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Two waves of the same frequency and
constant phase difference have intensities in
the ratio 36 : 4. These waves are superposed
and interference fringe pattern is obtained.
Calculate the ratio of the maximum to
minimum intensity.
How do filters in Drama lights work.
I had two ideas, would these work?
1.) spread a transition metal on the filter and certain frequencies of light are absorbed
2.) atoms of the filter are excited and energy levels are such that red light is emitted.
An object placed in front of a convex mirror of radius 20cm produces an erect image which is one-fifth the size of the object. How far is the object from the mirror
1.A 1.00 cm tall light bulb is placed a distance of 2.5 cm from a concave mirror having a focal length of 5 cm.

a.Draw Image Formation of the object and mention the characteristics of image.
b.Determine the image distance
c.Determine the image height.

2.A 1.00 cm tall light bulb is placed a distance of 7.5 cm from a concave mirror having a focal length of 5 cm.
a.Draw Image Formation of the object and mention the characteristics of image.
b.Determine the image distance
c.Determine the image height

35.A 1.00 cm tall light bulb is placed a distance of 5 cm from a convex mirror having a focal length of 5 cm.
a.Draw Image Formation of the object and mention the characteristics of image.
How is a coherence radiation produced from an optical resonator? With reference this coherence radiation explain Spatial and temporal coherence. Is it a good source to realize interference pattern, Justify your answer?
A concave shaving mirror has a focal length
of 36 cm.
a) Calculate the image position of a cologne
bottle placed in front of the mirror at a distance
of 76 cm. (Answer with −1000 if the
image does not exist.)
Answer in units of cm.
The focal length of the equivalent mirror of a equiconvex lens silvered at one side
Numerical aperture of fibre is 0.5 and the core refractive index is 1.48 find the refractive index of the cladding acceptance angle
Why the site of hologram recording should be free from vibrations
Why does the frequency not change when refraction occurs? It has to change as the wavelength becomes shorter, causing more waves to pass in a second. Right?