled, spherical fish bowl \(28.0 \mathrm{~cm}\) in diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_{1}=+12.0 \mathrm{~cm}\) and \(R_{2}=+28.0 \mathrm{~cm} .\) The index of refraction is \(1.60 .\) (a) Compute the position and size of the image of an object in the form of an arrow \(5.00 \mathrm{~mm}\) tall, perpendicular to the lens axis, \(45.0 \mathrm{~cm}\) to the left of the lens. (b) A second converging lens with the same focal length is placed \(3.15 \mathrm{~m}\) to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens \(45.0 \mathrm{~cm}\) to the right of the first.

To determine the focal length \(f\) of a converging thin lens, you place a \(4.00-\mathrm{mm}\) -tall object a distance \(s\) to the left of the lens and measure the height \(h^{\prime}\) of the real image that is formed to the right of the lens. You repeat this process for several values of \(s\) that produce a real image. After graphing your results as \(1 / h^{\prime}\) versus \(s\), both in \(\mathrm{cm}\), you find that they lie close to a straight line that has slope \(0.208 \mathrm{~cm}^{-2}\). What is the focal length of the lens?

A frog can see an insect clearly at a distance of \(10 \mathrm{~cm}\). At that point the effective distance from the lens to the retina is \(8 \mathrm{~mm} .\) If the insect moves \(5 \mathrm{~cm}\) farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus? (a) \(0.02 \mathrm{~cm}\), toward the retina; (b) \(0.02 \mathrm{~cm}\), away from the retina; (c) \(0.06 \mathrm{~cm}\), toward the retina; (d) \(0.06 \mathrm{~cm}\), away from the retina.
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