A student and his lab partner create a single slit by carefully aligning two razor blades to a separation of 0.530 mm. When a helium–neon laser at 543 nm illuminates the slit, a diffraction pattern is observed on a screen 1.55 m beyond the slit. Calculate the angle θdark to the first minimum in the diffraction pattern and the width of the central maximum.
Answers
Answer:
angle = 0.058699 degree
width of central maximum is 3.170566 × m
distance D = 1.55 m
wavelength w = 543 nm = 543× m
to find out
angle θ and width of the central maximum
solution
we know according to first condition first dark that mean
wavelength = dsinθ
so put value and find θ
543× = 0.530×
×sinθ
sinθ = 543× / 0.530×
sinθ = 1.02452 × [tex]10^{-3}
θ = 0.058699 degree
and
we can say
tanθ = y/D
here y is width of central maximum Y = 2y
put all value we get y
so y = D tanθ
y = 1.55 (tan0.0586)
y = 1.58528 × [tex]10^{-3} m =
so Y = 2 ( 1.58528 × [tex]10^{-3} )
so width of central maximum is 3.170566 × [tex]10^{-3} ) m
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Answer:
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Answers
Answer:
θ=π/2
Explanation:
The definition of work is W = → F ⋅ → d = q E c o s θ d W=F→⋅d→=qEcosθd. So if no work is done, the displacement must be in the direction perpendicular to the force ie c o s θ = 0 → θ = π / 2 cosθ=0→θ=π/2
Final answer:
A charged particle can be displaced without any external work done on it in a uniform electric field when its movement is perpendicular to the direction of the electric field.
Explanation:
In a uniform electric field, the electric force is the same in every direction. Therefore, if a charge were to be displaced perpendicular to the original direction of the electric field (i.e., in the y or z direction), it would not encounter any extra electric forces. This means there would be no external work being done on the charge. When a charge is moved perpendicular to an electric field, the field does not affect it, and hence, no work is done by the field.
In other words, a charge can be displaced in this field without any external work being done on it when it is moved in a direction perpendicular to the uniform electric field, either in y-axis or z-axis, assuming the electric field is constant in the x-axis direction.
Learn more about Uniform Electric Field here:
#SPJ3
Answers
Answer:
(a): should be pumped out of the room 18 L/min to keep the water level constant.
(b): should be pumped out of the room 78 L/min to reduce the water level by 4 cm/hr.
Explanation:
S= 90 m²
rate= 1.2 cm/hr = 0.012 m/hr = 0.0002 m/min
Water leak= S*rate= 90 m² * 0.0002 m/min
Water leak= 0.018 m³/min * 1000 L/m³
Water leak= 18 L/min (a) Water should be pumped out to keep the level constant.
By the rule of 3:
1.2 cm/hr ------------- 18 L/min
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(b) Find the magnitude of force F~2 that is acting on the block
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Answers
Answer:
Normal force=7.48 N
Explanation:
N+F~1 sinθ-mg=0
=>N=1.5*9.8-12 sin37◦
=>N=14.7-7.22=7.48 N
Answers
Answer:
listed
Explanation:
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Answers
Answer:
0.00124 V
Explanation:
Parameters given:
Initial circumference = 162 cm
Rate of decrease of circumference = 14 cm/s
Magnetic field, B = 0.5 T
Time, t = 8 secs
The magnitude of the EMF induced in the loop is given as:
V = (-NBA) / t
Where N = number of turns = 1
B = magnetic field
A = area of loop
t = time taken
First, we need to find the area of the loop.
To do this, we will find the radius after the loop circumference has decreased for 8 secs.
The rate of decrease of the circumference is 14 cm/s and 8 secs has passed, which means after 8 secs, it has decreased by:
14 * 8 = 112 cm
The new circumference is:
162 - 112 = 50 cm = 0.5 m
To get radius:
C = 2 * pi * r
r = C / (2 * pi)
r = 0.5 / (2 * 3.142)
r = 0.0796 m
The area is:
A = pi * r²
A = 3.142 * 0.0796²
A = 0.0199 m²
Therefore, the EMF induced is:
V = (-1 * 0.5 * 0.0199) / 8
V = -0.00124V
This is the EMF induced in the coil.
The magnitude is |-0.00124| V = 0.00124 V.