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marusya05 [52]
3 years ago
11

When light is incident normally on the interface between two transparent optical media, the intensity of the reflected light is

given by the expression(equation 1)In this equation S1 represents the average magnitude of the Poynting vector in the incident light (the incident intensity), S’1 is the reflected intensity, and n1 and n2 are the refractive indices of the two media. The fraction of incident intensity is 0.0246 or 2.46 precent for 589-nm light.For light normally incident on an interface between vacuum and a transparent medium of index n is given by S2/S1 = 4n/(n + 1)2 (equation 2). (a) Light travels perpendicularly through a diamond slab, surrounded by air, with parallel surfaces of entry and exit. Apply the tramsmission fraction in equation 2 to find the approximate overall transmission through the slab of diamond, as a percentage. Ignore light relected back and forth within the slab.(b) Light travels perpendicularly through a diamond slab, surrounded by air, with parallel surfaces of entry and exit. The intensity of the transmitted light is what fraction of the incident intensity? Include the effects of light reflected back and forth inside the slab.If someone could answer part (a) and (a) I would gladly appreciate it.
Physics
1 answer:
Alex777 [14]3 years ago
8 0

i need to ask sum so oycixgixigdtdidit

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a force

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Which kind of force do you exert on an object when you pull it toward you?
jarptica [38.1K]

By definition we have to:

Applied force: It is the external force that acts directly on a body.

Therefore, we can say that if you have an object and push it towards yourself, you are exerting an external force on the object.

This external force was not acting on the object previously, therefore, it is a force that you are applying at that moment.

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you exert an Applied Force on an object when you pull it towards you

A. Applied Force

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The main reason that most professional research telescopes are reflectors is that
GuDViN [60]
<h3><u>Answer;</u></h3>

Large mirrors are easier to build than large lenses.

<h3><u>Explanation;</u></h3>
  • <em><u>Reflector telescopes have a number of advantages as compared to refracting telescopes and other types of telescopes. </u></em>
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4 years ago
if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

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