مسائل في الضوء الرجاء المساعدة في حلها بالتفصيل

[قلوب ستار]

Determine the number of grating lines necessary to resolve the 589.59 nm and 589.00 nm sodium lines in second order.
a. 999 b. 680 c. 500
d. 340 e. 380


Two slits separated by 0.10 mm are illuminated with green light ( = 540 nm). Calculate the distance (in cm) from the central bright-region to the fifth bright band if the screen is 1.0 m away.
a. 2.3 b. 2.5 c. 2.7
d. 2.1 e. 2.0

11. The electric fields arriving at a point P from three coherent sources are described by E1 = E0 sin t, E2 = E0 sin (t + /4) and E3 = E0 sin (t + /2). Assume the resultant field is represented by Ep = ER sin (t + ). The amplitude of the resultant wave at P is
a. E0. b. 1.5E0. c. 1.7E0.
d. 2.7E0. e. 2.9E0.

12. Monochromatic light ( = 500 nm) is incident on a soap bubble (n = 1.40). What is the wavelength of the light (in nm) in the bubble film?
a. 255 b. 500 c. 700
d. 357 e. 422

13. In a Newton’s rings apparatus, find the phase difference (in radians) when an air wedge of 500 nm thickness is illuminated with red light ( = 640 nm).
a. 13 b. 11 c. 9
d. 7 e. 3

If you stand closer to a concave mirror than a distance of one focal length, the image you see is
a. real and inverted. b. real and upright.
c. virtual and inverted. d. virtual and upright.
e. none of the above because you do not get an image.

A light ray is partially reflected and partially refracted at a boundary between two media, the upper one having index of refraction n, the lower one having index of refraction n, as shown in the figure. The reflected ray is perpendicular to the refracted ray when

a. n = n tan incident. b. n = n cot incident. c. n = n .
d. n = n .


. How much electromagnetic energy is contained in each cubic meter near the Earth’s surface if the intensity of sunlight under clear skies is 1000 W/m2?
a. 3.3  10–6 J b. 3.3 J c. 0.003 J
d. 10–4 J e. 3.0  105 J



2. At a distance of 10 km from a radio transmitter, the amplitude of the E-field is 0.20 volts/meter. What is the total power emitted by the radio transmitter?
a. 10 kW b. 67 kW c. 140 kW
d. 245 kW e. 21 kW

[ف.نووي]

Determine the number of grating lines necessary to resolve the 589.59 nm and 589.00 nm sodium lines in second order.
a. 999 b. 680 c. 500
d. 340 e. 380
solution:

1- resolving power must a grating have if these wavelengths are to be distinguished
R=λ/Δλ=((λ_1+λ_2)⁄2)/(λ_2-λ_1 )=(589.30nm)/(589.59nm-589.00nm)=999
2- To resolve these lines in the second-order spectrum
N=R/m=999/2=500

[ف.نووي]

Monochromatic light ( = 500 nm) is incident on a soap bubble (n = 1.40). What is the wavelength of the light (in nm) in the bubble film?
a. 255 b. 500 c. 700
d. 357
e. 422
solution:

the wavelength of light in medium whose refraction index is n is
λ_n=λ/n=(500 nm)/(1.40)=357 nm

[ف.نووي]

How much electromagnetic energy is contained in each cubic meter near the Earth’s surface if the intensity of sunlight under clear skies is 1000 W/m2?
a. 3.3  10–6 J b. 3.3 J c. 0.003 J
d. 10–4 J e. 3.0  105 J

This energy is contained in a cylindrical volume of unit area and height h,
where h = c∆t. Light travels with speed c,
so it travels a distance of h = c∆t in a time interval ∆t.
If we set h = 1 m,
then ∆t = (1/3)*10-8 s.
So E = (S/(3*108 s))(1 m2) is the energy that is contained in a cubic meter.
E = ((1000 W/m2)/(3*108 s))(1 m2) = 3.33*10-6 J.
.

[ف.نووي]

At a distance of 10 km from a radio transmitter, the amplitude of the E-field is 0.20 volts/meter. What is the total power emitted by the radio transmitter?
a. 10 kW b. 67 kW c. 140 kW
d. 245 kW e. 21 kW
solution:

S=POWER/area=P/(4πr^2 )----->P=4πr^2 S

But S=(E_m^2)/(2Cµ_0 ) ---->→P=(2πr^2 E_m^2)/(Cµ_0 )=
(2πr^2 E_m^2)/(Cµ_0 )=(2π×((10^4)^2×(0.2)^2)/(4π×10^(-7)×3×10^8 )=67kW

[ف.نووي]

A light ray is partially reflected and partially refracted at a boundary between two media, the upper one having index of refraction n, the lower one having index of refraction n, as shown in the figure. The reflected ray is perpendicular to the refracted ray when

a. n' = n tanθ incident.
b. n' = n cot θincident.
c. n' = n .
d. n' = n .
e. n' = n sec θincident.
Solution:

If .. θ_r+ θ_i⁡= 90 >>>>> 1

Applying the law of refraction in this case::
n sin( θ_i) = n' sin (θ_r) >>>>> 2

Compensation of 1 in 2, we find that::

n sin( θ_i) = n' sin (90 - θ_i) >>>>> 3

It is well known that:

(sin (90 - θ_i) = cos (θ_i))

If you become a 3 on the following image::

n sin( θ_i) = n' cos (θ_i)

n'/n=(sinθ_i)/(cosθ_i )=tan⁡(θ_i
n'=n tan⁡(θ_i

[ف.نووي]

Two slits separated by 0.10 mm are illuminated with green light ( = 540 nm). Calculate the distance (in cm) from the central bright-region to the fifth bright band if the screen is 1.0 m away.
a. 2.3 b. 2.5 c. 2.7
d. 2.1 e. 2.0
solution:

y_bright=λLm/d=(540×〖10〗^(-6) m×1m×5)/(0.1×〖10〗^(-3) m)=2.7

[ف.نووي]

In a Newton’s rings apparatus, find the phase difference (in radians) when an air wedge of 500 nm thickness is illuminated with red light ( = 640 nm).
a. 13 b. 11 c. 9
d. 7 e. 3
solution:

Δ=2t+λ/2=2×500nm+640nm/2 =13

[ف.نووي]

9. If you stand closer to a concave mirror than a distance of one focal length, the image you see is
a. real and inverted. b. real and upright.
c. virtual and inverted. d. virtual and upright.
e. none of the above because you do not get an image.
الاختيار هوd لا يوجد طريقه للحل لكن من الفهم