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3 years ago

A solar eclipse will occur Group of answer choices

Physics

1 answer:

myrzilka [38]3 years ago

7 0

Answer:

3. at new Moon only when the Moon is on the ecliptic.

Explanation:

Solar eclipse is the condition when the moon comes in between the sun and the earth. In this condition the moon casts its shadow on the earth.

Whether the eclipse is a total solar eclipse, a partial solar eclipse or an annular solar eclipse depends on various factors, but the position of the moon must be on the same orbital plane as that of the earth's orbit around the sun.

The sun is about 400 times larger than the moon in size and the sun is almost 400 times farther from the earth than the moon is, this makes it possible for the moon to cover the sun completely leading to a complete solar eclipse.

As we know that the orbit of the earth around the sun and the orbit of the moon around the earth is elliptical which leads to a variation in the distance from their rotating centers, so not of every eclipse the moon covers the sun completely developing an annular eclipse.

When the moon is close enough to the earth on the ecliptic but not completely aligned in between the sun and the earth leads to a partial solar eclipse.

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3 years ago

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen

Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}$

$sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m$

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

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