That's 105 km that he flew, or 65.2 miles ! I'm absolutely positive
that the crow must have landed and gotten some rest when you
weren't looking. But that had no effect on his displacement when
he got where he was going, so we can continue to solve the problem:
The displacement is the distance and direction from the place
where the crow took off to the place where he landed.
-- It's distance is the hypotenuse of the right triangle whose legs
are 60 km and 45 km.
D² = (60 km)² + (45 km)²
= 3,600 km² + 2,025 km² = 5,625 km²
D = √(5625 km²) = 75 km .
-- It's direction is the angle whose tangent is (45 S / 60 W).
tan⁻¹ (45/60) = tan⁻¹ (0.75) = 36.9° south of west
= 53.1° west of south.
= not exactly southwest but close.
Answer:
Explanation:
1 )
Here
wave length used that is λ = 580 nm
=580 x 10⁻⁹
distance between slit d = .46 mm
= .46 x 10⁻³
Angular position of first order interference maxima
= λ / d radian
= 580 x 10⁻⁹ / .46 x 10⁻³
= 0.126 x 10⁻² radian
2 )
Angular position of second order interference maxima
2 x 0.126 x 10⁻² radian
= 0.252 x 10⁻² radian
3 )
For intensity distribution the formula is
I = I₀ cos²δ/2 ( δ is phase difference of two lights.
For angular position of θ1
δ = .126 x 10⁻² radian
I = I₀ cos².126x 10⁻²/2
= I₀ X .998
For angular position of θ2
I = I₀ cos².126x2x 10⁻²/2
= I₀ cos².126x 10⁻²
