A thin flake of mica ( n = 1.58 ) is used to cover one slit ofa double slit interference arrangement. The central point on thevi
1 answer:
Answer:
the thickness of the mica is 6.64μm
Explanation:
By definition we know that the phase between two light waves that are traveling on different materials (in this case also two) is given by the equation
Where
L = Thickness
n = Index of refraction of each material
Our values are given as
Replacing our values at the previous equation we have
the thickness of the mica is 6.64μm
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A) Moment of inertia about an axis passing through the point where the two segments meet :
B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends:
What is Moment of inertia?
The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.
A) The moment of inertia about an axis passing through the point where the two segments meet is given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.
B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends
First step: determine the distance between the ends ( d )
After applying Pythagoras theorem
Next step : determine distance between the two axis
After applying Pythagoras theorem
Final step : Calculate the value of
applying Parallel Axis Theorem
Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: , Moment of inertia passing through the point where the midpoint of the line connects its two ends:
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Answer:
Option c) are perpendicular to the electric field
Explanation:
Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90 to the equipotential surface.
Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.
Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.