/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 A thin double-convex glass lens ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 38

A thin double-convex glass lens (with an index of 1.56 ) while surrounded by air has a 10 -cm focal length. If it is placed under water (having an index of 1.33 ) \(100\) \(\mathrm{cm}\) beyond a small fish, where will the guppy's image be formed?

Short Answer

Expert verified
The guppy's image is formed 36.63 cm on the opposite side of the lens.

Step by step solution

01

Understanding Lens Maker's Formula in Air

The lens maker's formula helps to calculate the focal length of the lens based on the refractive indices of the lens and the surrounding medium:\[ \frac{1}{f} = \left( \frac{n_{lens}}{n_{air}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]Here, the focal length \( f \) is given as 10 cm, \( n_{lens} = 1.56 \), and \( n_{air} = 1.0 \).Let's express the power of the lens: \( P = \frac{1}{f} = \frac{1}{0.10} = 10 \text{ D (diopters)} \). The power of the lens which is the same because the curvature will not change.
02

Adjusting the Focal Length for a Submerged Lens

When the lens is submerged in water, the new focal length \( f' \) can be found using the modified lens maker's formula:\[ \frac{1}{f_{water}} = \left( \frac{n_{lens}}{n_{water}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]Using the relationship \( P_{water} = P \cdot \frac{n_{air}}{n_{water}} \), plugging in \( n_{water} = 1.33 \):\[ f' = \frac{f \cdot n_{water}}{n_{lens} - n_{water}} = \frac{10 \cdot 1.33}{1.56 - 1.33} \approx 57.826 \text{ cm} \]
03

Applying the Lens Formula to Find Image Distance Underwater

Use the lens formula \( \frac{1}{f_{water}} = \frac{1}{v} - \frac{1}{u} \), where \( u = -100 \text{ cm} \) is the object distance (negative because object is on the same side as the lens) and \( f_{water} \approx 57.826 \text{ cm} \):\[ \frac{1}{57.826} = \frac{1}{v} - \frac{1}{(-100)} \]Solving for \( v \) yields:\[ \frac{1}{v} = \frac{1}{57.826} + \frac{1}{100} = \frac{0.0173 + 0.01}{1} = 0.0273 \implies v \approx 36.630 \text{ cm} \]
04

Concluding the Image Location

From the above calculation, the image distance \( v \) comes out to be approximately \( 36.63 \text{ cm} \). This indicates that the image of the guppy is formed at \( 36.63 \text{ cm} \) on the opposite side of the lens when placed underwater.

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Key Concepts

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

Double-convex lens
A double-convex lens is a type of lens that has two outward-curving surfaces. It's designed to converge light rays to a focal point, hence also called a converging lens.
These lenses are commonly used in glasses, cameras, and microscopes. They work by bending light due to their shape and the differences in refractivity between the lens material and the surrounding medium.
  • When light passes through the lens, it slows down, causing a bend towards the normal (the imaginary line perpendicular to the surface).
  • The amount of bending depends on the lens shape and the refractive indices involved.
  • This bending helps focus parallel light rays to meet at a focal point, creating a sharp image.
Understanding the physical principles behind a double-convex lens is key in fields like optics and photography.
Refractive index
The refractive index is a measure of how much the speed of light is reduced inside a medium compared to a vacuum. It is denoted by the symbol "n".
This index is crucial for determining how much light bends when it enters a new medium.
  • The higher the refractive index, the slower the light travels inside that medium.
  • In our original exercise, the lens has a refractive index of 1.56, which means light travels slower inside the glass lens than in surrounding air, forming a focal point.
  • When placed in water, which has a refractive index of 1.33, the difference in speed changes, altering how the light bends.
Understanding this concept is crucial for applications in lens design and optical engineering.
Focal length calculation
Calculating the focal length of a lens is fundamental to understanding how lenses form images. The focal length is the distance from the lens to the focal point, where light rays converge.
Using the Lens Maker's Formula, we can find how the focal length changes depending on the medium:
  • By using the lens maker's formula, we originally find the focal length of the lens in air by using the equation \[ \frac{1}{f} = \left( \frac{n_{lens}}{n_{air}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
  • This formula accounts for the curvature radii of the lens surfaces and the refractive index difference.
  • For a lens submerged in water, the equation modifies with water's refractive index, leading to a new focal length calculation.
This calculation is essential in optics to determine how lenses will behave in different environments.
Image formation
Image formation with lenses involves the convergence of light rays at a certain point to create a visual representation of an object. This process can be calculated using the lens formula:
The formula \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] is used, where "u" is the object distance and "v" is the image distance.
  • In our exercise, the object's distance was placed at 鈭�100 cm, indicating it鈥檚 on the same side of the lens, thus considered negative.
  • After calculating with the given focal length for the submerged lens, the image distance "v" resulted at approximately 36.63 cm.
  • This shows that the image forms on the opposite side of the lens, indicating it is a real, inverted image.
Understanding image formation is vital for designing lenses in optical instruments, determining how clear and accurate an image will be formed based on the lens configuration.

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Most popular questions from this chapter

A thin positive lens of focal length \(f_{L}\) is positioned very close to and in front of a front-silvered concave spherical mirror of radius \(R_{M} .\) Write an expression approximating the effective focal length of the combination in terms of \(f_{L}\) and \(R_{M}\).

An LED 0.60 cm tall is on the central axis 30.0 \(\mathrm{cm}\) in front of a convex spherical mirror. If the radius of curvature of the mirror is \(12.0 \mathrm{cm}\) determine the location of the image, describe it, and draw a ray diagram. How big is the image?

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 ( \(S\) and \(P\) ) 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.

A point source \(S\) sitting on the central axis of a positive thin lens is located (to the left) between one and two focal lengths from the lens. A concave spherical mirror is to be positioned to the right of the lens so that the final real image also lies at point \(S\). Where should the mirror be placed? Where should a convex spherical mirror be located to accomplish the same feat?

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?
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