Ask question

Ask question

All categories

2 years ago

8

Light of wavelength 510 nm is used to illuminate normally two glass plates 24.3 cm in length that touch at one end and are separ

ated at the other by a wire of radius 0.023 mm. How many bright fringes appear along the total length of the plates.

1 answer:

bulgar [2K]2 years ago

3 0

Answer:

$m = 180 \ bright \ fringes$

Explanation:

From the question we are told that

The wavelength is $\lambda = 510 \ nm = 510 *10^{-9} \ m$

The length of the of the plate is $l = 24.3 \ cm = 0.243 \ m$

The radius of the wire is $r= 0.023 \ mm = 2.3 *10^{-5} \ m$

Generally the diameter of the wire which is the distance between the glass plates is mathematically represented as

$d = 2 * r$

=> $d = 2 * 2.3 *10^{-5}$

=> $d = 4.6 *10^{-5} \ m$

Generally the condition for constructive interference is mathematically represented as

$2 *d = m \lambda$

=> $m = \frac{2d}{\lambda}$

=> $m = \frac{2 * (4.6* 10^{-5})}{ 510 *10^{-9} }$

=> $m = 180 \ bright \ fringes$

You might be interested in

A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi

algol13

Answer:

The time taken by the wave to travel along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire $\rho_{s}= 7.86\times10^{3}\ kg/m^{3}$

Density of copper wire $\rho_{c}=8.92\times10^{3}\ kg/m^3$

We need to calculate the linear density of steel wire

Using formula of linear density

$\mu_{s}=\rho_{s}A$

$\mu_{s}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{s}=5.00\times10^{-3}\ kg/m$

We need to calculate the linear density of copper wire

Using formula of linear density

$\mu_{c}=\rho_{s}A$

$\mu_{c}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{c}=5.67\times10^{-3}\ kg/m$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}$

$v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}$

$v_{s}=170.3\ m/s$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}$

$v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}$

$v_{c}=159.9\ m/s$

We need to calculate the time taken by the wave to travel along the combination of two wires

$t=t_{s}+t_{c}$

$t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}$

Put the value into the formula

$t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}$

$t=0.458\ sec$

$t=458\ ms$

Hence, The time taken by the wave to travel along the combination of two wires is 458 ms.

4 0

3 years ago

When are magnets attracted and when are they repelled.

vova2212 [387]

Magnets are attracted when each of the different sides, most commonly known as "North" and "South", are facing each other. They repel when North and North, or South and South are facing each other.

3 0

3 years ago

Consider a variety of colors of visible light (say 400 nm to 700 nm) falling onto a pair of slits.

babymother [125]

Answer:

Explanation:

The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by

For constructive interference

$d\sin \theta =m\lambda$

For Destructive interference

$d\sin \theta =(m+\frac{1}{2})\lambda$

where $\lambda =wavelength$

$d=slit\ width$

m=order of maxima and minima

for second order maxima i.e. $m=2$

For smallest separation taking $\lambda =400 nm, \theta =90^{\circ}$

$d\sin 90=2\times 400\times 10^{-9}$

$d=0.8\times 10^{-6}$

$d=0.8\mu m$

6 0

3 years ago

If the mass of both weights is 225 gm, the first mass is located 20∘ north of east, the second mass is located 20∘ south of east

Montano1993 [528]

Answer:

The voltage is 2.114 V.

Explanation:

Given that,

Mass of both weights = 225 gm

Transducer sensitivity = 0.5 V/N

The first mass is located 20∘ north of east, the second mass is located 20∘ south of east,

We need to calculate the net equivalent force

Using formula of force

$F_{3}=m_{1}g\cos\theta+m_{2}g\cos\theta$

$F_{3}=2mg\cos\theta$

Put the value into the formula

$F_{3}=2\times0.225\times10\cos20^{\circ}$

$F_{3}=4.228\ N$

We need to calculate the voltage

Using formula of voltage

$Voltage =sensitivity\times F_{3}$

Put the value into the formula

$Voltage=0.5\times4.228$

$Voltage =2.114\ V$

Hence, The voltage is 2.114 V.

7 0

3 years ago

If a nucleus was as big as a nonpareil, an atom would be ____

Mamont248 [21]

Answer:

small marble

Explanation:

A nonpareil are confectionery tiny ball items that are made up of starch and sugar. It is very small of the size of sugar crystal or sand grains. They were the miniature version of the comfits. They are generally opaque white but bow available in all colors.

In the context, if the nucleus is compared to the size of a nonpareil then its atom would be of the size of small size marble. An atom is bigger in size than that of nucleus as the nucleus is located inside the atoms.

6 0

2 years ago