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2 years ago

The angle of incidence for a ray of light passing through the centre of curvature of a concave mirror is.

Physics

1 answer:

evablogger [386]2 years ago

8 0

The angle of incidence for a ray of light passing through the center of curvature of a concave mirror is 0°.

The angle of incidence is the angle between the surface's normal and the incident ray. For a concave mirror, the normal of the surface is along the center of the curvature, and a ray of light passed through a center of curvature passes through the normal of the surface.

The ray of light retreats its path making a zero angle of reflection. The law of reflection state that the angle of incidence is equal to the angle of reflection; therefore, the angle of incidence of a concave surface passed through the center of curvature is zero degrees.

Learn more about the angle of incidence here:

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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

Alexus [3.1K]

Answer:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = $\pi d^{2} /4$ = (3.142 x $(2*10^{-3} )^{2}$ )/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field is given as

E = $\sqrt{2I/ce_{0} }$

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

$e_{0}$ = 8.85 x 10^9 m kg s^-2 A^-2

E = $\sqrt{2*318.2/3*10^8*8.85*10^9}$ = 1.55 x 10^-8 v/m

6 0

3 years ago

describe what happens to the arrangement of water molecules as ice is\to liquid to vapor be sure to explain the temperature chan

alex41 [277]

Water freezes at the freezing point to ice then melts to the melting turning it to liquid and vapor causing gas in precipitation

8 0

3 years ago

1. write the meaning of the following terms:electrostatic,neutral, positively charged, negatively charged, coulomb,microcoulomb,

dybincka [34]

ELECTROSTATIC:

relating to stationary electric charges or fields as opposed to electric currents.

NEUTRAL:

nor negative nor positive/having no charge

POSITIVELY CHARGED:

positive charge occurs when the number of protons exceeds the number of electrons

NEGATIVELY CHARGED:

negative charge occurs when the number of electrons exceeds the number of protons.

COULOMB:

SI unit for electric charge. One coulomb is equal to the amount of charge from a current of one ampere flowing for one second.

MICROCOULOMB:

a unit of electrical charge equal to one millionth of a coulomb.

NANOCOULOMB:

Nanocoulombs are a unit of charge 1,000,000,000 times smaller than Coulomb.

CONSERVATION OF CHARGE:

constancy of the total electric charge in the universe or in any specific chemical or nuclear reaction

QUANTISATION OF CHARGE:

Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge.

5 0

2 years ago

Which will ensure laboratory safety during the experiment? Check all that apply. using beaker tongs to handle the hot beaker rea

Aleks [24]

Answer: • using beaker tongs to handle the hot beaker.

• checking the beaker for chips prior to heating on the hot plate.

• Turning off the hot plate after use

Explanation:

The options that will ensure laboratory safety during the experiment will be:

• using beaker tongs to handle the hot beaker.

• checking the beaker for chips prior to heating on the hot plate.

• Turning off the hot plate after use.

We should note that the beaker tongs are simply used in the holding of the beakers that have hot liquids in them. Also, it s vital for the hot plate to be turned off after its use so as to prevent accident.

7 0

3 years ago

Read 2 more answers

A box of volume V has a movable partition separating it into two compartments. The left compartment contains 3000 particles, the

storchak [24]

Answer:

a) V1 = 4V - V2/3 and V2 = 4V - 3V1

b) Δe = 4000V - 4000V2 + 9000V1

Explanation:

Let V represent volume of the box containing the two compartments

V1 represents compartment of the left compartment

V2 represents compartment of the right compartment

Momentum of the compartments before impact:

3000V1 + 1000V2

Momentum of the compartments after impact:

V(3000 + 1000) = 4000V

a) To obtain the volume of each compartment, that is, V1 and V2, we say:

Momentum before impact = Momentum after impact

3000V1 + 1000V2 = 4000V

∴ V1 = 4000V - 1000V2/3000 = 4V - V2/3

Also, V2 = 4000V - 3000V1/1000 = 4V - 3V1

b) Change in entropy,Δe = 4000V1 - 1000V2

By substituting the V1 and V2, we have:

4000(4V - V2)/3 - 1000(4V - 3V1)

16000V - 4000V2/3 - 4000V + 3000V1

16000V - 4000V2 - 12000V + 9000V1

∴ Δe = 4000V - 4000V2 + 9000V1

6 0

3 years ago