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The distance between slit and the screen is 1.214m.
To find the answer, we have to know about the width of the central maximum.
<h3>How to find the distance between slit and the screen?</h3>
- It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
- We have the expression for slit width w as,

where, d is the distance between slit and the screen, and a is the slit width.
- Thus, distance between slit and the screen is,

Thus, we can conclude that, the distance between slit and the screen is 1.214m.
Learn more about the width of the central maximum here:
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Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?
B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>
<span>Solution:
A) A charge q </span>under an electric field of intensity E will experience a force F equal to:

In our problem we have
and
, so we can find the magnitude of the electric field:

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.
B) The proton charge is equal to

Therefore, the magnitude of the force acting on the proton will be

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.
Answer : The power absorbed by the bulb is, 0.600 W
Explanation :
As we know that,
Power = Voltage × Current
Given:
Voltage = 3 V
Current = 200 mA = 0.200 A
Conversion used : (1 mA = 0.001 A)
Now put all the given values in the above formula, we get:
Power = Voltage × Current
Power = 3V × 0.200 A
Power = 0.600 W
Thus, the power absorbed by the bulb is, 0.600 W
This is a power problem which requires the rearranging of a formula. The lamps energy used is 5 N, and the TV’s usage is 116.7 N (rounded from 116.6666repeating). Here my work: