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Parallel rays from a distant object are traveling in air and then are incident on the concave end of a glass rod with a radius of curvature of \(15.0 \mathrm{~cm} .\) The refractive index of the glass is \(1.50 .\) What is the distance between the vertex of the glass surface and the image formed by the refraction at the concave surface of the rod? Is the image in the air or in the glass?

Short Answer

Expert verified
The distance between the vertex of the glass surface and the image formed by the refraction at the concave surface of the rod is 30 cm and the image is inside the glass rod.

Step by step solution

01

Determine the value of R.

Given that the glass rod has a concave end with a radius of curvature of 15.0 cm. Since the rays of light are moving from air into the glass, the centre of curvature lies on the side from which the light approaches the glass, implying that R is -15 cm.
02

Apply Lens Maker's Formula

We need to find v, the image distance. For lenses, when parallel rays incident on the lens, they appear to be coming from the object at infinity. So, u will be considered as infinity. Applying these values into the lens maker's formula 1/v -1/u = (n-1)/R, use the fact that 1/infinity is actually 0. This simplifies the formula to 1/v = (n-1)/R.
03

Calculate the Value of v

For the given problem, the refractive index of glass, n is 1.50 and R is -15.0 cm. Plug in these values to the simplified formula obtained from step 2: 1/v = (1.50-1)/-15 = -0.033 cm^-1. Solve for v by taking the reciprocal. The image distance v will then be -30 cm.
04

Identify the Position

Since the value of v is negative, it means that the image's position is on the same side as the object, i.e., in the glass rod.

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Key Concepts

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

Refractive Index
The refractive index, also known as the index of refraction, is a fundamental concept in optics that measures how much light bends, or refracts, as it passes from one medium into another. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the given medium. The greater the refractive index of a medium, the more it slows down light, causing the light to bend more upon entering the material.

For instance, in the given exercise, the refractive index of glass is stated as 1.50. This means that light travels 1.50 times slower in glass than it does in a vacuum. The refractive index is central to many equations and principles in optics, including Snell's Law and the lens maker's formula, which is used to calculate the image distance in this problem.
Radius of Curvature
The radius of curvature refers to the radius of the spherical surface that forms a lens or a mirror. It's a measure of how 'curved' a surface is. When dealing with lenses, like the glass rod in our exercise, the radius of curvature determines how the light rays will converge or diverge.

In optics, it's important to note the sign convention for radius of curvature. A positive radius indicates a surface that's curved outward, like a biconvex lens, while a negative radius indicates a surface that's curved inward, like the concave end of the glass rod in our scenario. This sign convention helps in determining the direction of the light's travel after refraction or reflection. The exercise specified a radius of curvature of -15 cm, indicating a negative value that plays a crucial role in the lens maker's formula to find the image distance.
Image Distance
In the realm of optics, the image distance is the distance from the lens or mirror to the point where an image is formed. Understanding this concept is vital for visualizing and calculating the properties of the image, such as its size, orientation, and type (real or virtual).

In the exercise, the lens maker's formula is used to determine the image distance, denoted by 'v'. The formula relates the refractive index of the material, the radius of curvature, and the distances of the object and the image from the lens. Once 'v' is calculated, its sign will indicate the location of the image. A positive 'v' would mean the image is formed on the opposite side of the light source, while a negative 'v' indicates that the image forms on the same side as the light source. The result of the calculation showed that 'v' is -30 cm, implying that the image formed is within the glass itself.
Optics
Optics is the branch of physics that deals with the behavior and properties of light, including its interactions with matter and the construction of instruments that use or detect it. Optics is divided into several subfields: geometric optics studies light as rays, wave optics as waves, and quantum optics as particles.

In the context of geometric optics, which the lens maker's formula falls under, the path of light can be represented as straight rays that bend (refract) or bounce (reflect) when crossing material boundaries or when passing through lenses and mirrors. This simplification allows us to solve complex light interactions using algebraic equations, such as the one used in the exercise to find the image distance formed by a concave glass surface.

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Most popular questions from this chapter

You wish to project the image of a slide on a screen \(9.00 \mathrm{~m}\) from the lens of a slide projector. (a) If the slide is placed \(15.0 \mathrm{~cm}\) from the lens, what focal length lens is required? (b) If the dimensions of the picture on a \(35 \mathrm{~mm}\) color slide are \(24 \mathrm{~mm} \times 36 \mathrm{~mm},\) what is the minimum size of the projector screen required to accommodate the image?

A converging lens with a focal length of \(12.0 \mathrm{~cm}\) forms a virtual image \(8.00 \mathrm{~mm}\) tall, \(17.0 \mathrm{~cm}\) to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

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?

When a camera is focused, the lens is moved away from or toward the digital image sensor. If you take a picture of your friend, who is standing \(3.90 \mathrm{~m}\) from the lens, using a camera with a lens with an \(85 \mathrm{~mm}\) focal length, how far from the sensor is the lens? Will the whole image of your friend, who is \(175 \mathrm{~cm}\) tall, fit on a sensor that is \(24 \mathrm{~mm} \times 36 \mathrm{~mm} ?\)

The overall angular magnification of a microscope is \(M=-178 .\) The eyepiece has focal length \(15.0 \mathrm{~mm}\) and the final image is at infinity. The separation between the two lenses is \(202 \mathrm{~mm}\). What is the focal length of the objective? Do not use the approximation \(s_{1} \approx f_{1}\) in the expression for \(M\).

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