Unpolarized light passes through two polarizers whose transmission axes are at an angle # with respect to each other. What shoul
1 answer:
Answer:
Explanation:
When unpolarized light passes through the first polarizer, the intensity of the light is reduced by a factor 1/2, so
(1)
where I_0 is the intensity of the initial unpolarized light, while I_1 is the intensity of the polarized light coming out from the first filter. Light that comes out from the first polarizer is also polarized, in the same direction as the axis of the first polarizer.
When the (now polarized) light hits the second polarizer, whose axis of polarization is rotated by an angle with respect to the first one, the intensity of the light coming out is
(2)
If we combine (1) and (2) together,
(3)
We want the final intensity to be 1/10 the initial intensity, so
So we can rewrite (3) as
From which we find
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Answer:
a) v₀ = 32.64 m / s , b) x = 59.68 m
Explanation:
a) This is a projectile launching exercise, we the distance and height of the cliff
x = v₀ₓ t
y = t - ½ g t²
We look for the components of speed with trigonometry
sin 43 = v_{oy} / v₀
cos 43 = v₀ₓ / v₀
v_{oy} = v₀ sin 43
v₀ₓ = v₀ cos 43
Let's look for time in the first equation and substitute in the second
t = x / v₀ cos 43
y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²
y = x tan 43 - ½ g x² / v₀² cos² 43
1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²
v₀² = g x² / [(x tan 43 –y) 2 cos² 43]
Let's calculate
v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]
v₀ = √ (35280 / 33.11)
v₀ = 32.64 m / s
.b) we use the vertical distance equation with the speed found
y = t - ½ g t²
.y = v₀ sin43 t - ½ g t²
25 = 32.64 sin 43 t - ½ 9.8 t²
4.9 t² - 22.26 t + 25 = 0
t² - 4.54 t + 5.10 = 0
We solve the second degree equation
t = (4.54 ±√(4.54 2 - 4 5.1)) / 2
t = (4.54 ± 0.46) / 2
t₁ = 2.50 s
t₂ = 2.04 s
The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled
x = v₀ₓ t
x = v₀ cos 43 t
x = 32.64 cos 43 2.50
x = 59.68 m
Answer:
For left = 0 N/C
For right = 0 N/C
At middle = N/C
Explanation:
Given data :-
б = C/ m²
Considering the two thin metal plates to be non conducting sheets of charges.
Electric field is given by
1) To the left of the plate
= 0 N/C.
2) To the right of them.
= 0 N/C.
3) Between them.
=
=
=
N/C