Physics Topics are often described using mathematical equations, making them precise and quantifiable.
Determine the Expression of Net Dispersion, in the Case of Dispersion without Deviation
‘We have already seen that polychromatic light gets dispersed as well as deviated while refracted through a prism.
Two prisms are used to get deviation without dispersion or dispersion without deviation. The prisms are placed side by side with their refracting angles in opposite directions.
Deviation without dispersion: The materials and the refractive angles of two prisms are so chosen that the dispersion due to one of the prisms can be cancelled by the other. It means white ray after passing through the prism gets refracted without any dispersion. But deviation may happen. Such type of combination of prisms is called achromatic combination [Fig.].
Let the refractive indices of the prisms are µ and µ’ and their refractive angles are A and A’ respectively. Vertices of two prisms are opposite and placed back to back.
If the combination is not to produce a net dispersion, then angular dispersion of one prism is equal and opposite to other prism.
δv – δr = \(\delta_v^{\prime}\) – \(\delta_r^{\prime}\) ….. (1)
or, (µv – µr)A = (\(\mu_v^{\prime}\) – \(\mu_r^{\prime}\))A’
or, \(\frac{\left(\mu_v-\mu_r\right)(\mu-1) A}{(\mu-1)}\) = \(\frac{\left(\mu_\nu^{\prime}-\mu_r^{\prime}\right)\left(\mu^{\prime}-1\right) A^{\prime}}{\left(\mu^{\prime}-1\right)}\)
or, ωδ = ω’δ’ …… (2)
This is the condition for no dispersion. Here ω and ω’ are the dispersive powers of the prisms and δ and δ’ are the mean deviations by the prisms respectively.
Total deviation: As the two deviations are opposite to each other, the total deviation,
δ – δ’ = (µ – 1)A – (µ’ – 1)A’
Now considering µv – µr = dµ and \(\mu_v^{\prime}\) – \(\mu_r^{\prime}\) = dµ’ we get from equation (1),
dµ ᐧ A = dµ’ ᐧ A’ or, A’ = \(\frac{d \mu}{d \mu^{\prime}}\) ᐧ A
∴ δ – δ’ = (µ – 1)A – (µ’ – 1) ᐧ \(\frac{d \mu}{d \mu^{\prime}}\) ᐧ A
= (µ – 1)A[1 – \(\frac{\left(\mu^{\prime}-1\right) d \mu}{(\mu-1) d \mu^{\prime}}\)]
= (µ – 1)A[1 – \(\frac{d \mu /(\mu-1)}{d \mu^{\prime} /\left(\mu^{\prime}-1\right)}\)]
= δ(1 – \(\frac{\omega}{\omega^{\prime}}\)) …. (3)
Although the dispersion of white light is acceptable during the experiment with the spectrum of light, It is a real problem in different optical instruments. Dispersion may cause the images so formed by such instruments coloured and blurred which is unwanted. It is therefore necessary to deviate the light without dispersing it and prisms (as well as lenses) that do this are called achromatic (Greek ‘without colour’).
Dispersion without mean deviation: in this case the polychromatic ray gets dispersed in different colours. Only the mean ray (i.e., yellow ray) remains parallel to the Incident ray.
Let the refractive indices of the prisms for yellow rays be µ and µ’ and their refracting angles are A and A’ respectively. Thé prisms are placed as shown in Fig. 4.6. Here µ’ > µ.
Let the deviations of the mean ray for two different prisms used be δ and δ’ respectively.
For no mean deviation by this combination
δ – δ’ = 0 or, (µ – 1)A = (µ’ – 1)A’
or, \(\frac{A}{A^{\prime}}\) = \(\frac{\mu^{\prime}-1}{\mu-1}\) …. (4)
This is the condition for no mean deviation. In this case, though no deviation occurs for the yellow ray, all the other rays deviate and hence a spectrum is formed.
Net dispersion: As angular dispersion due to one prism takes place in the opposite direction of the angular dispersion due to the other prism, so net dispersion:
= (δv – δr) – (\(\delta_v^{\prime}-\delta_r^{\prime}\)) = (µv – µr)A – (\(\mu_v^{\prime}-\mu_r^{\prime}\))A’
= dµ ᐧ A – dµ’ ᐧ A’ [∵ µv – µr = dµ, \(\mu_\nu^{\prime}-\mu_r^{\prime}\) = dµ’]
= \(\frac{d \mu}{\mu-1}\) ᐧ (µ – 1)A – \(\frac{d \mu^{\prime}}{\mu^{\prime}-1}\) ᐧ (µ’ – 1)A’ = ωδ – ω’δ’
Numerical Examples
Example 1.
A prism of 6° angle is made of crown glass. The refractive indices of the material of the prism for red and blue light are 1.514 and 1.532 respectively. Find the angular dispersion produced by the prism. Also calculate the dispersive power of the material of the prism.
Solution:
Angular dispersion = δb – δr = (µb – µr) × A
= (1.532 – 1.514) × 6°
= 0.018 × 6°= 0.1080
Mean refractive index, µ = \(\frac{\mu_b+\mu_r}{2}\) = \(\frac{1.532+1.514}{2}\) = 1.523
∴ Dispersive power of the material of the prism,
ω = \(\frac{\mu_b-\mu_r}{\mu-1}\) = \(\frac{1.532-1.514}{1.523-1}\) = \(\frac{0.018}{0.523}\)
= 0.034 (approx.)
Example 2.
The refractive Indices of quartz relative to air in the wavelengths 4500 A and 4600 A are 1.4725 and 1.4650 respectively. What is the angular dispersion in degree ᐧ angstrom-1 unit? Suppose, angle of incidence = 45°.
Solution:
µ1 = 1.4725 and µ2 = 1.4650.
Suppose, angle of refraction in case of light of wavelength 4500 A (= λ1) = r1 and angle of refraction in case of light of wavelength
4600A (=λ2) = r2
Therefore angular dispersion between the given wavelength range δ = r2 – r1.
So, angular dispersion for unit difference of wavelength
= \(\frac{\delta}{\Delta \lambda}\) = \(\frac{r_2-r_1}{\lambda_2-\lambda_1}\)
According to the Fig.
sin i = µ1sin r1 or, sin r1 = \(\frac{\sin 45^{\circ}}{1.4725}\) = 0.4802
∴ r1 = sin-1(0.4802) = 28.7°
Similarly, sin r2 = \(\frac{\sin 45^{\circ}}{1.4650}\) = 0.4827
∴ r2 = sin-1(0.4827) = 28.86°
∴ δ = r2 – r1 = 28.86° – 28.7° = 0.16°
∴ \(\frac{\delta}{\Delta \lambda}\) = \(\frac{0.16}{4600-4500}\) = 0.0016° A-1
Example 3.
The refractive indices of crown-glass for red and blue light are 1.517 and 1.523 respectively. Calculate the dispersive power of crown glass with respect to the two colours.
Solution:
Dispersive power of crown glass, ω = \(\frac{\mu_b-\mu_r}{\mu-1}\)
Here, μb = 1.523, μr = 1.517,
∴ μ = \(\frac{\mu_b+\mu_r}{2}\) = \(\frac{1.523+1.517}{2}\) = 1.520
∴ ω = \(\frac{1.523-1.517}{1.520-1}\) = \(\frac{0.006}{0.520}\) = 0.0115
Example 4.
The refractive indices of crown and flint glass for red light are 1.515 and 1.644. Again, refractive indices of crown and flint glass for violet light are 1.532 and 1.685 respectively. For making the lens of spectacles, which glass would be more suitable and why?
Solution:
Dispersive power of crown glass,
1.532 – 1.515 0.017
ωC = \(\frac{1.532-1.515}{\left(\frac{1.532+1.515}{2}\right)-1}\) = \(\frac{0.017}{0.523}\) = 0.0325
Dispersive power of flint glass,
ωF = \(\frac{1.685-1.644}{\left(\frac{1.685+1.644}{2}\right)-1}\) = \(\frac{0.041}{0.6645}\) = 0.0617
For spectacles, the material of the lens should have minimum dispersive power so as to minimise chromatic aberration, Hence crown glass is more suitable than flint glass as the material for spectacles.
Second method:
Angular dispersion for crown glass,
(µv – µr)A = (1.532 – 1.515)A = 0.017 A
Angular dispersion for flint glass,
(µv – µr)A = (1.685 – 1.644)A = 0.041 A
If it is considered that lens of a spectacle is the combination of prisms then A will be the refracting angle of any such prism. Angular dispersion should be minimum for spectacle lens. As crown glass has less angular dispersion than flint glass, for preparation of spectacle lens crown glass is more preferable.
Example 5.
Write down the expression for dispersive power. Refractive index of crown glass for violet and red colour are 1.523 and 1.513 respectively. Calculate the dispersive power of crown glass. [WBCHSE Sample Question]
Solution:
Refractive index of crown glass for violet light, µv = 1.523; refractive index of crown glass for red light µr = 1.513.
So, average refractive index, µ = \(\frac{1.523+1.513}{2}\) = 1.518
∴ Dispersive power, ω = \(\frac{\mu_\nu-\mu_r}{\mu-1}\) = \(\frac{1.523-1.513}{1.518-1}\) = 0.0193