Dispersion of Light Questions in English

(B) According to Cauchy's dispersion formula,the refractive index $\mu$ of a material is inversely proportional to the square of the wavelength $\lambda$ of light,expressed as $\mu \approx A + \frac{B}{\lambda^2}$.
Since the wavelength of violet light $(\lambda_v)$ is the shortest among the visible spectrum and the wavelength of red light $(\lambda_r)$ is the longest $(\lambda_v < \lambda_r)$,the refractive index is inversely related to the wavelength.
Therefore,the refractive index is greatest for the light with the shortest wavelength,which is violet light.

(B) The dispersive power $\omega$ of a lens is defined as $\omega = \frac{f_R - f_V}{f_y}$, where $f_R$ is the focal length for red light, $f_V$ is the focal length for violet (or blue) light, and $f_y$ is the mean focal length.
For a thin lens, the mean focal length is approximately $f_y = \sqrt{f_R f_V}$.
Given: $f_R = 100 \text{ cm}$, $f_V = 96.8 \text{ cm}$.
First, calculate the mean focal length: $f_y = \sqrt{100 \times 96.8} = \sqrt{9680} \approx 98.387 \text{ cm}$.
Now, calculate the dispersive power: $\omega = \frac{100 - 96.8}{98.387} = \frac{3.2}{98.387} \approx 0.0325$.
Thus, the correct option is $B$.

(B) The dispersive power $(\omega)$ of a material is defined as the ratio of the angular dispersion for the extreme colours (violet and red) to the mean deviation produced by the prism.
Mathematically,it is expressed as:
$\omega = \frac{\delta_v - \delta_r}{\delta_y}$
Since the deviation $\delta$ produced by a thin prism is given by $\delta = (\mu - 1)A$,where $A$ is the angle of the prism,we substitute this into the formula:
$\omega = \frac{(\mu_v - 1)A - (\mu_r - 1)A}{(\mu_y - 1)A}$
Simplifying the expression by cancelling $A$:
$\omega = \frac{\mu_v - 1 - \mu_r + 1}{\mu_y - 1}$
$\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$
Therefore,the correct option is $B$.

(B) The dispersive power $\omega$ of a prism is defined as the ratio of the angular dispersion to the mean deviation produced by the prism.
Mathematically,$\omega = \frac{\mu_v - \mu_r}{\mu - 1}$,where $\mu_v$ and $\mu_r$ are the refractive indices for violet and red light,and $\mu$ is the mean refractive index.
Since $\mu_v, \mu_r,$ and $\mu$ are properties of the material of the prism,the dispersive power $\omega$ depends only on the nature of the material of the prism and is independent of the prism's angle or shape.

(A) According to Cauchy's equation,the refractive index $\mu$ is given by $\mu = a + \frac{b}{\lambda^2}$,where $a$ and $b$ are constants.
Dispersion $D$ is defined as the magnitude of the rate of change of refractive index with respect to wavelength: $D = |\frac{d\mu}{d\lambda}|$.
Calculating the derivative: $\frac{d\mu}{d\lambda} = \frac{d}{d\lambda}(a + b\lambda^{-2}) = -2b\lambda^{-3}$.
Thus,$D = | -\frac{2b}{\lambda^3} | = \frac{2b}{\lambda^3}$.
This shows that $D \propto \frac{1}{\lambda^3}$.
For a new wavelength $\lambda' = 2\lambda$,the new dispersion $D'$ is given by $\frac{D'}{D} = (\frac{\lambda}{\lambda'})^3 = (\frac{\lambda}{2\lambda})^3 = \frac{1}{8}$.
Therefore,$D' = \frac{D}{8}$.

(A) The dispersive power $(\omega)$ of a material is defined as the ratio of the angular dispersion to the mean deviation produced by a prism.
Mathematically,it is given by $\omega = \frac{\mu_v - \mu_r}{\mu - 1}$,where $\mu_v, \mu_r$ are refractive indices for violet and red light,and $\mu$ is the mean refractive index.
Flint glass has a higher refractive index and a greater difference between the refractive indices of extreme colors compared to crown glass.
Therefore,the dispersive power of flint glass is higher than that of crown glass.

(A) The dispersive power $\omega$ is given by the formula $\omega = \frac{n_v - n_r}{n_y - 1}$,where $n_v, n_r$,and $n_y$ are the refractive indices for violet,red,and yellow colours respectively.
For crown glass:
$\omega_{\text{crown}} = \frac{1.5318 - 1.5140}{1.5170 - 1} = \frac{0.0178}{0.5170} \approx 0.034$
For flint glass:
$\omega_{\text{flint}} = \frac{1.6852 - 1.6434}{1.6499 - 1} = \frac{0.0418}{0.6499} \approx 0.064$
Thus,the dispersive powers are $0.034$ and $0.064$ respectively.

(B) Dispersion without deviation is achieved by using a combination of two prisms (usually crown and flint glass) such that the net deviation produced is zero for the mean colour.
In the visible spectrum,the mean colour is taken to be yellow.
Therefore,the net deviation for the yellow ray is zero,which means the emergent yellow ray is parallel to the incident ray.
Thus,the correct option is $B$.

(B) The angular dispersion produced by a prism is given by the formula: $\theta = (\mu_v - \mu_r)A$.
Given:
Refractive index for violet light,$\mu_v = 1.54$.
Refractive index for red light,$\mu_r = 1.52$.
Angle of the prism,$A = 10^{\circ}$.
Substituting the values into the formula:
$\theta = (1.54 - 1.52) \times 10^{\circ}$.
$\theta = 0.02 \times 10^{\circ}$.
$\theta = 0.2^{\circ}$.
Therefore,the angular dispersion is $0.2^{\circ}$.

(C) The dispersive power $\omega$ of a prism is defined as the ratio of the angular dispersion between the extreme colours (violet and red) to the mean deviation produced by the prism.
Mathematically,$\omega = \frac{\mu_V - \mu_R}{\mu_Y - 1}$,where $\mu_V$,$\mu_R$,and $\mu_Y$ are the refractive indices for violet,red,and yellow colours respectively.
Given: $\mu_R = 1.61$,$\mu_Y = 1.63$,and $\mu_V = 1.65$.
Substituting these values into the formula:
$\omega = \frac{1.65 - 1.61}{1.63 - 1}$.
Thus,the correct option is $C$.

(A) The dispersive power $(\omega)$ of a material is defined as the ratio of the angular dispersion for extreme colors to the mean deviation.
It is given by the formula: $\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$
Given:
Refractive index for violet light,$\mu_v = 1.64$
Refractive index for red light,$\mu_r = 1.52$
Refractive index for yellow light,$\mu_y = 1.60$
Substituting the values in the formula:
$\omega = \frac{1.64 - 1.52}{1.60 - 1}$
$\omega = \frac{0.12}{0.60}$
$\omega = 0.2$
Thus,the dispersive power of the material is $0.2$.

(A) Dispersion occurs because the refractive index of a medium depends on the wavelength of light.
Since the refractive index $n$ is related to the speed of light $v$ in the medium by $n = c/v$,where $c$ is the speed of light in a vacuum,a variation in refractive index for different wavelengths implies that light of different wavelengths propagates at different speeds in the medium.
For example,in glass,the refractive index for red light is lower than that for violet light,meaning red light travels faster than violet light.
Therefore,the correct condition for a medium to be dispersive is that light of different wavelengths propagates at different speeds.

(A) The dispersive power $\omega$ of a material is defined as the ratio of the angular dispersion between the extreme colors to the mean deviation.
Mathematically,it is given by the formula: $\omega = \frac{\mu_F - \mu_C}{\mu_D - 1}$
Given:
Refractive index for blue $F$ line,$\mu_F = 1.6333$
Refractive index for red $C$ line,$\mu_C = 1.6161$
Refractive index for yellow $D$ line,$\mu_D = 1.622$
Substituting the values into the formula:
$\omega = \frac{1.6333 - 1.6161}{1.622 - 1}$
$\omega = \frac{0.0172}{0.622}$
$\omega = 0.02765... \approx 0.0276$
Thus,the correct option is $A$.

(A) When white light passes through a glass prism,it splits into its constituent colours. This phenomenon is known as dispersion. Dispersion occurs because the refractive index $(\mu)$ of the glass prism depends on the wavelength $(\lambda)$ of the light. Since different colours have different wavelengths,they undergo different amounts of refraction,causing them to separate. Therefore,the splitting of white light is primarily due to refraction.

(C) The dispersive power $(\omega)$ of a medium is defined as the ratio of the angular dispersion to the mean deviation.
It is given by the formula: $\omega = \frac{\mu_v - \mu_R}{\mu_y - 1}$
Given values:
Refractive index for violet light,$\mu_v = 1.62$
Refractive index for red light,$\mu_R = 1.52$
Refractive index for yellow light,$\mu_y = 1.55$
Substituting these values into the formula:
$\omega = \frac{1.62 - 1.52}{1.55 - 1}$
$\omega = \frac{0.10}{0.55}$
$\omega = \frac{10}{55} = \frac{2}{11} \approx 0.1818$
Rounding to two decimal places,we get $\omega = 0.18$.

(A) The dispersive power $\omega$ of a medium is defined as the ratio of the angular dispersion to the mean deviation.
Mathematically,it is given by the formula: $\omega = \frac{\mu_V - \mu_R}{\mu_Y - 1}$
Given:
Refractive index for violet light,$\mu_V = 1.62$
Refractive index for red light,$\mu_R = 1.42$
Refractive index for yellow light,$\mu_Y = 1.50$
Substituting these values into the formula:
$\omega = \frac{1.62 - 1.42}{1.50 - 1}$
$\omega = \frac{0.20}{0.50}$
$\omega = 0.4$
Thus,the dispersive power of the medium is $0.4$.

(B) The dispersive power $(\omega)$ of a prism is defined as the ratio of the angular dispersion to the mean deviation.
It is given by the formula: $\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$
Where:
$\mu_v = 1.69$ (refractive index for violet colour)
$\mu_r = 1.65$ (refractive index for red colour)
$\mu_y = 1.66$ (refractive index for mean colour)
Substituting the values into the formula:
$\omega = \frac{1.69 - 1.65}{1.66 - 1}$
$\omega = \frac{0.04}{0.66}$
$\omega = \frac{4}{66} = \frac{2}{33} \approx 0.0606$
Rounding to two decimal places,we get $\omega = 0.06$.

(A) The dispersive power $(\omega)$ of a prism is defined as the ratio of the angular dispersion between the extreme colors (violet and red) to the mean deviation (yellow color).
Formula: $\omega = \frac{\delta_V - \delta_R}{\delta_Y}$
Given:
$\delta_V = 3.72^\circ$
$\delta_R = 2.84^\circ$
$\delta_Y = 3.28^\circ$
Substituting the values:
$\omega = \frac{3.72 - 2.84}{3.28}$
$\omega = \frac{0.88}{3.28}$
$\omega = 0.268$
Thus,the dispersive power of the prism material is $0.268$.

(A) Dispersion of light is the phenomenon of splitting of white light into its constituent colors when it passes through a prism.
This occurs because the refractive index of the material of the prism depends on the wavelength of the light.
According to Cauchy's formula,the refractive index $n$ is related to the wavelength $\lambda$ by $n(\lambda) = A + B/\lambda^2 + C/\lambda^4 + \dots$.
Since different colors have different wavelengths,they travel at different speeds in the medium and undergo different amounts of deviation,leading to dispersion.
Therefore,the correct option is $A$.

(B) The phenomenon of splitting white light into its constituent colours is called dispersion. The refractive index $\mu$ of the prism material depends on the wavelength $\lambda$ of the light,as given by Cauchy's equation $\mu(\lambda) = A + B/\lambda^2 + ...$. Since different colours have different wavelengths,they experience different refractive indices $\mu$ within the prism. Consequently,each colour undergoes a different angle of deviation according to the prism formula $\delta = (\mu - 1)A$. Thus,the correct reason is that $\mu$ is different for different $\lambda$.

(A) Cauchy's dispersion formula describes the relationship between the refractive index $n$ of a transparent medium and the wavelength $\lambda$ of light passing through it.
It is empirically given by the expression: $n(\lambda) = A + \frac{B}{\lambda^2} + \frac{C}{\lambda^4} + \dots$
Where $A$,$B$,and $C$ are constants specific to the material.
Thus,the correct formula is $n = A + B\lambda^{-2} + C\lambda^{-4}$.

(D) Dispersive power is an intrinsic property of the material of the optical element. It depends only on the refractive indices of the material for different wavelengths of light (specifically for violet,red,and yellow light). Since all the given objects (convex lens,glass slab,glass prism,and solid sphere) are made of the same glass,they all possess the same dispersive power. Therefore,the property of dispersion exists in all four objects.

(C) The dispersive power $\omega$ of a lens is defined as the ratio of the difference in focal lengths for red and blue light to the mean focal length (yellow light).
Given:
Focal length for blue light,$f_v = 0.20 \ m$
Focal length for yellow light,$f_y = 0.205 \ m$
Focal length for red light,$f_r = 0.214 \ m$
The formula for dispersive power is $\omega = \frac{f_r - f_v}{f_y}$.
Substituting the values:
$\omega = \frac{0.214 - 0.200}{0.205} = \frac{0.014}{0.205} = \frac{14}{205}$.

(A) The dispersive power $\omega$ of a material is defined as the ratio of the angular dispersion to the mean deviation.
It is given by the formula: $\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$
Given values:
Refractive index for blue light (violet end),$\mu_v = 1.67$
Refractive index for red light,$\mu_r = 1.63$
Refractive index for yellow light (mean),$\mu_y = 1.65$
Substituting these values into the formula:
$\omega = \frac{1.67 - 1.63}{1.65 - 1}$
$\omega = \frac{0.04}{0.65}$
$\omega = \frac{4}{65} \approx 0.0615$
Thus,the dispersive power of the material is $0.0615$.

(C) The formula for dispersive power is $\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$.
First,we calculate the mean refractive index $\mu_y$ using the formula $\mu_y = \frac{\mu_v + \mu_r}{2}$.
Substituting the given values: $\mu_y = \frac{1.52 + 1.51}{2} = \frac{3.03}{2} = 1.515$.
Now,substitute $\mu_y$,$\mu_v$,and $\mu_r$ into the dispersive power formula:
$\omega = \frac{1.52 - 1.51}{1.515 - 1} = \frac{0.01}{0.515} \approx 0.0194$.
Rounding to the given options,the correct value is $0.019$.

(C) The longitudinal chromatic aberration $(LCA)$ of a thin lens is given by the product of its dispersive power $(\omega)$ and its focal length $(f)$.
$LCA = \omega \times f$
Given:
Dispersive power $(\omega) = 0.08$
Focal length $(f) = 20 \ cm$
Therefore, $LCA = 0.08 \times 20 \ cm = 1.6 \ cm$.
Thus, the correct option is $C$.

(C) The longitudinal chromatic aberration $(LCA)$ of a thin lens is given by the formula:
$LCA = \omega \times f$
where $\omega$ is the dispersive power and $f$ is the focal length of the lens.
Given:
Focal length $(f)$ = $20 \ cm$
Dispersive power $(\omega)$ = $0.08$
Substituting the values:
$LCA = 0.08 \times 20 \ cm$
$LCA = 1.6 \ cm$
Therefore,the longitudinal chromatic aberration is $1.6 \ cm$.

(A) The dispersive power $\omega$ of a prism is given by the formula: $\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$,where $\mu_v, \mu_r, \mu_y$ are the refractive indices for violet,red,and yellow colors respectively.
For the first prism:
$\omega_1 = \frac{1.5318 - 1.5140}{1.5170 - 1} = \frac{0.0178}{0.5170} \approx 0.034$.
For the second prism:
$\omega_2 = \frac{1.6852 - 1.6434}{1.6499 - 1} = \frac{0.0418}{0.6499} \approx 0.064$.
Thus,the dispersive powers are $0.034$ and $0.064$ respectively.

(B) The angular dispersion is the difference in the angles of deviation for the two wavelengths,which is $(\theta + d\theta) - \theta = d\theta$.
The change in wavelength is given as $d\lambda$.
The dispersive power $\omega$ of a medium or prism is defined as the ratio of the angular dispersion produced by the two wavelengths to the deviation of the mean ray.
Mathematically,$\omega = \frac{\text{Angular dispersion}}{\text{Deviation in mean ray}} = \frac{d\theta}{d\lambda}$.

(D) The refractive index $(n)$ of a material depends on the wavelength $(\lambda)$ of the incident light, a phenomenon known as dispersion. According to Cauchy's equation, $n(\lambda) = A + B/\lambda^2 + C/\lambda^4 + ...$, where $A, B, C$ are constants.
Since the focal length $(f)$ of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$, it follows that $f$ also depends on the wavelength $\lambda$ because $n$ depends on $\lambda$.
Power $(P)$ of a lens is defined as $P = \frac{1}{f}$. Since $f$ depends on $\lambda$, the power $P$ also depends on the wavelength $\lambda$.
Therefore, all the given quantities (refractive index, focal length, and power) depend on the wavelength of the incident light.

(B) Dispersive power $(\omega)$ is defined as the ratio of angular dispersion to the mean deviation.
Mathematically, $\omega = \frac{\delta_2 - \delta_1}{\delta_{avg}}$, where $\delta_2$ and $\delta_1$ are the deviations of the extreme wavelengths and $\delta_{avg}$ is the deviation of the mean wavelength.
Visually, the dispersive power is proportional to the angular spread between the extreme rays ($\lambda_1$ and $\lambda_2$) relative to the deviation of the mean ray $(\lambda_{avg})$.
In diagram $B$, the angular separation between the rays corresponding to $\lambda_1$ and $\lambda_2$ is the largest relative to the deviation of the mean ray $\lambda_{avg}$.
Therefore, diagram $B$ represents the case with the maximum dispersive power.

(B) The dispersive power $\omega$ is given by the formula: $\omega = \frac{\mu_{v} - \mu_{R}}{\mu_{y} - 1}$.
Here,$\mu_{v} = 1.52$ and $\mu_{R} = 1.51$.
The mean refractive index $\mu_{y}$ is calculated as: $\mu_{y} = \frac{\mu_{v} + \mu_{R}}{2} = \frac{1.52 + 1.51}{2} = 1.515$.
Now,substituting these values into the formula for dispersive power:
$\omega = \frac{1.52 - 1.51}{1.515 - 1} = \frac{0.01}{0.515}$.
$\omega \approx 0.0194$.
Rounding to three decimal places,we get $\omega = 0.019$.

(A) The dispersive power $\omega$ is given by the formula $\omega = \frac{\mu_V - \mu_R}{\mu_y - 1}$.
Given: $\mu_V = 1.6333$ (blue) and $\mu_R = 1.6161$ (red).
The mean refractive index $\mu_y$ is calculated as $\mu_y = \frac{\mu_V + \mu_R}{2} = \frac{1.6333 + 1.6161}{2} = 1.6247$.
Now,substitute the values into the formula:
$\omega = \frac{1.6333 - 1.6161}{1.6247 - 1} = \frac{0.0172}{0.6247} \approx 0.02753$.
Rounding to the nearest provided option,we get $\omega = 0.0276$.

(C) The relationship between the refractive index $\mu$ and the wavelength $\lambda$ of light is given by Cauchy's dispersion formula,which is approximately $\mu = A + \frac{B}{\lambda^2}$,where $A$ and $B$ are constants.
This equation shows that as the wavelength $\lambda$ increases,the refractive index $\mu$ decreases.
Graphically,this represents a curve that starts at a high value of $\mu$ for small $\lambda$ and decreases as $\lambda$ increases,asymptotically approaching a constant value $A$ as $\lambda \rightarrow \infty$.
Among the given options,the curve in option $C$ represents this inverse relationship correctly.
Therefore,option $C$ is the most suitable.

(C) The dispersive power $\omega$ is defined as the ratio of the angular dispersion to the mean deviation.
It is given by the formula: $\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}$,where $\mu_y \approx \frac{\mu_v + \mu_r}{2}$.
Given: $\mu_v = 1.5230$ and $\mu_r = 1.5145$.
First,calculate the mean refractive index: $\mu_y = \frac{1.5230 + 1.5145}{2} = \frac{3.0375}{2} = 1.51875$.
Now,substitute the values into the formula: $\omega = \frac{1.5230 - 1.5145}{1.51875 - 1} = \frac{0.0085}{0.51875}$.
Calculating the final value: $\omega \approx 0.016385$,which rounds to $0.0164$.

(N/A) The speed of light in a vacuum,i.e.,$3 \times 10^{8} \ m/s$ (approximately),is a universal constant. It is not affected by the motion of the source,the observer,or both. Hence,none of the given factors affect the speed of light in a vacuum.
$(b)$ The speed of light in a medium depends on the wavelength of light in that medium (due to the phenomenon of dispersion).

It has been known for a long time that when a narrow beam of sunlight,usually called white light,is incident on a glass prism,the emergent light is seen to consist of several colours.
There is actually a continuous variation of colour,but broadly,the different component colours that appear in sequence are: violet,indigo,blue,green,yellow,orange,and red (given by the acronym $VIBGYOR$).
The red light bends the least,while the violet light bends the most.
The phenomenon of the splitting of light into its component colours is known as dispersion.
The pattern of colour components of light is called the spectrum of light.

(N/A) Each colour is associated with a specific wavelength of light.
In the visible spectrum, red light is at the long-wavelength end $(\sim 700 \, nm)$, while violet light is at the short-wavelength end $(\sim 400 \, nm)$.
Dispersion occurs because the refractive index of a medium is different for different wavelengths (colours).
For example, the bending (deviation) of the red component of white light is the least, while it is the most for the violet component.
Equivalently, red light travels faster than violet light in a glass prism.
The table below provides the refractive indices for different wavelengths for crown glass and flint glass:

Colour	Wavelength $(nm)$	Crown glass	Flint glass
Violet	$396.9$	$1.533$	$1.663$
Blue	$486.1$	$1.523$	$1.639$
Yellow	$589.3$	$1.517$	$1.627$
Red	$656.3$	$1.515$	$1.622$

(N/A) medium is called $dispersive$ if the refractive index of the medium depends on the wavelength of light. In such a medium,different colors (wavelengths) travel at different speeds,causing light to disperse into its constituent colors.
$A$ medium is called $non-dispersive$ if the refractive index of the medium is independent of the wavelength of light. In such a medium,all colors travel at the same speed.
$1$. $Non-dispersive$ medium: Vacuum (or air,approximately) is a $non-dispersive$ medium. This is why sunlight reaches us as white light rather than separated colors.
$2$. $Dispersive$ medium: Glass or water are examples of $dispersive$ media,as they cause the separation of white light into its spectrum (dispersion).

(C) White light is composed of seven distinct colors,which are collectively known as the spectrum. These colors are arranged in the order of increasing frequency and decreasing wavelength,commonly remembered by the acronym $VIBGYOR$.
The constituent colors are:
$1$. Violet
$2$. Indigo
$3$. Blue
$4$. Green
$5$. Yellow
$6$. Orange
$7$. Red
Therefore,the correct sequence of colors in white light is Violet,Indigo,Blue,Green,Yellow,Orange,and Red.

(B) Dispersion of light is the phenomenon in which white light is split into its constituent colors (spectrum) when it passes through a refractive medium like a glass prism.
This occurs because the refractive index of the material depends on the wavelength of the light.
As a result,different colors travel at different speeds within the medium and bend by different angles,causing them to separate.

(N/A) dispersive medium is a medium in which the phase velocity of a wave depends on its frequency. In the context of optics,a dispersive medium is one where the refractive index $n$ varies with the wavelength $\lambda$ of light. As a result,different colors (wavelengths) of light travel at different speeds through the medium,causing them to refract at different angles. $A$ classic example of a dispersive medium is glass or water,which causes white light to split into its constituent colors (dispersion) when passing through a prism.

(C) The dispersive power $\omega$ of a prism is defined as $\omega = \frac{\delta_v - \delta_r}{\delta_y}$,where $\delta_v$ and $\delta_r$ are the deviations of violet (or blue) and red rays,and $\delta_y$ is the mean deviation,given by $\delta_y = \frac{\delta_v + \delta_r}{2}$.
For the first prism: $\delta_v = 10^{\circ}$,$\delta_r = 6^{\circ}$.
$\delta_{y1} = \frac{10^{\circ} + 6^{\circ}}{2} = 8^{\circ}$.
$\omega_1 = \frac{10^{\circ} - 6^{\circ}}{8^{\circ}} = \frac{4}{8} = 0.5$.
For the second prism: $\delta_v = 8^{\circ}$,$\delta_r = 4.5^{\circ}$.
$\delta_{y2} = \frac{8^{\circ} + 4.5^{\circ}}{2} = 6.25^{\circ}$.
$\omega_2 = \frac{8^{\circ} - 4.5^{\circ}}{6.25^{\circ}} = \frac{3.5}{6.25} = 0.56$.
The ratio of dispersive powers is $\frac{\omega_1}{\omega_2} = \frac{0.5}{0.56} \approx 0.89$.

(B) When light passes through a thick lens,it behaves similarly to a prism.
Different wavelengths (colors) of light travel at different speeds within the lens material,leading to different refractive indices for each color.
This phenomenon,where white light splits into its constituent colors due to the variation of refractive index with wavelength,is known as dispersion.
Consequently,the lens focuses different colors at slightly different points,resulting in colored fringes or blurred images,a defect known as chromatic aberration.
Therefore,the formation of colored images due to thick lenses is explained by the property of dispersion.

(C) In a vacuum,all electromagnetic waves,regardless of their frequency or wavelength,travel at the same speed,which is $c \approx 3 \times 10^8 \ m/s$.
Therefore,the statement that red light travels faster than other colours in a vacuum is incorrect,as all colours of white light travel at the same speed in a vacuum.
Option $A$ is correct because violet light has the shortest wavelength and experiences the most deviation.
Option $B$ is correct because the refractive index for red light is lower than that for violet light in any material medium,implying $v = c/n$ is higher for red light.
Option $D$ is correct because flint glass has a higher dispersive power and generally higher refractive indices than crown glass.

(A) The dispersive power $(\omega)$ of a prism is defined as the ratio of the angular dispersion $(\delta_v - \delta_r)$ to the mean deviation $(\delta_y)$.
Mathematically, $\omega = \frac{\delta_v - \delta_r}{\delta_y} = \frac{(\mu_v - 1)A - (\mu_r - 1)A}{(\mu_y - 1)A} = \frac{\mu_v - \mu_r}{\mu_y - 1}$.
From this formula, it is clear that the dispersive power depends only on the refractive indices $(\mu_v, \mu_r, \mu_y)$ of the material of the prism for different colors.
It is independent of the angle of the prism $(A)$, the shape, or the size of the prism.
Therefore, the correct option is $A$.