I Review | Constants | Periodic Table Part A The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular reflection). However, you've likely been told that when you look at something, you are seeing light reflected from the object that you are looking at. This is reflection of a different sort: diffuse reflection. In this problem, you will see how diffuse reflection actually arises from the same law of reflection that you are accustomed to for reflections from mirrors. Consider a spotlight shining onto a horizontal mirror (Figure 1). If the light from the spotlight strikes the mirror at an angle 0, to the normal, what angle 0, to the normal would you expect for the reflected rays? Express your answer in terms of 0a. Submit Request Answer This simple rule of reflection no longer seems to hold for diffuse reflection. Consider the same spotlight but now reflecting from the surface of a table (Figure 2). Unlike the light reflected from the mirror, the light reflected from the table seems to go in all directions. If it didn't, then you'd only be able to see tables when you were at a specific angle to the lights above you! To understand why the light reflects in all directions, you must first look at a slightly simpler problem. Consider a flat surface, inclined downward from the horizontal by an angle a (Figure 3). The red line represents the surface and the red dotted line indicates the normal to this surface (the normal line). The two blue dashed lines represent horizontal and vertical. The angle between the incoming ray and the vertical is 0. Throughout this problem, assume that 0, is larger than a but smaller than 2a. (If you wish, you can determine the correct sign rules to generalize your results later.) • Part B Igure < 1 of 3 > Find the angle 0, between the reflected ray and the vertical. Express the angle between the reflected ray and the vertical in terms of a and 0. > View Available Hint(s) Submit Mirror

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Chapter36: Applications Of The Wave Model
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please help with both parts

Diffuse Reflection
3 of 24
I Review | Constants | Periodic Table
Part A
The law of reflection is quite useful for mirrors and other flat, shiny surfaces.
(This sort of reflection is called specular reflection). However, you've likely
been told that when you look at something, you are seeing light reflected
from the object that you are looking at. This is reflection of a different sort:
diffuse reflection. In this problem, you will see how diffuse reflection actually
arises from the same law of reflection that you are accustomed to for
reflections from mirrors.
Consider a spotlight shining onto a horizontal mirror (Figure 1). If the light from the spotlight strikes the mirror at an angle 0a to the normal, what angle 0, to the normal would you
expect for the reflected rays?
Express your answer in terms of 0a.
O, =
Submit
Request Answer
This simple rule of reflection no longer seems to hold for diffuse reflection. Consider the same spotlight but now reflecting from the surface of a table (Figure 2). Unlike the light reflected
from the mirror, the light reflected from the table seems to go in all directions. If it didn't, then you'd only be able to see tables when you were at a specific angle to the lights above you! To
understand why the light reflects in all directions, you must first look at a slightly simpler problem.
Consider a flat surface, inclined downward from the horizontal by an angle a (Figure 3). The red line represents the surface and the red dotted line indicates the normal to this surface
(the normal line). The two blue dashed lines represent horizontal and vertical. The angle between the incoming ray and the vertical is Oa. Throughout this problem, assume that 0a is
larger than a but smaller than 2a. (If you wish, you can determine the correct sign rules to generalize your results later.)
Part B
Figure
1 of 3
Find the angle 0, between the reflected ray and the vertical.
Express the angle between the reflected ray and the vertical in terms of a and Oa.
• View Available Hint(s)
ΑΣΦ
?
O, =
Submit
Mirror
Transcribed Image Text:Diffuse Reflection 3 of 24 I Review | Constants | Periodic Table Part A The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular reflection). However, you've likely been told that when you look at something, you are seeing light reflected from the object that you are looking at. This is reflection of a different sort: diffuse reflection. In this problem, you will see how diffuse reflection actually arises from the same law of reflection that you are accustomed to for reflections from mirrors. Consider a spotlight shining onto a horizontal mirror (Figure 1). If the light from the spotlight strikes the mirror at an angle 0a to the normal, what angle 0, to the normal would you expect for the reflected rays? Express your answer in terms of 0a. O, = Submit Request Answer This simple rule of reflection no longer seems to hold for diffuse reflection. Consider the same spotlight but now reflecting from the surface of a table (Figure 2). Unlike the light reflected from the mirror, the light reflected from the table seems to go in all directions. If it didn't, then you'd only be able to see tables when you were at a specific angle to the lights above you! To understand why the light reflects in all directions, you must first look at a slightly simpler problem. Consider a flat surface, inclined downward from the horizontal by an angle a (Figure 3). The red line represents the surface and the red dotted line indicates the normal to this surface (the normal line). The two blue dashed lines represent horizontal and vertical. The angle between the incoming ray and the vertical is Oa. Throughout this problem, assume that 0a is larger than a but smaller than 2a. (If you wish, you can determine the correct sign rules to generalize your results later.) Part B Figure 1 of 3 Find the angle 0, between the reflected ray and the vertical. Express the angle between the reflected ray and the vertical in terms of a and Oa. • View Available Hint(s) ΑΣΦ ? O, = Submit Mirror
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