Answer:
a) 0.0130 m
b') w' = =6.46*10^{-3] m
Explanation:
given data:
\lambda of light = 633 nm
width of siit a =0.360 mm
distance from screen = 3.75 m
a) the first minima is located at
=
with of central fringe = 2y_1 = 2*6.54 *10^{-3} = 0.0130 m
b)
width of the first bright fringe on either side of the central one =
calculation for y_2
=
w' = =6.46*10^{-3] m
Answer:
r = 41.1 10⁹ m
Explanation:
For this exercise we use the equilibrium condition, that is, we look for the point where the forces are equal
∑ F = 0
F (Earth- probe) - F (Mars- probe) = 0
F (Earth- probe) = F (Mars- probe)
Let's use the equation of universal grace, let's measure the distance from the earth, to have a reference system
the distance from Earth to the probe is R (Earth-probe) = r
the distance from Mars to the probe is R (Mars -probe) = D - r
where D is the distance between Earth and Mars
M_earth (D-r)² = M_Mars r²
(D-r) = r
r ( ) = D
r =
We look for the values in tables
D = 54.6 10⁹ m (minimum)
M_earth = 5.98 10²⁴ kg
M_Marte = 6.42 10²³ kg = 0.642 10²⁴ kg
let's calculate
r = 54.6 10⁹ / (1 + √(0.642/5.98) )
r = 41.1 10⁹ m