It's bright during most of the show ... although they do dim it somewhat
during the trapeze lady ... and then it's dark after the audience is gone.
central maximum is double the with of any other fringe. all other fringes(dark and bright) are of the same width.
Of Course! You don't need to have bright blue eyes to be beautiful. Dark brown eyes are gorgous!
Jupiter, Saturn, Uranus, and Neptune all have rings. Jupiter's ring is thin and dark, and cannot be seen from Earth. Saturn's rings are the most magnificent; they are bright, wide, and colorful. Uranus has nine dark rings around it, and Neptune's rings are also dark, but contain a few bright arcs.
It depends if light can travel in it or not. If not, then it will be dark and nothing can go in. If so, then it will look like space, or its surroundings, however bright or dark they are.
stars
Because the intensity is maximum at the centre. Therefore the central spot is bright and not dark.
central maximum is double the with of any other fringe. all other fringes(dark and bright) are of the same width.
Because the path difference or the phase difference between two waves is zero
Ring of Bright Water was created in 1969-01.
The duration of Ring of Bright Water is 1.78 hours.
Ring of Bright Water is an autobiographical book by Gavin Maxwell
An antonym for bright is dark.
Dark, but with a bright look.
The Esperanto words for dark and bright are malhela and brila.
We see it precisely because it is bright. If it were dark, we wouldn't see it.We see it precisely because it is bright. If it were dark, we wouldn't see it.We see it precisely because it is bright. If it were dark, we wouldn't see it.We see it precisely because it is bright. If it were dark, we wouldn't see it.
the newton's rings are formed due to the phenomenon of thin film interference. here, the condition for constructive interference(the ring appearing bright) is that the optical path difference between interfering waves should be an integral multiple of the wavelength. the optical path difference is given by 2t-(l/2) if t is the thickness of the air film at that point and l is the wavelength of light. at the central point, the lens touches the surface so thickness t=0. thus the optical path difference is simply l/2, which is the condition for destructive interference, not constuctive interference. so the central spot has to always be dark.
Saturn