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Goshia [24]
3 years ago
9

Laser light of wavelength lambda passes through a thin slit of width a and produces its first dark fringes at angles of +/- 30 d

egree with the original direction of the beam. The slit is then reduced in size to a circle of diameter a. When the same laser light is passed through the circle, the first dark fringe occurs at
A) +/- 66.9 degree.
B) +/- 45.0 degree.
C) +/- 37.6 degree.
D) +/- 36.6 degree.
E) +/- 15.0 degree.
Physics
1 answer:
Scilla [17]3 years ago
7 0

Answer:

C- ±37.6°

Explanation:

This is the case of a single slit diffraction. In this case, the dark fringe occurs at

sin θ = ± mλ/a

where m = 1

θ = ± 30°

Hence we have sin ± 30° = ± λ/a

±0.5 = ± λ/a

Hence, we have a = 2λ

For a circular aperture, the condition for the first dark fringe is

D = diameter of circle = a = 2λ

so we have sin θ1 = 1.22λ/a

sin θ1 = 1.22λ/2λ

hence sin θ1 = 0.61

θ1 = sin⁻¹0.61 = ±37.6°

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Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV
Korolek [52]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.

How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Give your answer in centimeters, to two significant figures.

Answer:

 The radius  of  the dish is R = 18cm

Explanation:

  From the question we are told that

     The radius of the orbit is  = R = 35,000km = 35,000 *10^3 m

    The power output of the power is  P = 1 kW = 1000W

   The electric vector amplitude is given as E = 0.1 mV/m = 0.1 *10^{-3}V/m

    The area of thereciever  is   A_R = 5cm^2

Generally the intensity of the dish is mathematically represented as

         I = \frac{P}{A}

Where A is the area orbit which is a sphere so this is obtained as

          A = 4 \pi r^2

              = (4 * 3.142 * (35,000 *10^3)^2)

              =1.5395*10^{16} m^2

  Then substituting into the equation for intensity

          I_s  =  \frac{1000}{1.5395*10^{16}}

            = 6.5*10^ {-14}W/m2

 Now the intensity received by the dish can be mathematically evaluated as

              I_d = \frac{1}{2}  * c \epsilon_o E_D ^2

  Where c is thesped of light with a constant value  c = 3.0*10^8 m/s

              \epsilon_o is the permitivity of free space  with a value  8.85*10^{-12} N/m

              E_D is the electric filed on the dish

So  since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

      Now making the eletric field intensity the subject of the formula

                  E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }

substituting values

                 E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }

                       = 7*10^{-6} V/m

The incident power on the dish is what is been reflected to the receiver

                P_D = P_R

Where P_D is the power incident on the dish which is mathematically represented as

              P_D = I_d A_d

                   = \frac{1}{2}  c \epsilon_o E_D^2  (\pi R^2)

And  P_R is the power incident on the dish which is mathematically represented as

                 P_R = I_R A_R

                       = \frac{1}{2} c \epsilon_o E_R^2 A_R

Now equating the two

                \frac{1}{2}  c \epsilon_o E_D^2  (\pi R^2) =  \frac{1}{2} c \epsilon_o E_R^2 A_R

   Making R the subject we have

                   R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }

Substituting values

                   R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }

                     R = 18cm

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3 years ago
What is the pressure of a 300 lb. object on a 100 sq. in. area?
denis23 [38]

Answer:

D. 3 psi

Explanation:

Pressure is defined as force acting per unit area and is numerically expressed as:

P= \frac {F}{A}

where P represent pressure, F is force and A is area where the force acts. Substituting 300 lb for force and 100 sq.in for area then pressure,

P=\frac {300 lb}{100 in^{2}}= 3 \ psi

Therefore, from the choices given, option D, 3psi is the right choice.

3 0
3 years ago
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