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xxMikexx [17]
3 years ago
15

Coherent light that contains two wavelengths, 660 nm and 470 nm , passes through two narrow slits with a separation of 0.280 mm

and an interference pattern is observed on a screen which is a distance 5.50 m from the slits.
Required:
What is the disatnce on the screen between the first order bright fringe for each wavelength?
Physics
1 answer:
bekas [8.4K]3 years ago
7 0

Answer:

λ1 = 0.0129m = 1.29cm

λ2 = 0.00923m = 0.92 cm

Explanation:

To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:

y_m=\frac{m\lambda D}{d}    (1)

m: order of the bright fringe = 1

λ: wavelength of the light = 660 nm, 470 nm

D: distance from the screen = 5.50 m

d: distance between slits = 0.280mm = 0.280 *10^⁻3 m

ym: height of the m-th fringe

You replace the values of the variables in the equation (1) for each wavelength:

For λ = 660 nm = 660*10^-9 m

y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm

For λ = 470 nm = 470*10^-9 m

y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm

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The distance between the earth and sun is 1.5 x 108 kilometers and the speed of light is 3.00 x 108 meters per second. Calculate
butalik [34]

Answer:

time = 8.3333 minutes.

Explanation:

distance between earth and sun = 1.5 * 10^{8}km

speed of light = 3* 10^{8}m/s

convert the distance unit from km to m so we can have uniform units.

distance between earth and sun = 1.5 *10^{8}*1000m

distance between earth and sun = 1.5 * 10^{11}m

speed = distance /time

time = distance / speed

time = \frac{1.5*10^{11} }{3*10^{8} }

= 0.5*10^{3}

time =500 sec

time = 500/60 minutes

time = 8.3333 minutes.

3 0
3 years ago
I need help ASAP!!!! Here’s the question. Stephanie is going to buy an ice cream cone. She can pick a sugar cone or a waffle con
PolarNik [594]

Answer:

Six

Explanation:

Assuming she can only have ONE flavor in her cone

  2 cone types   *    3 flavors   =   6 combos

7 0
2 years ago
150c passes through cell in 30 seconds. cell has a potential difference of 12v. what is current in the circuit
zzz [600]

The current in the circuit is 5 A

Explanation:

The intensity of current is given by the equation:

I=\frac{q}{t}

where

I is the current

q is the amount of charge passing through a given point of the circuit in a time interval of t

For the cell in this problem, we have

q = 150 C is the charge

t = 30 s is the time interval

Substituting into the equation, we f ind

I=\frac{150}{30}=5 A

Learn more about current:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

6 0
3 years ago
What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface
kipiarov [429]

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman = \frac{95.08}{100 } = 0.9508

The density of the woman is calculate as;

Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman  = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

4 0
3 years ago
During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips i
Crank

Answer:

The mass of the mud is 3040000 kg.

Explanation:

Given that,

length = 2.5 km

Width = 0.80 km

Height = 2.0 m

Length of valley = 0.40 km

Width of valley = 0.40 km

Density = 1900 Kg/m³

Area = 4.0 m²

We need to calculate the mass of the mud

Using formula of density

\rho=\dfrac{m}{V}

m=\rho\times V

Where, V = volume of mud

\rho = density of mud

Put the value into the formula

m=1900\times4.0\times0.40\times10^{3}

m =3040000\ kg

Hence, The mass of the mud is 3040000 kg.

4 0
3 years ago
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