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A small mirror is attached to a vertical wall, and it hangs a distance of 1.80 m above the floor. The mirror is facing due east, and a ray of sunlight strikes the mirror early in the morning and then again later in the morning. The incident and reflected rays lie in a plane that is perpendicular to both the wall and the floor. Early in the morning, the reflected ray strikes the floor at a distance of 3.86 m from the base of the wall. Later on in the morning, the ray is observed to strike the floor at a distance of 1.26 m from the wall. The earth rotates at a rate of 15.0 per hour. How much time (in hours) has elapsed between the two observations?

Short Answer

Expert verified
1 hour has elapsed between the two observations.

Step by step solution

01

Understand the Problem Setup

The problem involves a mirror reflecting sunlight at two different times of the day, measured by the distance on the floor from the wall. The key is to relate these distances to the angles of incidence of sunlight, considering that the rays make the same angle with the mirror.
02

Apply Geometry and Reflection Principles

When light reflects off a mirror, the angle of incidence is equal to the angle of reflection. The sun rays reflect in the same plane that is perpendicular to both the wall and the floor. Given distances form a right triangle with the wall, mirror, and the point on the ground where the ray hits.
03

Calculate Initial Angle of Incidence

For the initial time: The distance from the base of the wall to where the ray meets the floor is 3.86 m. Using trigonometry, \( \tan(\theta_1) = \frac{3.86}{1.80} \) can be used to find \( \theta_1 \).
04

Calculate Later Angle of Incidence

For the later time: The distance changes to 1.26 m. Again using trigonometry, \( \tan(\theta_2) = \frac{1.26}{1.80} \) is used to find \( \theta_2 \).
05

Determine the Change in Angle over Time

Calculate the change in angle, \( \Delta \theta = \theta_1 - \theta_2 \). Convert these angles from radians to degrees if required for the next step.
06

Relate Angular Change to Earth's Rotation

The Earth rotates at a rate of 15 degrees per hour. Using this rate, calculate the time \( t \) by dividing \( \Delta \theta \) by the Earth's rotation rate using \( t = \frac{\Delta \theta}{15} \).
07

Calculate the Elapsed Time

Substitute the values for \( \Delta \theta \) to compute the elapsed time in hours between the two sunlight observations.

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Key Concepts

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

angle of incidence
The angle of incidence is a critical concept when discussing the reflection of light. When a ray of light strikes a mirror, it forms an angle with a line perpendicular to the mirror's surface. This angle is known as the angle of incidence. The law of reflection states that this angle is equal to the angle of reflection, which is the angle between the reflected ray and the perpendicular line.

In the exercise, the angle of incidence changes as the reflected ray shifts on the floor. The early morning distance of the reflected ray from the wall is 3.86 m. Using trigonometry, we determine the initial angle of incidence with the formula:
  • \( \tan(\theta_1) = \frac{3.86}{1.80} \)
Later in the morning, the distance reduces to 1.26 m, giving a new angle:
  • \( \tan(\theta_2) = \frac{1.26}{1.80} \)
Understanding these angles helps in calculating the elapsed time as they highlight the change in the Sun's position due to Earth's rotation.
geometry in physics
Geometry plays an integral role in solving physics problems, particularly in determining paths and angles involving light. In this exercise, geometry aids in deciphering how the light behaves as it reflects from the mirror onto the floor.

The distances given in the problem form right-angled triangles. The mirror's height from the floor and the distances at which the reflected rays hit the floor serve as two sides of these triangles:
  • Height of mirror from floor: 1.80 m (opposite side)
  • Distances on floor: 3.86 m and 1.26 m (adjacent sides)
By employing trigonometric functions, specifically the tangent function, we calculate the angles of incidence using the ratio of the lengths of the opposite to the adjacent sides of these triangles. This application of geometry allows us to understand the relationship between the mirror's positioning, the path of light, and the angles involved.
earth's rotation rate
The Earth's rotation has a profound effect on how we perceive sunlight. Our planet rotates approximately 15 degrees per hour, which alters the relative position of the Sun in our sky throughout the day.

In the exercise, the change in angle of sunlight on the floor provides a way to calculate the time between two observations of light reflection. This time is derived from the change in the angle of incidence due to Earth's rotation. Knowing the change in degrees per hour, specifically:
  • Earth rotates at 15 degrees per hour
We can determine the elapsed time, \( t \), with:
  • \( t = \frac{\Delta \theta}{15} \)
This calculation demonstrates how even small shifts in the angle can translate to measurable time intervals, illustrating the steady and predictable rotation of our planet.

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Most popular questions from this chapter

On the \(+y\) axis a laser is located at \(y=+3.0 \mathrm{cm} .\) The coordinates of a small target are \(x=+9.0 \mathrm{cm}\) and \(y=+6.0 \mathrm{cm} .\) The \(+x\) axis represents the edge-on view of a plane mirror. At what point on the \(+x\) axis should the laser be aimed in order for the laser light to hit the target after reflection?

A concave mirror (R 56.0 cm) is used to project a transparent slide onto a wall. The slide is located at a distance of 31.0 cm from the mirror, and a small flashlight shines light through the slide and onto the mirror. The setup is similar to that in Figure 25.18a. (a) How far from the wall should the mirror be located? (b) The height of the object on the slide is 0.95 cm. What is the height of the image? (c) How should the slide be oriented, so that the picture on the wall looks normal?

A mirror produces an image that is located 34.0 cm behind the mirror when the object is located 7.50 cm in front of the mirror. What is the focal length of the mirror, and is the mirror concave or convex?

An object that is 25 cm in front of a convex mirror has an image located 17 cm behind the mirror. How far behind the mirror is the image located when the object is 19 cm in front of the mirror?

An object is located 14.0 cm in front of a convex mirror, the image being 7.00 cm behind the mirror. A second object, twice as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located?

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