Solve Ray Optics Problem to Find delta y

In summary, to find the spot on the wall where the reflected light from a tilted mirror will hit, you need to add 5 degrees to the original angle of reflection, which is now 37 degrees, giving a new angle of 42 degrees. Using this angle in the equation, the spot on the wall can be calculated to be 24.45 cm higher. However, when submitting the answer in WebAssign, the angle used should be labeled differently from the original angle of reflection.
  • #1
FlipStyle1308
267
0
Here is the problem I am working on:

Sunlight enters a room at an angle of theta = 32° above the horizontal and reflects from a small mirror lying flat on the floor. The reflected light forms a spot on a wall that is d = 1.9 m behind the mirror, as shown in Figure 26-42. If you now place a pencil under the edge of the mirror nearer the wall, tilting it upward by 5.0°, how much higher on the wall (delta y) is the spot?

26-42alt.gif


Here is what I have done:

delta y = yf - yi = d tan thetaf - d tan thetai = (1.9m)(tan37 - tan32) = 24.45 cm

In WebAssign, I submitted both 24.45 cm and 24 cm, but neither was accepted. What am I doing wrong? My instructor said my answer is not close to that in WebAssign.
 
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  • #2
FlipStyle1308 said:
Here is what I have done:

delta y = yf - yi = d tan thetaf - d tan thetai = (1.9m)(tan37 - tan32) = 24.45 cm

In WebAssign, I submitted both 24.45 cm and 24 cm, but neither was accepted. What am I doing wrong? My instructor said my answer is not close to that in WebAssign.
You have the wrong angle. When the mirror's flat, the normal line to its surface points straight up. But when it's tilted, where does the normal line point? Now apply the law of reflection about that new normal.
 
  • #3
Draw a separate diagram for the second situation, and put in the angles carefully.

Edit: Doc beat me to it.
 
  • #4
Doc Al said:
You have the wrong angle. When the mirror's flat, the normal line to its surface points straight up. But when it's tilted, where does the normal line point? Now apply the law of reflection about that new normal.

When the mirror's tilted, the normal line points upward, away from the wall, right? So does this mean the angle I should replace is supposed to be 85 degrees?
 
  • #5
FlipStyle1308 said:
When the mirror's tilted, the normal line points upward, away from the wall, right? So does this mean the angle I should replace is supposed to be 85 degrees?
When you tilt the mirror, the angle θ changes, so the reflection angle changes. But these angles are relative to the plane of the mirror, and the mirror is no longer flat on the table.
 
  • #6
OlderDan said:
When you tilt the mirror, the angle θ changes, so the reflection angle changes. But these angles are relative to the plane of the mirror, and the mirror is no longer flat on the table.

Okay, so I drew myself a picture. Originally the normal line is straight up, as Doc Al said. As the mirror is tilted 5 degrees, the normal line is now pointing 5 degrees in the upper left direction. Theta changes and increases to 37 degrees. But on the other side of the normal line, theta is 42 degrees, is that right? So I replace 37 with 42, or am I still doing something wrong?
 
  • #7
FlipStyle1308 said:
Okay, so I drew myself a picture. Originally the normal line is straight up, as Doc Al said. As the mirror is tilted 5 degrees, the normal line is now pointing 5 degrees in the upper left direction. Theta changes and increases to 37 degrees. But on the other side of the normal line, theta is 42 degrees, is that right? So I replace 37 with 42, or am I still doing something wrong?
I would give the angle on the right a name other than theta, since the angle you need to locate the spot is now theta + the 5 degree rotation, but yes, 42 degrees is the angle between the reflected beam and the floor.
 
  • #8
OlderDan said:
I would give the angle on the right a name other than theta, since the angle you need to locate the spot is now theta + the 5 degree rotation, but yes, 42 degrees is the angle between the reflected beam and the floor.

Thank you!
 
Last edited:

Related to Solve Ray Optics Problem to Find delta y

1. What is Ray Optics?

Ray Optics is a branch of optics that deals with the behavior of light as it travels through different media, such as air, water, or glass. It focuses on the study of the propagation of light in the form of rays and how they interact with various surfaces and objects.

2. What is delta y in Ray Optics?

In Ray Optics, delta y refers to the displacement of a ray of light as it passes through a medium with a change in its direction due to refraction or reflection. It is measured as the perpendicular distance between the incident ray and the refracted or reflected ray.

3. How do you solve Ray Optics problems to find delta y?

Solving Ray Optics problems to find delta y involves using the laws of reflection and refraction, namely Snell's law and the law of reflection, along with the properties of light rays, such as the angle of incidence and the angle of refraction or reflection. By applying these laws and principles, we can calculate the value of delta y in a given scenario.

4. What factors affect the value of delta y in Ray Optics?

The main factors that affect the value of delta y in Ray Optics are the angle of incidence, the refractive index of the medium, and the angle of refraction or reflection. The angle of incidence determines the direction of the incident ray, while the refractive index of the medium affects the speed and direction of light as it passes through it. The angle of refraction or reflection is also crucial in finding the value of delta y as it determines the change in direction of the light ray.

5. Can Ray Optics problems be solved using mathematical calculations?

Yes, Ray Optics problems can be solved using mathematical calculations based on the laws and principles of reflection and refraction. These calculations involve basic trigonometry and geometry, and they can be solved using equations such as Snell's law and the law of reflection. However, some problems may require a more complex approach, such as using the concept of total internal reflection, to find the value of delta y.

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