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GuDViN [60]
3 years ago
11

Two slits of width 2μm, each in an opaque material, are separated by a center-to-center distance of 6μm. A monochromatic light o

f wavelength 450 nm is incident on the double-slit. One finds a combined interference and diffraction pattern on the screen.A) How many peaks of the interference will be observed in the central maximum of the diffraction pattern?B) How many peaks of the interference will be observed if the slit width is doubled while keeping the distance between the slits same?C) How many peaks of interference will be observed if the slits are separated by twice the distance, that is, 12 μm, while keeping the widths of the slits same?D) What will happen in (a) if instead of 450-nm light another light of wavelength 680 nm is used?E) What is the value of the ratio of the intensity of the central peak to the intensity of the next bright peak in (a)?F) Does this ratio depend on the wavelength of the light?G) Does this ratio depend on the width or separation of the slits?
Physics
1 answer:
Alona [7]3 years ago
6 0

Answer:

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Explanation:

You might be interested in
Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just
bagirrra123 [75]

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

5 0
4 years ago
A 920-kg compact car moving at 92 m/s has approximately 3,893,440 Joules of kinetic energy. What is the change in kinetic energy
tensa zangetsu [6.8K]

Answer:

Change in kinetic energy = 3297280 J

Explanation:

Given that,

Mass, m = 920 kg

Speed of a car, v = 92 m/s

Kinetic energy, K = 3,893,440 J

If the speed of a car, V = 36 m/s

Net kinetic energy is given by :

K_n=\dfrac{1}{2}mV^2\\\\=\dfrac{1}{2}\times 920\times (36)^2\\\\K_n=596160\ J

The change in kinetic energy = 3,893,440 - 596160

= 3297280 J

So, the change in kinetic energy of the car is 3297280 J.

3 0
3 years ago
HELP HELP HELP help pls
GalinKa [24]
Answer is A.) 40 m/km

7 0
3 years ago
Describe what happens when heat is removed from a system
OleMash [197]
This happens<span> at the boiling </span>temperature<span> of every substance that can vaporize. At the boiling </span>temperature<span>, adding </span>heat<span> energy converts the liquid into a gas WITHOUT RAISING THE </span>TEMPERATURE<span>. Adding </span>heat<span> to a boiling liquid is an important exception to general rule that more </span>heat<span> makes a higher </span>temperature<span>.</span>
8 0
3 years ago
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