derivation of snell law sin i/sinr=n2/n1 clearly
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Like with reflection, refraction also involves the angles that the incident ray and the refracted ray make with the normal to the surface at the point of refraction. Unlike reflection, refraction also depends on the media through which the light rays are travelling. This dependence is made explicit in Snell's Law via refractive indices, numbers which are constant for given media1.
Snell's Law is given in the following diagram.
Basic illustration of Snell's Law.
As in reflection, we measure the angles from the normal to the surface, at the point of contact. The constants n are the indices of refraction for the corresponding media.
Tables of refractive indices for many substances have been compiled.
n for Light of Wavelength 600 nm
Substance Refractive Index, n
Air (1 atmosphere pressure, 0 degrees C) 1.00029
Water (20 degrees C) 1.33
Crown Glass 1.52
Flint Glass 1.66
Say, in our simple example above, that we shine a light of wavelength 600 nm from water into air, so that it makes a 30o angle with the normal of the boundary. Suppose we wish to find the angle x that the outgoing ray makes with the boundary. Then, Snell's Law gives
1.33 sin 30o = 1.00029 sin x
x = 41o
Refraction certainly explains why fishing with a rod is a sport, while fishing with a spear is not2.
A more complicated illustration of Snell's Law proves something that seems intuitively correct, but is not obvious directly. If you stand behind a window made of uniform glass, then you know by now that the images of the things on the other side of the window have been refracted. Assuming that the air on both sides of your window have the same refractive indices, we have the following situation:
Two refractions.
We find that the incoming and outgoing light beams are actually parallel.
Rearranging Snell's Law, with i and r being the incident and refracted angles,
n1sin(i) = n2sin(r)
(n1/n2)sin(i) = sin(r)
a qualitative description of refraction becomes clear. When we are travelling from an area of higher index to an area of lower index, the ratio n1/n2 is greater than one, so that the angle r will be greater than the angle i; i.e. the refracted ray is bent away from the normal. When light travels from an area of lower index to an area of higher index, the ratio is less than one, and the refracted ray is smaller than the incident one; hence the incident ray is bent toward the normal as it hits the boundary.
Of course, refraction can also occur in a non-rectangular object (indeed, the objects that we are interested in, lenses, are not rectangular at all). The calculation of the normal direction is harder under these circumstances, but the behaviour is still predicted by Snell's Law.
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Answer:HEY MATE here is your answer
Explanation:
Consider two triangle’s ABC and AEC :-
In triangle ABC sin i= (BC/AC) and sin r=(AE/AC)
By dividing (sin i/sin r) = (BC/AC) x (AC/AE)
Therefore (sin i/sin r) = (BC/AE) =(v1τ) /(v2τ)
=> (sin i/sin r) =(v1/v2)
Refractive index n; =(c/V)
Where c = velocity of light in vacuum and V=velocity of light in medium.
Therefore (sin i/sin r) = (c/n1)/(c/n2)
Where n1 and n2 are the refractive index in medium 1 and 2 respectively
Snell s law (sin i/sin r) = (n1/ n2). Hence proved.
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