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aliina [53]
3 years ago
5

Light incident on a surface at an angle of 45° undergoes diffused reflection. At what angle will it reflect?

Physics
2 answers:
jok3333 [9.3K]3 years ago
8 0
<span>The correct answer is option D. i.e. at any angle between -90 and 90 degrees. In the diffused reflection ,the surface is rough and the direction of light in not the same as the incident light. It can be reflected at any angle. </span>
Korvikt [17]3 years ago
3 0

Answer:

D. at any angle between -90° and 90°

Explanation:

First of all, we need to understand the difference between specular reflection and diffuse reflection:

- Specular reflection occurs when the surface hit by the light is perfectly smooth. In this case, the law of reflection states that the angle at which the light is reflected is equal to the angle of incidence of the light (measured relative to the normal to the surface)

- Diffuse reflection occurs when the surface hit by the light is rough. In this case, the law of reflection is still valid for every ray of light hitting the surface; however, due to the imperfections of the surface, each ray of light hits the surface at a different angle of incidence. This means that the angle of reflection for each ray will be different, so light is reflected in any direction between -90 degrees and 90 degrees, which corresponds to the minimum and the maximum angle of reflection measured relative the normal to the surface.

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Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawi
Nina [5.8K]

Answer:

T_2 = 0.592

Explanation:

Given

T_1 = 1.48s

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}

Take LCM

\frac{1}{C_{up}} = \frac{1+1}{C}

\frac{1}{C_{up}}= \frac{2}{C}

Cross Multiply

C_{up} * 2 = C * 1

C_{up} * 2 = C

Make C_{up} the subject

C_{up} = \frac{1}{2}C

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

C_1 = 2 * C_{up}

C_1 = 2 * \frac{1}{2}C

C_1 = C

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

C_p = C+C

C_p = 2C

So, the total capacitance (C2) is:

\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}

\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}

Take LCM

\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}

\frac{1}{C_2} = \frac{5}{2C}

Inverse both sides

C_2 = \frac{2}{5}C

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

T\ \alpha\ C

Convert to equation

T\ =kC

Make k the subject

k = \frac{T}{C}

k = \frac{T_1}{C_1} = \frac{T_2}{C_2}

\frac{T_1}{C_1} = \frac{T_2}{C_2}

Make T2 the subject

T_2 = \frac{T_1 * C_2}{C_1}

Substitute values for T1, C1 and C2

T_2 = \frac{1.48 * \frac{2}{5}C}{C}

T_2 = \frac{1.48 * \frac{2}{5}}{1}

T_2 = \frac{0.592}{1}

T_2 = 0.592

Hence, the time constance of (b) is 0.592 s

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