Answer:
a) 9.72 mm
b) 4.86 mm
Explanation:
wave length of light λ is 580 nm = 580 \times 10⁻⁹ m
Width of slit d = 0.215\times 10⁻³ m
Distance of screen D = 1.8 m.
Width of one fringe = 
Putting the values we get fringe width
= 
=4.86 mm.
a) Width of central maxima = 2 times width of one fringe
= 2 times 4.86
=9.72 mm
b) width of each fringe except central fringe is same , no matter what the order is.Only brightness changes .
So width of either of the two first order bright fringe will be same and it will be
= 4.86 mm.
What’s the question? Is it true or false?
Answer:
it loaded and it is C. buddy sorry about that :)
Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s
To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s
Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m
This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.
But anyways, good luck!