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3 years ago

according to the law of reflection what is the relationship between the angle 9f incidence and the angle of reflection

Physics

2 answers:

ryzh [129]3 years ago

7 0

According to the law of reflection, the angle of incidence and the angle of reflection are equal angles.

Alja [10]3 years ago

4 0

Answer:

Angle of incidence is equal to the angle of reflection.

Explanation:

Reflection is defined as the repropagation of light rays incident on a surface. An incident light is a light striking the surface of a plane mirror at an angle to the mirror. This incident light will reflect out at the same angle it strikes the plane. This angle is known as the angle of glance (g). The ray perpendicular to the plane surface is called the normal ray.

Therefore, the angle between the incident ray and the normal ray will be the angle of incidence (i) and the angle between the reflected ray and the normal ray will be the angle of reflection (r). Since the angle the incident ray made with the plane surface is equal to the angle that the reflected ray made with the plane surface then our angle of incidence will also be equal to the angle of reflection according to the second law of reflection i.e i=r.

i = r = 90°-g

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A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan

Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

$E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C$

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

$E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C$

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

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A spring that is stretched by hanging a 5kg mass. It's equilibrium length was 0.5 meters. Now the length of the spring is 1.6 me

gtnhenbr [62]

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2 years ago

A ray of light strikes a mirror with an angle of incident of 48º. Calculate: i) the angle of reflection and ii). the angle of de

Arada [10]

Answer:

I) angle of reflection = 48 degrees

II) angle of deviation = 0

Explanation:

Given that the:

angle of incident = 48º.

According to law of reflection which state that the angle of incident is equal to the angle of reflection.

I.) Therefore,

angle of reflection = 48º.

II.) There can only be deviation if the mirror has a diffused surface. But for a smooth plane mirror, there can never be any deviation.

Therefore ,

The angle of deviation = 0

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Answer:

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