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arlik [135]
3 years ago
8

The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular re

flection). 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 of diffuse reflection.
Suppose that the spotlight shines so that different parts of the beam reflect off of different two surfaces, one inclined at an angle alpha (from the horizontal) and one inclined at an angle beta. What would the angular separation between the rays reflected from the two surfaces?
Physics
1 answer:
Sever21 [200]3 years ago
5 0

Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

You might be interested in
In a Young’s interference experiment, the two slits are separated by 0.150 mm and the incident light includes two wavelengths: l
dezoksy [38]

Answer:

2.52 × 10⁻² cm

Explanation:

The distance of bright fringe from the center of the screen is given by the formula

                y = \frac{m\lambda D}{d}

Here, wavelength is λ, Distance of the screen from the slits is D, seperation between the

slits is d.

   

Separation between the slits, d = 0.15 mm

                                                     = 0.15 * 10^{-3} m

Distance of the screen from the slits = 1.40 m

We have a wavelength, λ1 = 540 nm

                                           = 540 * 10^{-9} m

By substituting all these values in the above equation we get

                y1 = mλD/d

                y1 = m(540 \times 10^{-9} m)(1.40 m)/(0.15 \times 10^{-3} m)

                y1 = m(5.04 * 10^{-3} m)

We have a wavelength, λ2 = 450 nm

                                           = 450 * 10^-9 m

By substituting all these values in the above equation we get

                y_2 = \frac{m\lambda D}{d}

                y_2 = m(450 * 10^{-9} m)(1.40 m)/(0.15 * 10^{-3} m)

                y_2 = m'(4.20 * 10^{-3} m)

According to the problem, these two distance are coincides with each other.

So,

                           y_1 = y_2

m(5.04 * 10^{-3} m) = m'(4.20 * 10^{-3} m)

by testing values, the above equation is satisfied only when, m = 5 and m' = 6

Then from the above we have

                           y1 = y2 = 0.0252 m

                                         = 2.52 × 10⁻² cm

8 0
3 years ago
What are all the different energy transformations and what are examples of them being used? PLEASE HELP ;)
Snezhnost [94]

There are many different forms of energy such as electrical, thermal, nuclear, mechanical, electromagnetic, sound, and chemical.

6 0
3 years ago
There are 30 students in Ms. Andrews’ fourth grade class. There are 6 times as many students in the entire fourth grade as there
KatRina [158]

Answer:

30×6=n

Explanation:

30 students in her class× 6 = number of students in the entire 4th grade.

6 0
3 years ago
NEED ANSWER PLEASE!!!!
olganol [36]

Answer:A, Concave

Explanation:

3 0
4 years ago
Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the
Svet_ta [14]

Answer:

b)  x_{cm} = 4.88 cm , c) x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃) and d)

x_{cm}’= 1.88 cm

Explanation:

The definition of mass center is

    x_{cm} = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

    D₂ = d₁ + d₂

    D₂ = 1.1 + 1.9

    D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

    D₃ = D₂ + d₂

    D₃ = 3.0 + 3.9

    D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

    M = m₁ + m₂ + m₃

    M = 24 + 12 + 56

    M = 92 g

    x_{cm} = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

    x_{cm} = 1/92 (448.8)

    x_{cm} = 4,878 cm

    x_{cm} = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at   d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

    x_{cm} ’= 1/M  (m₁ d₁ + m₂ d₂ + m₃ d₃)

   

   x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

    x_{cm}’= 1/92 ((24 (-1.9) +56 3.9)

    x_{cm}’= 1/92 (172.8)

    x_{cm}’= 1.88 cm

This distance is to the right of the central marble

3 0
3 years ago
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