The human eye can respond to as little as 10^−18 J of light energy. For a wavelength at the peak of visual sensitivity, 550 nm,
2 answers:
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Answer:
E. d and O
Explanation:
"Light passing through a single slit forms a diffraction pattern somewhat different from those formed by double slits or diffraction gratings".
According to Huygens’s principle, "for each element of the wavefront in the slit emits wavelets. These are like rays that start out in phase and head in all directions. (Each ray is perpendicular to the wavefront of a wavelet.) Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel".
The destructive interference for a single slit is given by:
Where
d is the slit width
is the light's wavelength
is the angle relative to the original direction of the light
m is the order od the minimum
I represent the intensity
When the intensity and the wavelength are incident normally the angular as we can see on the expression above the angular separation just depends of the distance d and the wavelength O.
Answer:
819.78 m
Explanation:
<u>Given:</u>
- OA = range of initial position of the airplane from the point of observation = 375 m
- OB = range of the final position of the airplane from the point of observation = 797 m
= angle of the initial position vector from the observation point =
= angle of the final position vector from the observation point =
= displacement vector from initial position to the final position
A diagram has been attached with the solution in order to clearly show the position of the plane.
Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:
The magnitude of the displacement vector =
Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.
