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amid [387]
4 years ago
13

What is the difference between a total solar eclipse, an annular solar eclipse, and a partial solar eclipse.

Physics
2 answers:
bekas [8.4K]4 years ago
5 0

hii! In a total eclipse, the disk of the Sun is fully obscured by the Moon. In partial and annular eclipses, only part of the Sun is obscured. ... Total solar eclipses are rare at any particular location because totality exists only along a narrow path on the Earth's surface traced by the Moon's full shadow or umbra.

lozanna [386]4 years ago
4 0

The total solar eclipse is the eclipse in which the Sun gets completely covered by the Moon and Solar Corona can be seen whereas in case of partial solar eclipse, the Sun is partially covered by the Moon and Solar Corona can’t be seen whereas, in case of annular solar eclipse, the moon is further away from the Earth than in case of Total solar eclipse.

<span>Corona is visible in case of the Total solar eclipse.</span>

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1)Light of wavelength 588.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 55.5 cm from the slit
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These are Diffraction Grating Questions.

Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:  

Given as  

y = nDλ/w                                                       Eqn 1

where  

w = width of slit  

D = distance to screen  

λ = wavelength of light  

n = order number  

Making x the subject of the formula gives,  

w = nDλ/y  

Given  

y = 0.0149 m  

D = 0.555 m  

λ = 588 x 10-9 m  

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w = 6.6x10⁻⁵m

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Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen

i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx

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given, D = 27cm = 0.27m  

λ = 632 x 10-9 m  

w = 0.1mm = 1.0x10⁻⁴m

For the 9th order, n = 9,

y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m

Similarly, for n = 5,

y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m

Recall,  Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m

Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm

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