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3 years ago

What happens to parallel light rays that strike a concave lens?

Physics

1 answer:

torisob [31]3 years ago

5 0

Answer:

They diverge on refraction

Explanation:

When parallel light rays strike a concave lens, they will diverge that is they spread out .

Concave lens is also known as diverging lens, which means that when parallel rays of light strike on it, the lens spreads out the light rays ( that is it diverges the rays of light) that are refracted through it.

At the middle of concave lens is thinner.

When light is passes through the lens they diverge it or spread out.

The concave lens causes light rays to bend away or diverge from its axis since the concave lens is a diverging lens.

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A national college researcher reported that 67%of students who graduated from high school in 2012 enrolled in college. Twenty-ni

dsp73

Answer:

Explanation:

Theorem of Binomial Distribution will apply here.

n = 29 , p = .67 , q = 0.33

mean = np = 29 x .67 = 19.43

Standard Deviation = √npq

= √29 x .67 x .33

= √6.4

= 2.53

4 0

3 years ago

Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien

jeka94

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis $\theat =45^{\circ}$

$S_1=S_0\cos ^2\theta$

here $S_0=\frac{S}{2}$

$S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}$

$S_1=\frac{S}{4}$

When it is passed through third Polarizer with its axis $90^{\circ}$ to first but $\theta =45^{\circ}$ to second thus $S_2$

$S_2=S_0\cos ^2\theta$

$S_2=\frac{S}{4}\times \frac{1}{2}$

$S_2=\frac{S}{8}$

When middle sheet is absent then Final Intensity will be zero

3 0

3 years ago

Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with

Ede4ka [16]

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is $2.728\times 10^{-3}\,\frac{m}{s^{2}}$ ( $\frac{2.784}{10000}\cdot g$ , where $g = 9.8\,\frac{m}{s^{2}}$ ).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration ( $g$ ), in meters per square second, is directly proportional to the mass of the Earth ( $M$ ), in kilograms, and inversely proportional to the distance from the center of the Earth ( $r$ ), in meters:

$g = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$G$ - Gravitational constant, in cubic meters per kilogram-square second.

$M$ - Mass of the Earth, in kilograms.

$r$ - Distance from the center of the Earth, in meters.

If we know that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972\times 10^{24}\,kg$ and $r = 382.26\times 10^{6}\,m$ , then the free-fall acceleration at the orbit of the Moon is: