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3 years ago

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Light of wavelength 500nm shines on a pair of slits. This produces an interference pattern on a screen 2.00m behind the slits. M

easuring from the center of the pattern, we find that the fifth dark spot is 4.00cm away. What is the spacing between the slits

1 answer:

grandymaker [24]3 years ago

7 0

Answer:

The value is $d = 0.000125 \ m$

Explanation:

From the question we are told that

The wavelength is $\lambda = 500 \ nm = 500 *10^{-9 } \ m$

The distance is of the screen is $D = 2.0 \ m$

The width of the fifth dark spot is $y = 4.00 \ cm = 0.04 \ m$

Generally the width of a fringe is mathematically represented as

$y = \frac{ n * \lambda * D }{d }$

=> $0.04 = \frac{ 5 * 500 *10^{-9} * 2 }{d}$

=> $d = \frac{ 5 * 500 *10^{-9} * 2 }{0.04}$

=> $d = 0.000125 \ m$

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Answer: 27.21 V

Explanation:

The <u>electric potential</u> $V_{E}$ due to a point charge is expressed as:

$V_{E}=k\frac{q}{r}$

Where:

$k=9(10)^{9}\frac{Nm^{2}}{C^{2}}$ is the <u>electric constant</u>

$q=1.6(10)^{-19}C$ is the <u>electric charge of the hydrogen nucleus</u>, which is positive

$r=5.292(10)^{-11}m$ is the <u>distance</u>

Rewritting the equation with the known values:

$V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}$

Finally:

$V_{E}=27.21 V$

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3 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

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What do MRI and ultrasound have in common as diagnostic imaging techniques? Check all that apply.

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Answer:

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Answer:

The (s) indicates that the state of matter for NaHCO3 is solid.

Explanation:

When a chemical reaction is written, the state of matter for each components of the reactants and products are mentioned in brackets along with their names or formulas.

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Light with a wavelength of 400 nm strikes the surface of cesium in a photocell, and the maximum kinetic energy of the electrons

Firdavs [7]

Answer:

The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

Explanation:

Given that,

Wavelength = 400 nm

Energy $E=1.54\times10^{-19}\ J$

We need to calculate the longest wavelength of light that is capable of ejecting electrons from that metal

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$\lambda=\dfrac{hc}{E}$

Put the value into the formula

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