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Elza [17]
3 years ago
8

Present the ways to observe interference patterns on thin films. Why the thickness of thin films should be in scale of wavelengt

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Physics
2 answers:
frez [133]3 years ago
6 0

the president of The proposal ahr stay Sheba SBR Abba and and I send svrvs you svrvs

antoniya [11.8K]3 years ago
5 0

Answer:

A colorful interference pattern is observed when light is reflected from the top and bottom boundaries of a thin oil film. The different bands form as the film's thickness diminishes from a central run-off-poin

Explanation:

<h3>hope it help brainliest pls</h3>
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SO MANY POINTS! WILL MARK BRAINIEST!!!! PLZ BE FASSTT!
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Answer:  Point A is the answer for the potential energy. Point D is the answer for the kinetic energy.

Explanation:

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Are you better off in an accident where the hood is
horsena [70]

Answer:

you would be better off if the car bounced backwards

Explanation:

because if the hood was dismembered than you have a high chance of very bad injury but if it is just bounced back you would have less chance of getting hurt if properly sitting and seat belted.

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How do salt and water form a solution?
iogann1982 [59]
It would be D. The solute, salt, must dissolve IN the solvent, water (:
8 0
3 years ago
Read 2 more answers
Two metal plates 15mm apart have a potential difference of 750v between them. The force on a small charged sphere placed between
Romashka [77]

Answer:

50,000 V/m

Explanation:

The electric field between two charged metal plates is uniform.

The relationship between potential difference and electric field strength for a uniform field is given by the equation

\Delta V=Ed

where

\Delta V is the potential difference

E is the magnitude of the electric field

d is the  distance between the plates

In this problem, we have:

\Delta V=750 V is the potential difference between the plates

d = 15 mm = 0.015 m is the distance between the plates

Therefore, rearranging the equation we find the strength of the electric field:

E=\frac{\Delta V}{d}=\frac{750}{0.015}=50,000 V/m

6 0
3 years ago
You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with
Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

           

Explanation:

For this exercise we use the constructor equation or Gaussian equation

        \frac{1}{f}  = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

          m = \frac{h'}{h} = - \frac{q}{p}

             h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

           \frac{1}{q} = \frac{1}{f }  - \frac{1}{p}

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

          q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

           h ’= h q / p

where p has a high value and is the same for all systems

           h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

 

f = 12 cm h ’= fo 12  cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0
3 years ago
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