Angular dispersion through a prism? Rather

[rs4orce]

(a) The index of refraction for violet light in silica flint glass is 1.65, and that for red light is 1.61. What is the angle of deviation for the red ray passing through a prism of apex angle 58.4o if the angle of incidence is 51.4o? Units of o (degrees)

(b) What is the angular dispersion of visible light with the same angle of incidence? Units of o (degrees)




I am using n1sin(theata1)=n2sin(theata2) with no avail. I broke the equation into its 2 parts, but derived 4.275o (under the impression that angle of deviation=angle of dispersion), to no avail. I've got a 3rd problem which would also be great if someone could take a look, as I believe it is along similar lines to the above, which must mean that I'm making the same mistake on both:

A light ray is incident normally on one face at 25o-65-90o block of dense flint glass (prism) that immersed in water. Find the exit angle (theata4) on the light ray. n2=1.77, n1= 1.333.

I'm working on a relatively (very) brief time line, so any help would be greatly appreciated. Thanks in advance.

 

[physixguru]

I understand the urgency.But we at PF appreciate some work to be shown.
IF WE JUST TELL YA THE ANSWERS IT SHALL NEVER HELP YA IN THE LONG RUN.

 

[Moetasim]

(a) To find the angle of deviation for the red ray, we can use Snell's law: n1sin(theata1) = n2sin(theata2). We know that n1 = 1.61 for red light and n2 = 1.65 for violet light. The angle of incidence (theata1) is given as 51.4o. We can rearrange the equation to solve for the angle of deviation (theata2): theata2 = sin^-1(n1sin(theata1)/n2). Plugging in the values, we get theata2 = 49.13o.

(b) Angular dispersion is the difference in the angle of deviation for different wavelengths of light. In this case, we are looking at visible light, which has a range of wavelengths from approximately 400nm (violet) to 700nm (red). Since we already calculated the angle of deviation for red light, we can use the same equation to find the angle of deviation for violet light: theata2 = sin^-1(n1sin(theata1)/n2). Plugging in the values, we get theata2 = 50.15o. Therefore, the angular dispersion for visible light is 50.15o - 49.13o = 1.02o.

For your third problem, we can use the same equation: n1sin(theata1) = n2sin(theata2). Since the light ray is incident normally (perpendicular) to the first face, the angle of incidence (theata1) is 90o. The index of refraction for water is 1.33, so we can solve for the angle of refraction (theata2): theata2 = sin^-1(n1/n2) = sin^-1(1.33/1.77) = 40.17o. Therefore, the exit angle (theata4) is also 40.17o.