Dispersion and Spectra

Question 1:

Question 2:

Dispersive power depends on angular deviation, and angular deviation is valid only for a small refracting angle and a small angle of incidence. Therefore, dispersive power is not valid for a prism of large refracting angle. It is also not valid for a glass slab or a glass sphere, as it has a large refracting angle.

Question 3:

No, it cannot be negative, as the refractive index for violet light is always greater than that for red light. Also, refractive index is inversely proportional to ${\lambda }^2$. The sign of ω will be positive, as $\mu$ is still greater than 1 and as ${\mu }_v>{\mu }_r$.

Question 4:

No, it is not possible even when prisms are be combined with their refractive angle reversed with respect to each other. There will be at least a net deviation and dispersion equal to the dispersion and deviation produced by a single prism.

Question 5:

No, monochromatic light cannot be used to produce a pure spectrum. A spectrum is produced when a light of different wavelengths is deviated through different angles and gets separated. Monochromatic light, on the other hand, has a single wavelength.

Question 6:

Yes, the focal length of a lens depends on the colour of light.
According to lens-maker's formula,
$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$
Here, f is the focal length, μ is the refractive index, R is the radius of curvature of lens.
The refractive index (μ) depends on the inverse of square of wavelength.

The focal length of a mirror is independent of the colour of light.

Question 1:

A rainbow can be produced using a prism. Another way of producing a rainbow is to dip a mirror inside water, keeping it inclined along the wall of a tumbler. The light coming from water after reflecting from the mirror will give a rainbow.

Question 8:

(a) increases if the average refractive index increases

If μ is the average refractive index and A is the angle of prism, then the angular dispersion produced by the prism is given by $\delta =(\mu -1)A$.

Question 1:

For the crown glass, we have:
Refractive index for red rays = μ_r
Refractive index for yellow rays = μ_y
Refractive index for violet rays = μ_v
For the flint glass, we have:
Refractive index for red rays = μ'_r
Refractive index for yellow rays = μ'_y
Refractive index for violet rays = μ'_v
Let δ_cy and δ_fy be the angles of deviation produced by the crown and flint prisms for the yellow light.
Total deviation produced by the prism combination for yellow rays:
δ_y = δ_cy − δ_fy
    = 2δ_cy − δ_fy
    =2(μ_cy + 1)A − (μ_fy − 1)A'
Angular dispersion produced by the combination is given by
δ_v − δ_r = [(μ_vc − 1)A − (μ_vf − 1)A' + (μ_vc − 1)A − $\left[\left({\mu }_{rc}-1\right)A-\left({\mu }_{rf}-1\right)A\text{'}+\left({\mu }_{rc}-1\right)A\right]$
Here,
μ_{vc =} Refractive index for the violet colour of the crown glass
μ_vf = Refractive index for the violet colour of the flint glass 
${\mu }_{rc}$ = Refractive index for the red colour of the crown glass 
${\mu }_{rf}$ = Refractive index for the red colour of the flint glass  
On solving, we get: 
δ_v − δ_r = 2(μ_vc −1)A − (μ_vf − 1)A'
(a) For zero angular dispersion, we have:
δ_t − δ_t = 0 = 2(μ_vc −1)A − (μ_vf − 1)A'
$$\begin{gathered} \Rightarrow \frac{A\text{'}}{A}=\frac{2({\mu }_{\mathrm{vf}}-1)}{({\mu }_{vc}-1)} \\ =\frac{2({\mu }_r-{\mu }_r)}{({\mu }_r-\mu )} \end{gathered}$$
(b) For zero deviation in the yellow ray, δ_y = 0.
⇒ 2(μ_cy − 1)A = (μ_fy − 1)A
$$\begin{gathered} \Rightarrow \frac{A\text{'}}{A}=\frac{2({\mu }_{\mathrm{cy}}-1)}{({\mu }_{\mathrm{fy}}-1)} \\ =\frac{2({\mu }_y-1)}{(\mu {\text{'}}_y-1)} \end{gathered}$$

Question 2:

Given:
Refractive index of the flint glass, μ_f = 1.620
Refractive index of the crown glass, μ_c = 1.518
Refractive angle of the flint prism, A_f = 6°
Now,
Let the refractive angle of the crown prism be A_c.
For the net deviation of the mean ray to be zero,
Deviation by the flint prism = Deviation by the crown prism
i.e., (μ_f − 1)A_f = (μ_c − 1)A_e
$\Rightarrow A_c=\left(\frac{{\mu }_{\mathrm{f}}-1}{{\mu }_{\mathrm{e}}-1}\right)A_{\mathrm{f}}$
$\Rightarrow A_{\mathrm{c}}=\left(\frac{1.620-1}{1.518-1}\right)\times 6.0°=7.2°$

Thus, the refracting angle of the crown prism is 7.2$°$.

Question 3:

Given:
Refractive index for red light, μ_r = 1.56
Refractive index for yellow light, μ_y = 1.60
Refractive index for violet light, μ_v = 1.68
Angle of prism, A = 6°

(a)  Dispersive power $\left(\omega \right)$ is given by

$$\begin{gathered} \omega =\frac{{\mu }_v-u_r}{{\mu }_y-1} \\ \mathrm{On} \mathrm{substituting} \mathrm{the} \mathrm{values} \mathrm{in} \mathrm{the} \mathrm{above} \mathrm{formula}, \mathrm{we} \mathrm{get}: \\ \omega =\frac{\left(1.68-1.56\right)}{\left(1.60-1\right)} \end{gathered}$$
   $=\frac{0.12}{0.60}=0.2$

(b) Angular dispersion = (μ_v− μ_r)A
                                     =(0.12) × 6° = 0.72°
Thus, the angular dispersion produced by the thin prism is 0.72°.

Question 4:

Focal lengths of the convex lens:
For red rays, $f_r=100 \mathrm{cm}$
For yellow rays, $f_y=98 \mathrm{cm}$
For violet rays, $f_v=96 \mathrm{cm}$
Let:
${\mu }_r$ = Refractive index for the red colour
${\mu }_y$ = Refractive index for the yellow colour
${\mu }_v$ = Refractive index for the violet colour
Focal length of a lens $\left(f\right)$ is given by
$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Here, $\mu$ is the refractive index and R₁ and R₂ are the radii of curvatures of the lens.
Thus, we have:

$$\begin{gathered} \left(\mu -1\right)=\frac{1}{f}\times \frac{1}{\left({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}}\right)} \\ \Rightarrow \left(\mu -1\right)=\frac{k}{f} \left[k=\frac{1}{\left({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}}\right)}\right] \end{gathered}$$

For red rays, ${\mu }_r-1=\frac{k}{100}$
For yellow rays, ${\mathrm{\mu }}_y-1=\frac{k}{98}$
For violet rays, ${\mathrm{\mu }}_v-1=\frac{k}{96}$
Dispersive power (ω) is given by

$$\begin{gathered} \omega =\frac{{\mathrm{\mu }}_v-{\mathrm{\mu }}_r}{{\mathrm{\mu }}_y-1} \\ \mathrm{Or},\omega =\frac{({\mathrm{\mu }}_v-1)-({\mathrm{\mu }}_r-1)}{({\mathrm{\mu }}_y-1)} \end{gathered}$$
Substituting the values, we get:

$$\begin{gathered} \omega =\frac{{\displaystyle \frac{k}{96}}-{\displaystyle \frac{k}{100}}}{{\displaystyle \frac{k}{98}}}=\frac{98\times 4}{9600} \\ \Rightarrow \omega =0.0408 \end{gathered}$$

Thus, the dispersive power of the material of the lens is 0.048.

Question 5:

Given:
Difference in the refractive indices of violet and red lights = 0.014
Let μ_v and μ_r be the refractive indices of violet and red colours.
Thus, we have:
μ_v − μ_r = 0.014
Now,
Real depth of the newspaper = 2.00 cm
Apparent depth of the newspaper = 1.32 cm
$$\begin{gathered} \mathrm{Refractive} \mathrm{index}=\frac{\mathrm{Real} \mathrm{depth}}{\mathrm{Apparent} \mathrm{depth}} \\ \therefore \mathrm{Refractive} \mathrm{index} \mathrm{for} \mathrm{yellow} \mathrm{light} \left({\mu }_{\mathrm{y}}\right) \mathrm{is} \mathrm{given} \mathrm{by} \\ {\mu }_y=\frac{2.00}{1.32}=1.515 \end{gathered}$$
$$\begin{gathered} \mathrm{Also}, \\ \mathrm{Dispersive} \mathrm{power}, \omega =\frac{{\mu }_v-{\mu }_r}{{\mu }_y-1} \\ =\frac{0.014}{1.515-1} \end{gathered}$$

$\mathrm{Or}, \omega =\frac{0.014}{0.515}=0.027$
Thus, the dispersive power of the material is 0.027.

Question 6:

The refractive indices for red and yellow lights are μ_r = 1.61 and μ_y = 1.65, respectively.
Dispersive power, ω = 0.07
Angle of minimum deviation, δ_y = 4°
$$\begin{gathered} \mathrm{Now}, \mathrm{using} \mathrm{the} \mathrm{relation} \omega =\frac{{\mu }_v-{\mu }_r}{{\mu }_y-1}, \mathrm{we} \mathrm{get}: \\ \Rightarrow 0.07=\frac{1.65-1.61}{{\mu }_y-1} \end{gathered}$$
$\Rightarrow {\mu }_y-1=\frac{0.04}{0.07}=\frac{4}{7}$
Let the angle of the prism be A.
Angle of minimum deviation, δ = (μ − 1)A
$\Rightarrow A=\frac{{\delta }_y}{{\mu }_y-1}=\frac{4}{\left({\displaystyle \frac{4}{7}}\right)}=7°$
Thus, the angle of the prism is 7$°$.

Question 7:

Given:
Minimum deviations suffered by
Red beam, δ_r = 38.4°
Yellow beam, δ_y = 38.7° 
Violet beam, δ_v = 39.2°
If A is the angle of prism having refractive index μ, then the angle of minimum deviation is given by
$\delta =(\mu -1)A$
$\Rightarrow$$\left(\mu -1\right)=\frac{\delta }{A}$    ...(1)
Dispersive power $\left(\omega \right)$ is given by
$$\begin{gathered} \omega =\frac{{\mu }_v-{\mu }_r}{{\mu }_y-1} \\ =\frac{({\mu }_v-1)-({\mu }_r-1)}{({\mu }_y-1)} \end{gathered}$$
From equation (1), we get:
$\omega =\frac{{\displaystyle \frac{{\delta }_v}{A}-\frac{{\delta }_r}{A}}}{{\displaystyle \frac{{\delta }_y}{A}}}$
$\Rightarrow \omega =\frac{{\delta }_v-{\delta }_r}{{\delta }_y}=\frac{(39.2)-(38.4)}{(38.7)}$
$\Rightarrow \omega =\frac{(0.8)}{38.7}=0.0206$
So, the dispersive power of the medium is 0.0206.

Question 2:

Let A be the angle of the prisms.
Refractive indices of the prisms for violet light, μ₁ = 1.52 and μ₂ = 1.62
Angle of deviation, δ = 1.0°
As the prisms are oppositely directed, the angle of deviation is given by
δ = (μ₂ − 1)A − (μ₁ − 1)A
δ = (μ₂ −μ₁ )A
$$\begin{gathered} A=\frac{\delta }{{\mu }_2-{\mu }_1}=\frac{1}{(1.62)-(1.52)}=\frac{1}{0.1} \\ \Rightarrow A=10° \end{gathered}$$
So, the angle of the prisms is 10^∘.

Question 3:

(b) decreases

If μ is the refractive index and A is the angle of prism, then the angular dispersion produced by the prism will be given by $\delta =(\mu -1)A$.
Because the relative refractive index of glass with respect to water is small compared to the refractive of glass with respect to air, the dispersive power of the glass prism is more in air than that in water.

Question 4:

(b) δ

In combination (refractive angles of prisms reversed with respect to each other), the deviations through two prisms cancel out each other and the net deviation is due to the third prism only.

Question 5:

(d) Both A and B are correct.

Because line spectra contain wavelengths that are absorbed by atoms and band spectra contain bunch wavelengths that are absorbed by molecules, both statements are correct.

Question 1:

(c) f_v < f_r

Focal length is inversely proportional to refractive index and refractive index is inversely proportional to ${\lambda }^2$. So, keeping other parameters the same, we can say:
$f\propto \frac{1}{{\lambda }^2} (\because {\lambda }_r<{\lambda }_v)$
∴ f_v < f_r

Question 2:

(b) The emergent beam is white.
(c) The light inside the slab is split into different colours.

White light will split into different colours inside the glass slab because the value of refractive index is different for different wavelengths of light; thus, they suffer different deviations. But the emergent light will be white light. As the faces of the glass slide are parallel, the emerging lights of different wavelengths will reunite after refraction.

Question 3:

(a) have dispersion without average deviation
(b) have deviation without dispersion
(c) have both dispersion and average deviation

Consider the case of prisms combined such that the refractive angles are reversed w.r.t. each other. Then, the net deviation of the yellow ray will be
${\delta }_y=({\mu }_y-1)A-({\mu }_y\text{'}-1)A\text{'}$
And, the net angular dispersion will be
${\delta }_y-{\delta }_r=\left({\mu }_y-1\right)A\left(\omega -\omega \text{'}\right)$
Thus, by choosing appropriate conditions, we can have the above mentioned cases.

Question 4:

(d) allows a more parallel beam when it passes through the lens

To produce a pure spectrum, a parallel light beam is required to be incident on the dispersing element. So, the incident light is passed through a narrow slit placed in the focal plane of an achromatic lens.

Question 5:

(a) Power
(b) Focal length
(c) Chromatic aberration

The focal length, power and chromatic aberration are dependent on the refractive index of the lens, which itself is dependent on the wavelength of the light.

Question 9:

(b) The focal length of a converging lens
(d) The focal length of a diverging lens

The focal length of a lens is inversely proportional to the refractive index of the lens and the refractive index of the lens is inversely proportional to the square of wavelength. Therefore, the focal length is directly dependent on wavelength; it increases when the wavelength is increased.

Question 10:

For crown glass, we have:
Refractive index for red colour, μ_cr = 1.515
Refractive index for violet colour, μ_cv = 1.525
For flint glass, we have:
Refractive index for red colour, μ_fr = 1.612
Refractive index for violet colour,μ_fv = 1.632
Refracting angle, A = 5°
Let:
δ_c = Angle of deviation for crown glass
δ_f  = Angle of deviation for flint glass
As prisms are similarly directed and placed in contact with each other, the total deviation produced $\left(\delta \right)$ is given by
δ = δ_c + δ_f
   = (${\mu }_c$ – 1)A + (${\mu }_f$ – 1)A
   = (${\mu }_c$ + ${\mu }_f$– 2)A
For violet light, δ_v = (μ_cv + μ_fv – 2)A
For red light, δ_r= (μ_cr + μ_fr – 2)A
Now, we have:
Angular dispersion of the combination:
δ_v – δ_r= (μ_cv + μ_fv – 2)A – (μ_cr + μ_fr – 2)A
            = (μ_cv + μ_fv – μ_cr – μ_fr) A
            = (1.525 + 1.632 – 1.515 – 1.612)5
            = 0.15°
So, the angular dispersion produced by the combination is 0.15°.

Question 11:

Given:
For the first prism,
Angle of prism, A' = 6°
Angle of deviation, ω' = 0.07
Refractive index for yellow colour, μ'_y = 1.50
For the second prism,
Angle of deviation, ω = 0.08
Refractive index for yellow colour, μ_y= 1.60
Let the angle of prism for the second prism be A.
The prism must be oppositely directed, as the combination produces no deviation in the mean ray.
(a) The deviation of the mean ray is zero.
Thus, we have:
δ_y = (μ_y – 1)A – (μ'_y – 1)A' = 0
$\therefore$ (1.60 – 1)A = (1.50 – 1)A'
⇒ A = $\frac{0.50\times 6°}{0.60}=5°$



(b) Net angular dispersion on passing a beam of white light:
 (μ_y – 1)ωA – (μ_y – 1)ω'A'
⇒ (1.60 – 1)(0.08)(5°) – (1.50 – 1)(0.07)(6°)
⇒ 0.24° – 0.21° = 0.03°



(c) For the prisms directed similarly, the net deviation in the mean ray is given by
  δ_y = (μ_y – 1)A + (μ_y – 1)A'
      = (1.60 – 1)5° + (1.50 – 1)6°
      = 3° + 3° = 6°

(d) For the prisms directed similarly, angular dispersion is given by
δ_v – δ_r = (μ_y – 1)ωA – (μ_y – 1)ω'A'
           = 0.24° + 0.21°   
           = 0.45°