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#### Page No 441:

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.

#### Page No 441:

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 ${\mu }_{v}>{\mu }_{r}$.

#### Page No 441:

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.

#### Page No 441:

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.

#### Page No 441:

Yes, the focal length of a lens depends on the colour of light.
According to lens-maker's formula,
$\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)$
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.

#### Page No 441:

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.

#### Page No 441:

(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 $\delta =\left(\mu -1\right)A$.

#### Page No 442:

For the crown glass, we have:
Refractive index for red rays = μr
Refractive index for yellow rays = μy
Refractive index for violet rays = μv
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 δcy and δfy 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:
δy = δcyδfy
= 2δcyδfy
=2(μcy + 1)A − (μfy − 1)A'
Angular dispersion produced by the combination is given by
δv δr = [(μvc − 1)A − (μvf − 1)A' + (μvc − 1)A$\left[\left({\mu }_{rc}-1\right)A-\left({\mu }_{rf}-1\right)A\text{'}+\left({\mu }_{rc}-1\right)A\right]$
Here,
μvc = Refractive index for the violet colour of the crown glass
μvf = 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:
δv δr = 2(μvc −1)A (μvf 1)A'
(a) For zero angular dispersion, we have:
δt δt = 0 = 2(μvc −1)A (μvf 1)A'

(b) For zero deviation in the yellow ray, δy = 0.
⇒ 2(μcy − 1)A = (μfy − 1)A

#### Page No 442:

Given:
Refractive index of the flint glass, μf = 1.620
Refractive index of the crown glass, μc = 1.518
Refractive angle of the flint prism, Af = 6°
Now,
Let the refractive angle of the crown prism be Ac.
For the net deviation of the mean ray to be zero,
Deviation by the flint prism = Deviation by the crown prism
i.e., (μf − 1)Af = (μc − 1)Ae
$⇒{A}_{c}=\left(\frac{{\mu }_{\mathrm{f}}-1}{{\mu }_{\mathrm{e}}-1}\right){A}_{\mathrm{f}}$
$⇒{A}_{\mathrm{c}}=\left(\frac{1.620-1}{1.518-1}\right)×6.0°=7.2°$

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

#### Page No 442:

Given:
Refractive index for red light, μr = 1.56
Refractive index for yellow light, μy = 1.60
Refractive index for violet light, μv = 1.68
Angle of prism, A = 6°

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

$=\frac{0.12}{0.60}=0.2$

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

#### Page No 442:

Focal lengths of the convex lens:
For red rays,
For yellow rays,
For violet rays,
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:

For red rays, ${\mu }_{r}-1=\frac{k}{100}$
For yellow rays,
For violet rays,
Dispersive power (ω) is given by

$\omega =\frac{{\mathrm{\mu }}_{v}-{\mathrm{\mu }}_{r}}{{\mathrm{\mu }}_{y}-1}\phantom{\rule{0ex}{0ex}}\mathrm{Or},\omega =\frac{\left({\mathrm{\mu }}_{v}-1\right)-\left({\mathrm{\mu }}_{r}-1\right)}{\left({\mathrm{\mu }}_{y}-1\right)}$
Substituting the values, we get:

$\omega =\frac{\frac{k}{96}-\frac{k}{100}}{\frac{k}{98}}=\frac{98×4}{9600}\phantom{\rule{0ex}{0ex}}⇒\omega =0.0408$

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

#### Page No 442:

Given:
Difference in the refractive indices of violet and red lights = 0.014
Let μv and μr be the refractive indices of violet and red colours.
Thus, we have:
μvμr = 0.014
Now,
Real depth of the newspaper = 2.00 cm
Apparent depth of the newspaper = 1.32 cm

Thus, the dispersive power of the material is 0.027.

#### Page No 442:

The refractive indices for red and yellow lights are μr = 1.61 and μy = 1.65, respectively.
Dispersive power, ω = 0.07
Angle of minimum deviation, δy = 4°

$⇒{\mu }_{y}-1=\frac{0.04}{0.07}=\frac{4}{7}$
Let the angle of the prism be A.
Angle of minimum deviation, δ = (μ − 1)A
$⇒A=\frac{{\delta }_{y}}{{\mu }_{y}-1}=\frac{4}{\left(\frac{4}{7}\right)}=7°$
Thus, the angle of the prism is 7$°$.

#### Page No 442:

Given:
Minimum deviations suffered by
Red beam, δr = 38.4°
Yellow beam, δy = 38.7°
Violet beam, δv = 39.2°
If A is the angle of prism having refractive index μ, then the angle of minimum deviation is given by
$\delta =\left(\mu -1\right)A$
$⇒$$\left(\mu -1\right)=\frac{\delta }{A}$    ...(1)
Dispersive power $\left(\omega \right)$ is given by

From equation (1), we get:

$⇒\omega =\frac{{\delta }_{v}-{\delta }_{r}}{{\delta }_{y}}=\frac{\left(39.2\right)-\left(38.4\right)}{\left(38.7\right)}\phantom{\rule{0ex}{0ex}}$
$⇒\omega =\frac{\left(0.8\right)}{38.7}=0.0206$
So, the dispersive power of the medium is 0.0206.

#### Page No 442:

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 = (μ2μ1 )A
$A=\frac{\delta }{{\mu }_{2}-{\mu }_{1}}=\frac{1}{\left(1.62\right)-\left(1.52\right)}=\frac{1}{0.1}\phantom{\rule{0ex}{0ex}}⇒A=10°$
So, the angle of the prisms is 10.

#### Page No 442:

(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 $\delta =\left(\mu -1\right)A$.
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.

#### Page No 442:

(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.

#### Page No 442:

(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.

#### Page No 442:

(c) fv < fr

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:

fv < fr

#### Page No 442:

(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.

#### Page No 442:

(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
${\delta }_{y}=\left({\mu }_{y}-1\right)A-\left({\mu }_{y}\text{'}-1\right)A\text{'}$
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.

#### Page No 442:

(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.

#### Page No 442:

(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.

#### Page No 442:

(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.

#### Page No 443: For crown glass, we have:
Refractive index for red colour, μcr = 1.515
Refractive index for violet colour, μcv = 1.525
For flint glass, we have:
Refractive index for red colour, μfr = 1.612
Refractive index for violet colour,μfv = 1.632
Refracting angle, A = 5°
Let:
δc = Angle of deviation for crown glass
δf  = 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 = δc + δf
= (${\mu }_{c}$ – 1)A + (${\mu }_{f}$ – 1)A
= (${\mu }_{c}$${\mu }_{f}$ – 2)A
For violet light, δv = (μcv + μfv – 2)A
For red light, δr = (μcr + μfr – 2)A
Now, we have:
Angular dispersion of the combination:
δvδr = (μcv + μfv – 2)A – (μcr + μfr – 2)A
= (μcv + μfvμcrμfr) A
= (1.525 + 1.632 – 1.515 – 1.612)5
= 0.15°
So, the angular dispersion produced by the combination is 0.15°.

#### Page No 443:

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, μy= 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:
δy = (μy – 1)A – (μ'y – 1)A' = 0
$\therefore$ (1.60 – 1)A = (1.50 – 1)A'
⇒ A = $\frac{0.50×6°}{0.60}=5°$ (b) Net angular dispersion on passing a beam of white light:
(μy – 1)ωA – (μy – 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
δy = (μy – 1)A + (μy – 1)A'
= (1.60 – 1)5° + (1.50 – 1)6°
= 3° + 3° = 6°

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

#### Page No 443:

If μ'v and μ'r are the refractive indices of material M1, then we have:
μ'vμ'r = 0.014
If μv and μr are the refractive indices of material M2, then we have:
μvμr = 0.024
Now,
Angle of prism for M1, A' = 5.3°
Angle of prism for M2, A = 3.7°
(a) When the prisms are oppositely directed, angular dispersion $\left({\delta }_{1}\right)$ is given by
δ1 = (μvμr)A – (μ'vμ'r)A'
On substituting the values, we get:
δ1 = 0.024 × 3.7° – 0.014 × 5.3°
= 0.0146°
So, the angular dispersion is 0.0146°.

(b) When the prisms are similarly directed, angular dispersion$\left({\delta }_{2}\right)$ is given by
δ2 = (μvμr)A + (μ'vμ'r)A'
On substituting the values, we get:
δ2 = 0.024 × 3.7° + 0.014 × 5.3°
= 0.163°
So, the angular dispersion is 0.163°.

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