Physics SS 3 Week 4
Topic: Glass Prism
Spectrum from Prism
Sunlight is often called white light, since it is a combination of all the visible colors. Since the index of refraction is different for each color, the angle of refraction will be different for each color when the light passes from air into glass or other transparent material. This is according to Snell’s Law.
Now if the piece of glass has parallel sides, such as with a window, the light will return in the same direction that it entered the material. But if the material is shaped like a prism, the angles for each color will be exaggerated, and the colors will be displayed as a spectrum of light.
Prism spreading white light into a spectrum of light
The visible colors are in the order of the spectrum. You can remember the order by the name ROY G. BIV. Note that in the illustration above, the colors are distinctly separated. In realty, they blend into each other, such that there are colors in between. For example, there is red-orange in between red and orange.
Tiny droplets of water refract the white light from the Sun and create a spectrum of colors similar to what happens in a prism. Since the droplets are spheres, the light is reflected internally in the droplets and the spectrum or rainbow returns toward the direction of the light. That is why the Sun will always be behind you when you see a rainbow.
Chromatic dispersion can be a problem in optical equipment like cameras, microscopes and telescopes. Since a lens is similar to a prism, it will disperse the light into a spectrum. This can be exaggerated when there are several lenses in the system. But you don’t want a blurry image where the system focuses different colors of light at different spots. Such blurring is called chromatic aberration.
The solution to that problem is to combine a positive lens that has one index of refraction with a negative lens of a different type of glass and a different index of refraction, such that the dispersion of the two negate each other.
Deviation, measured in degrees, is the angle an incident ray is turned through after passing through a prism (or other optical component).
This deviation is a minimum for a prism when the path of a light ray is symmetrical about its axis of symmetry.
Derivation of Minimum Deviation D
D = ∠RPQ + ∠PQR ……………… (i)
(D external angle of ∆PQR)
around points P and Q respectively,
i = ∠RPQ + r i = ∠PQR + r
∠RPQ = i – r ∠PQR = i – r
substituting into equation (i)
∴ D = 2(i – r) …………… (ii)
A + ∠OPQ + ∠PQO = 1800
A + (900 – r) + (900 – r) = 1800
A = 2r ……………….(iii)
adding together equations (ii & (iii,
A + D = 2r + 2(i – r)
= 2r + 2i -2r
I = A + D/2………………..(iv)
rearranging equation (iii)
r = A/2
Using Snell’s Law equation,
n1 sin(i) = n2 sin(r)
and substituting for i and r from equations (iv & (v above,
n1 sin (A + D/2) = n2 sin (A/2)
n1 is the refractive index of air which approximates to 1. Hence the equation becomes:
sin (A + D/2) = n2 sin (A/2)
So for a 60o equilateral prism made of glass(n = 1.5 approx.), the minimum deviation angle D is given by:
sin (300 + D/2) = 1.5 x sin (300)
= 1.5 x 0.5
It follows that:
300 + D/2 = sin-1(0.75)
remembering that sin-1(x) means the angle whose sine is ‘x’,
300 + D/2 = 48.590
D/2 = 48.590 – 300
D = 2 x18 x 590
D = 37.180
Minimum angle of dispersion
This is measure of the angle of ‘spread’ of a spectrum when it leaves a prism. For minimum angular dispersion, the angle is derived from the difference in deviation between red and violet rays of light.
Replacing n2with nr for red, and nv for violet light in the minimum deviation equation below,
sin (A + D/2) = n2 sin (A/2)
we can calculate values of deviation for each colour. Subtracting the angles gives the minimum dispersion angle for white light.
Apparent and Real Depth
A pond of clear water appears to be shallower than it really is. Similarly, the pebbles lying at the bottom of the clear water of a river appear to be raised up above their actual position.
Point to Ponder
Why does the apparent depth of the pond look less than its real depth?
Relation of refractive index to real and apparent depth
In immediate fig. above, o is a bright point below a glass slab. A ray OA from O perpendicular to the surface of the medium passes straight into air. We consider another ray OB from O which is refracted along BC at the point ‘b’ because air is a rarer medium so angle of refraction is greater than the angle of incidence. When the ray CB is produced backwards, it meets the first ray OA at the point I. for an observer’s eye the object O appears as it were at I above O. thus I is the virtual image of object O.
Suppose refractive index of glass is ‘n’ with respect to air. It will have this value in case when light is entering into the glass from air, but in the present case, the light is entering from glass into the air. In this case, we can find the refractive index of air with respect to glass by using snell’s law, its value with be 1/n. Hence
sin i/sinr = ½ or n = sinr/sini
= AB/IB/AB/OB = OB/IB =
When B is very close to A i.e., observer’s eye is looking both the rays O A and OB, then IB = IA and OB = OA
n = OA/IA
n = Real Depth/Apparent Depth
Example 14.3: the real depth of a swimming pool is 2m. What is the apparent depth of the pool if the refractive index of water is 1.33?
Real depth =2m n= 1.33 , apparent depth =?
Refractive index of water = n = Real depth/Apparent depth
1.33 = 2m/Apparent depth
Thus apparent depth = 2m/1.33
= 1. 5 m