Answer:
A) first laser
B) 0.08m
C) 0.64m
Explanation:
To find the position of the maximum you use the following formula:
[tex]y=\frac{m\lambda D}{d}[/tex]
m: order of the maximum
λ: wavelength
D: distance to the screen = 4.80m
d: distance between slits
A) for the first laser you use:
[tex]y_1=\frac{(1)(d/20)(4.80m)}{d}=0.24m\\[/tex]
for the second laser:
[tex]y_2=\frac{(1)(d/15)(4.80m)}{d}=0.32m[/tex]
hence, the first maximum of the first laser is closer to the central maximum.
B) The difference between the first maximum:
[tex]\Delta y=y_2-y_1=0.32m-0.24m=0.08m=8cm[/tex]
hence, the distance between the first maximum is 0.08m
C) you calculate the second maximum of laser 1:
[tex]y_{m=2}=\frac{(2)(d/20)(4.80m)}{d}=0.48m[/tex]
and for the third minimum of laser 2:
[tex]y_{minimum}=\frac{(m+\frac{1}{2})(\lambda)(D)}{d}\\\\y_{m=3}=\frac{(3+\frac{1}{2})(d/15)(4.80m)}{d}=1.12m[/tex]
Finally, you take the difference:
[tex]1.12m-0.48m=0.64m[/tex]
hence, the distance is 0.64m
