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N76 [4]
2 years ago
11

Explain diffraction at a single slit (light)

Physics
1 answer:
Leto [7]2 years ago
4 0

Answer:

At some point on say, the receiving screen, light emanating from the left side of the slit will be out of phase (a difference of 1/2 wavelengths) from   light coming from the center of the slit.

Thus for every point that is left of the center of the slit, there will be a point on the right side of the slit that is out of phase,

There will be no light on the screen at that particular point and thus there will be a dark fringe there.

That is the basic explanation for the appearance of dark and bright fringes on the receiving screen.

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A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the e
gayaneshka [121]

Answer:

  V_inside = 36 V

Explanation:

<u>Given  </u>

We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.  

<u>Required</u>

We are asked to calculate the potential at the centre of the sphere  

<u>Solution</u>

The potential energy due to the sphere is given by equation

V = (1/4*π*∈o) × (q/r)                                          (1)

Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V  

V ∝ 1/r

The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next

V_1/V_2=r_2/r_1

V_inside/V_outside = r/R

V_inside = (r/R)*V_outside                               (2)

Now we can plug our values for r, R and V_outside into equation (2) to get  V_inside

V_inside = (1.2 m )/(0.600)*18

               = 36 V

  V_inside = 36 V

7 0
3 years ago
Read 2 more answers
A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wal
lisov135 [29]

Answer:

S=2.693m

Explanation:

Let point O be the origin of the room and the fly is crawling at point P as shown in the attached photo:

#Distance of the fly from the corner of the room is defined as:

OP=S=\sqrt{{(x_1-x_0)^2+(y_1+y_0)^2}

#Substitute the coordinate values in the above equation:

S=\sqrt{{(2.6m-0m)^2+(0.7m-0m)^2}}\\=\sqrt{7.25}\\=2.693m

Hence, the distance of the fly from the corner of the room is 2.693m

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If you want your investment or savings account to be easily turned into cash without losing its value, you should consider the?
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