Answer: The focal length would be four inches
hope this helps
Answer:
It is a change in velocity over a period of time.
Explanation:
- It is measured in metres/second2<em><u> </u></em><em><u>u</u></em><em><u>s</u></em><em><u>i</u></em><em><u>n</u></em><em><u>g</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u> </u></em><em><u> </u></em><em><u>accelerometer</u></em><em><u>.</u></em>
Answer:
a.3.20m
b.0.45cm
Explanation:
a. Equation for minima is defined as: ![sin \theta=\frac{m\lambda}{\alpha}](https://tex.z-dn.net/?f=sin%20%5Ctheta%3D%5Cfrac%7Bm%5Clambda%7D%7B%5Calpha%7D)
Given
,
and
:
#Substitute our variable values in the minima equation to obtain
:
![\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad](https://tex.z-dn.net/?f=%5Ctheta%3Dsin%5E-%5E1%20%28%5Cfrac%7B3%5Ctimes%206.33%5Ctimes%2010%5E-%5E7%7D%7B0.00015%7D%29%5C%5C%5C%5C%5Ctheta%3D0.01266rad)
#draw a triangle to find the relationship between
and
.
#where ![y=4.05cm](https://tex.z-dn.net/?f=y%3D4.05cm)
![L=y/tan(\theta)=3.20](https://tex.z-dn.net/?f=L%3Dy%2Ftan%28%5Ctheta%29%3D3.20)
Hence the screen is 3.20m from the split.
b. To find the closest minima for green(the fourth min will give you the smallest distance)
#Like with a above, the minima equation will be defined as:
, where
given that it's the minima with the smallest distance.
![sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad](https://tex.z-dn.net/?f=sin%20%5Ctheta%3D%5Cfrac%7B4%5Clambda%7D%7B%5Calpha%7D%5C%5C%5Ctheta%3Dsin%5E-%5E1%20%28%5Cfrac%7B4%5Ctimes%206.33%5Ctimes%2010%5E-%5E7%7D%7B0.00015%7D%29%5C%5C%5C%5C%5Ctheta%3D0.01688rad)
#we then use
to calculate
=4.5cm
Then from the equation subtract
from
:
![4.50cm-4.05cm=0.45cm](https://tex.z-dn.net/?f=4.50cm-4.05cm%3D0.45cm)
Hence, the distance
is 0.45cm