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serg [7]
3 years ago
12

Monochromatic light falls on two very narrow slits 0.048 mm apart. successive fringes on a screen 5.00 m away are 6.5 cm apart n

ear the center of the pattern. determine the frequency of the light.
Physics
1 answer:
atroni [7]3 years ago
4 0
In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
y= \frac{m \lambda D}{d}
where D=5.00 m is the distance of the screen from the slits, and 
d=0.048 mm=0.048 \cdot 10^{-3}m is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
\lambda =  \frac{yd}{mD}= \frac{(0.065 m)(0.048 \cdot 10^{-3}m)}{(1)(5.00 m)}=  6.24 \cdot 10^{-7}m

And from the relationship between frequency and wavelength, c=\lambda f, we can find the frequency of the light:
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{6.24 \cdot 10^{-7}m}=4.81 \cdot 10^{14}Hz
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If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he
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The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q= mC_s \Delta T
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Read 2 more answers
You toss a rock up vertically at an initial speed of 39 feet per second and release it at an initial height of 6 feet. The rock
3241004551 [841]

Answer:

2.583 s, 29.77 ft and 1.219 s

Explanation:

Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground

H = 6 ft, the height from which it is tossed

V₁ = V + gt = V - gt

at maximum height the body came to rest momentarily V₁ = 0

0 = V - gt

-V = -gt

- 39 / -32 = t

t time to reach maximum height = 1.219 s

To Maximum height reached can be calculated with the formula

V₁² = V² + 2g( y - H) where H is the initial height reached by the tossed rock

where V₁ is the final velocity at maximum height which = 0

0 = V² - 2g(y-H) where y is the distance traveled from the ground

-V² = -2g(y-H)

₋V² / -2g = y-H

(V²/2g) + H = y in ft

(39² / (2 × 32)) + 6

y = 29.77 ft

The total time it will be in air can be calculated with the formula below

y = H + Vt - 0.5gt² from y-H = ut + 0.5at²

0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)

0.5gt² - Vt - H = 0

using quadratic formula

- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)

(V ± √ (V² + 2gH)) ÷ g

substitute the values into the expression

t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32

t = (39 + √(1521 +384))/32 = (39 + √1905) / 32  = 2.583 s

t = (39 - √1905) / 32 =  -0.15 s

The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds

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motikmotik

Answer:wave travel

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