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julia-pushkina [17]

2 years ago

13

A double slit apparatus is held 1.2 m from a screen. [___/4] (a) When red light (λ = 600 nm) is sent through the double slit, th

e interference pattern on the screen shows a distance of 12.5 cm between the first and tenth dark fringes. What is the separation of the slits? (b) What will be the difference in path length for the waves travelling from each slit to the tenth nodal line?

1 answer:

12345 [234]2 years ago

5 0

A. To solve for part A, we use the formula of Young’s double slit equation:

x = λ L m / d<span>

</span>Where,

x = distance between adjacent dark lines or fringes = 12.5 cm = 0.125 m

λ = wavelength of light = 600 × 10^-9 m

L = distance from the two sources of light to the screen = 1.2 m

m = number of fringes = 10 (tenth) – 1 (first) = 9

d = separation of slits = unknown

Rearranging the equation in terms of d and plugging in the values:

d = λ L m / x

d = (600 × 10^-9 m) (1.2 m) (9) / 0.125 m

d = 5.184 × 10^-5 m

d = 51.84 μm

B. The formula for path difference is:

path difference = (2 m + 1) (λ / 2)<span>
path difference = (2 * 9 + 1) (</span>600 × 10^-9 m / 2)

<span>path difference = 5.7 × 10-6 m</span>
<span>path difference = 5.7 μm</span>

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Answer:

2.7x10⁻⁸ N/m²

Explanation:

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<u>Where:</u>

$p_{rad}$ : is the radiation pressure

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c: is the speed of light = 3.00x10⁸ m/s

Hence, the radiation pressure is:

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I hope it helps you!

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2 years ago

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The two hot-air balloons in the drawing are 48.2m and 61.0 m above the ground.A person in the left balloon observes that the rig

mafiozo [28]

Answer:

The horizontal distance x between the two balloons is 54.15 m

Explanation:

The diagram described as obtained online is presented in the image attached to this solution.

Let the horizontal distance between the two balloons be x

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Using trigonometric relations, it is evident that

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x = 12.8/tan 13.3° = 12.8/0.2364 = 54.15 m

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2 years ago

The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan

miv72 [106K]

Answer:

mass of the planet X = 5.6 × 10²³ kg.

Explanation:

According to Newtons law of universal gravitation,

F = GM₁M₂/r²

Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X

making M₂ The subject of the equation above,

M₂ = Fr²/GM₁ .......................... equation 2

Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴ × 1000 m

r = 1.08 × 10⁷ m, G = 6.67 × 10 ⁻¹¹ Nm²/kg², M₁ = 75 kg

Substituting this values in equation 2,

M₂ = 24.13(1.08 × 10⁷ )²/75( 6.67 × 10 ⁻¹¹)

M₂ = 24.13 × 1.17 × 10¹⁴/500.25 × 10⁻¹¹

M₂ = (28.23 × 10¹⁴)/(500.25 × 10⁻¹¹)

M₂ = 0.056 × 10²⁵

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8 0

2 years ago

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2 years ago

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

yarga [219]

Answer

The rate at which the magnetic field is changing is $[\frac{dB}{dt} ] = 0.000467 T/s$

Explanation

From the question we are told that

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The radius is $r = 1.5 \ m$

The rate of change of the magnetic field is mathematically represented as

$\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

$\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

$A = \pi r^2$

So

$E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )$

$E L = ( \pi r^2) (\frac{dB}{dt} )$

where L is the circumference of the circle which is mathematically represented as

$L = 2 \pi r$

So

$E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]$

$E = \frac{r}{2} [\frac{dB}{dt} ]$

$[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

substituting values

$[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }$

$[\frac{dB}{dt} ] = 0.000467 T/s$

8 0

2 years ago