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Scorpion4ik [409]
3 years ago
5

Monochromatic light of wavelength λ illuminates a pair of thin parallel slits of width a separated by a distance d at normal inc

idence, producing an interference pattern on a distant screen. a) (10 points) Explain two modifications to the experiment which would cause the maxima (bright spots) of the diffraction pattern to move away from the center. b) (10 points) Explain two modifications to the experiment which would cause the minima (dark spots) of the interference pattern to move towards the center.
Physics
1 answer:
Gwar [14]3 years ago
4 0

Answer:

Explanation:

a ) If x be the position of n the bright fringe on the screen , following formula holds .

x = n (λD / d) ; λ is wave length , D is screen distance and d is slit separation .

If we increase the value of λ or  wave length, x will increase so central fringe along with all the fringes will shift away from the centre .

If we increase the value of D or screen distance , it will also increase x ,  so fringes along with central fringe  will shift away from the center.

b ) Fringes ,whether bright or dark , both shift together , either towards or away from the center .

So to move the dark spots towards the center , we need to do the opposite to what we did in the first case , ie decrease the wavelength or decrease the screen distance .

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Answer:

1 millions times stronger

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3 years ago
The magnitude of electrical force between a pair of charged particles is ____ as much when the particles are moved half as far a
Gnesinka [82]

The magnitude of the electrical force between a pair of charged particles is 4 Times as much when the particles are moved half as far apart.

This can be easily understood by Columb's law,

F_{new} = \frac{kQ_{1}Q_{2}}{r^{2}}

which state's that the amount of electrical force experienced by two charged particles is inversely proportional to the square of the distance between them.

∴ \frac{F_{new} }{F_{old} } = \frac{Distance_{new}^{2}  }{Distance_{old}^{2}  }

Now, we know the new distance is half the original distance,

F_{new} = \frac{kQ_{1}Q_{2}}{\frac{r}{2}^{2} } \\F_{new} = 4\frac{kQ_{1}Q_{2}}{r^{2}}

F_{new} = 4F_{old}

The electrical force of attraction or electrostatic force of attraction between two charged particles refers to the amount of attractive or repulsive force that exists between the two charges. This can be calculated by Columb's Law.

A charged particle in physics is a particle that has an electric charge. It might be an ion, such as a molecule or atom having an excess or shortage of electrons in comparison to protons. The same charge is thought to be shared by an electron, a proton, or another primary particle.

Learn more about electrical force here

brainly.com/question/2526815

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8 0
2 years ago
In which condition is relative velocity known by adding the velocity od the first body and that of the second body ?​
Softa [21]

Answer:

when a two bodies A and B are moving at an angle 180° with each other then the relative velocity is the sum of bodies the velocity .i.e,

when the bodies move the opposite direction

then their relative velocity is the sum of individual velocities.

3 0
3 years ago
The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

4 0
3 years ago
Read 2 more answers
Which of the following brain imaging techniques produce the most detailed picture of brain structure? (2 points)
Inga [223]

Answer:

MRI

Explanation:

Magnetic Resonance Imaging, furnishes the most detailed picture of brain structure. MRI produces the most point by point picture of the cerebrum. MRI examines by using attractive fields and radio waves to create PC produced pictures that recognize the structures inside the mind. It gives the most point by point and a detailed image of the brain structure.

6 0
3 years ago
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