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#### Page No 441:

#### Question 1:

#### Answer:

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:

#### 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.

#### Answer:

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:

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

#### Answer:

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:

#### 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.

#### Answer:

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:

#### 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.

#### Answer:

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.

#### Page No 441:

#### 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.

#### Answer:

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:

#### 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.

#### Answer:

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

#### Page No 442:

#### 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 $\delta =(\mu -1)A$.

#### Answer:

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}

= 2

*δ*

_{cy}−

*δ*

_{fy}

=2

*(*

*μ*

_{cy}+ 1)

*A*− (

*μ*− 1)

_{fy}*A'*

Angular dispersion produced by the combination is given by

*δ*−

_{v}

*δ*

_{r}= [(

*μ*

_{vc}− 1)

*A*− (

*μ*

_{vf}− 1)

*A'*+ (

*μ*

_{v}

_{c}− 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'*

$\Rightarrow \frac{A\text{'}}{A}=\frac{2({\mu}_{\mathrm{vf}}-1)}{({\mu}_{vc}-1)}\phantom{\rule{0ex}{0ex}}=\frac{2({\mu}_{r}-{\mu}_{r})}{({\mu}_{r}-\mu )}$

(b) For zero deviation in the yellow ray,

*δ*

_{y}= 0.

⇒ 2(μ

_{cy}− 1)

*A*= (μ

_{fy}− 1)

*A*

$\Rightarrow \frac{A\text{'}}{A}=\frac{2({\mu}_{\mathrm{cy}}-1)}{({\mu}_{\mathrm{fy}}-1)}\phantom{\rule{0ex}{0ex}}=\frac{2({\mu}_{y}-1)}{(\mu {\text{'}}_{y}-1)}$

#### Page No 442:

#### 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}

= 2

*δ*

_{cy}−

*δ*

_{fy}

=2

*(*

*μ*

_{cy}+ 1)

*A*− (

*μ*− 1)

_{fy}*A'*

Angular dispersion produced by the combination is given by

*δ*−

_{v}

*δ*

_{r}= [(

*μ*

_{vc}− 1)

*A*− (

*μ*

_{vf}− 1)

*A'*+ (

*μ*

_{v}

_{c}− 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'*

$\Rightarrow \frac{A\text{'}}{A}=\frac{2({\mu}_{\mathrm{vf}}-1)}{({\mu}_{vc}-1)}\phantom{\rule{0ex}{0ex}}=\frac{2({\mu}_{r}-{\mu}_{r})}{({\mu}_{r}-\mu )}$

(b) For zero deviation in the yellow ray,

*δ*

_{y}= 0.

⇒ 2(μ

_{cy}− 1)

*A*= (μ

_{fy}− 1)

*A*

$\Rightarrow \frac{A\text{'}}{A}=\frac{2({\mu}_{\mathrm{cy}}-1)}{({\mu}_{\mathrm{fy}}-1)}\phantom{\rule{0ex}{0ex}}=\frac{2({\mu}_{y}-1)}{(\mu {\text{'}}_{y}-1)}$

#### Answer:

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}

$\Rightarrow {A}_{c}=\left(\frac{{\mu}_{\mathrm{f}}-1}{{\mu}_{\mathrm{e}}-1}\right){A}_{\mathrm{f}}$

$\Rightarrow {A}_{\mathrm{c}}=\left(\frac{1.620-1}{1.518-1}\right)\times 6.0\xb0=7.2\xb0$

Thus, the refracting angle of the crown prism is 7.2$\xb0$.

#### Page No 442:

#### 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}

$\Rightarrow {A}_{c}=\left(\frac{{\mu}_{\mathrm{f}}-1}{{\mu}_{\mathrm{e}}-1}\right){A}_{\mathrm{f}}$

$\Rightarrow {A}_{\mathrm{c}}=\left(\frac{1.620-1}{1.518-1}\right)\times 6.0\xb0=7.2\xb0$

Thus, the refracting angle of the crown prism is 7.2$\xb0$.

#### Answer:

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

$\omega =\frac{{\mu}_{v}-{u}_{r}}{{\mu}_{y}-1}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{formula},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\omega =\frac{\left(1.68-1.56\right)}{\left(1.60-1\right)}$

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

(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°.

#### Page No 442:

#### 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

$\omega =\frac{{\mu}_{v}-{u}_{r}}{{\mu}_{y}-1}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{formula},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\omega =\frac{\left(1.68-1.56\right)}{\left(1.60-1\right)}$

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

(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°.

#### Answer:

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* **R*_{1} and *R*_{2} are the radii of curvatures of the lens.

Thus, we have:

$\left(\mu -1\right)=\frac{1}{f}\times \frac{1}{\left({\displaystyle \frac{1}{{R}_{1}}}-{\displaystyle \frac{1}{{R}_{2}}}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\mu -1\right)=\frac{k}{f}\left[k=\frac{1}{\left({\displaystyle \frac{1}{{R}_{1}}}-{\displaystyle \frac{1}{{R}_{2}}}\right)}\right]$

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

$\omega =\frac{{\mathrm{\mu}}_{v}-{\mathrm{\mu}}_{r}}{{\mathrm{\mu}}_{y}-1}\phantom{\rule{0ex}{0ex}}\mathrm{Or},\omega =\frac{({\mathrm{\mu}}_{v}-1)-({\mathrm{\mu}}_{r}-1)}{({\mathrm{\mu}}_{y}-1)}$

Substituting the values, we get:

$\omega =\frac{{\displaystyle \frac{k}{96}}-{\displaystyle \frac{k}{100}}}{{\displaystyle \frac{k}{98}}}=\frac{98\times 4}{9600}\phantom{\rule{0ex}{0ex}}\Rightarrow \omega =0.0408$

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

#### Page No 442:

#### 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* **R*_{1} and *R*_{2} are the radii of curvatures of the lens.

Thus, we have:

$\left(\mu -1\right)=\frac{1}{f}\times \frac{1}{\left({\displaystyle \frac{1}{{R}_{1}}}-{\displaystyle \frac{1}{{R}_{2}}}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\mu -1\right)=\frac{k}{f}\left[k=\frac{1}{\left({\displaystyle \frac{1}{{R}_{1}}}-{\displaystyle \frac{1}{{R}_{2}}}\right)}\right]$

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

$\omega =\frac{{\mathrm{\mu}}_{v}-{\mathrm{\mu}}_{r}}{{\mathrm{\mu}}_{y}-1}\phantom{\rule{0ex}{0ex}}\mathrm{Or},\omega =\frac{({\mathrm{\mu}}_{v}-1)-({\mathrm{\mu}}_{r}-1)}{({\mathrm{\mu}}_{y}-1)}$

Substituting the values, we get:

$\omega =\frac{{\displaystyle \frac{k}{96}}-{\displaystyle \frac{k}{100}}}{{\displaystyle \frac{k}{98}}}=\frac{98\times 4}{9600}\phantom{\rule{0ex}{0ex}}\Rightarrow \omega =0.0408$

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

#### Answer:

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

$\mathrm{Refractive}\mathrm{index}=\frac{\mathrm{Real}\mathrm{depth}}{\mathrm{Apparent}\mathrm{depth}}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Refractive}\mathrm{index}\mathrm{for}\mathrm{yellow}\mathrm{light}\left({\mu}_{\mathrm{y}}\right)\mathrm{is}\mathrm{given}\mathrm{by}\phantom{\rule{0ex}{0ex}}{\mu}_{y}=\frac{2.00}{1.32}=1.515$

$\mathrm{Also},\phantom{\rule{0ex}{0ex}}\mathrm{Dispersive}\mathrm{power},\omega =\frac{{\mu}_{v}-{\mu}_{r}}{{\mu}_{y}-1}\phantom{\rule{0ex}{0ex}}=\frac{0.014}{1.515-1}$

$\mathrm{Or},\omega =\frac{0.014}{0.515}=0.027$

Thus, the dispersive power of the material is 0.027.

#### Page No 442:

#### 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

$\mathrm{Refractive}\mathrm{index}=\frac{\mathrm{Real}\mathrm{depth}}{\mathrm{Apparent}\mathrm{depth}}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Refractive}\mathrm{index}\mathrm{for}\mathrm{yellow}\mathrm{light}\left({\mu}_{\mathrm{y}}\right)\mathrm{is}\mathrm{given}\mathrm{by}\phantom{\rule{0ex}{0ex}}{\mu}_{y}=\frac{2.00}{1.32}=1.515$

$\mathrm{Also},\phantom{\rule{0ex}{0ex}}\mathrm{Dispersive}\mathrm{power},\omega =\frac{{\mu}_{v}-{\mu}_{r}}{{\mu}_{y}-1}\phantom{\rule{0ex}{0ex}}=\frac{0.014}{1.515-1}$

$\mathrm{Or},\omega =\frac{0.014}{0.515}=0.027$

Thus, the dispersive power of the material is 0.027.

#### Answer:

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°

$\mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{relation}\omega =\frac{{\mu}_{v}-{\mu}_{r}}{{\mu}_{y}-1},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\Rightarrow 0.07=\frac{1.65-1.61}{{\mu}_{y}-1}$

$\Rightarrow {\mu}_{y}-1=\frac{0.04}{0.07}=\frac{4}{7}$

Let the angle of the prism be

*A*.

Angle of minimum deviation, δ = (

*μ*− 1)

*A*

$\Rightarrow A=\frac{{\delta}_{y}}{{\mu}_{y}-1}=\frac{4}{\left({\displaystyle \frac{4}{7}}\right)}=7\xb0$

Thus, the angle of the prism is 7$\xb0$.

#### Page No 442:

#### 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°

$\mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{relation}\omega =\frac{{\mu}_{v}-{\mu}_{r}}{{\mu}_{y}-1},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\Rightarrow 0.07=\frac{1.65-1.61}{{\mu}_{y}-1}$

$\Rightarrow {\mu}_{y}-1=\frac{0.04}{0.07}=\frac{4}{7}$

Let the angle of the prism be

*A*.

Angle of minimum deviation, δ = (

*μ*− 1)

*A*

$\Rightarrow A=\frac{{\delta}_{y}}{{\mu}_{y}-1}=\frac{4}{\left({\displaystyle \frac{4}{7}}\right)}=7\xb0$

Thus, the angle of the prism is 7$\xb0$.

#### Answer:

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

$\omega =\frac{{\mu}_{v}-{\mu}_{r}}{{\mu}_{y}-1}\phantom{\rule{0ex}{0ex}}=\frac{({\mu}_{v}-1)-({\mu}_{r}-1)}{({\mu}_{y}-1)}$

From equation (1), we get:

$\omega =\frac{{\displaystyle \frac{{\delta}_{v}}{A}-\frac{{\delta}_{r}}{A}}}{{\displaystyle \frac{{\delta}_{y}}{A}}}$

$\Rightarrow \omega =\frac{{\delta}_{v}-{\delta}_{r}}{{\delta}_{y}}=\frac{(39.2)-(38.4)}{(38.7)}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \omega =\frac{(0.8)}{38.7}=0.0206$

So, the dispersive power of the medium is 0.0206.

#### Page No 442:

#### 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

$\omega =\frac{{\mu}_{v}-{\mu}_{r}}{{\mu}_{y}-1}\phantom{\rule{0ex}{0ex}}=\frac{({\mu}_{v}-1)-({\mu}_{r}-1)}{({\mu}_{y}-1)}$

From equation (1), we get:

$\omega =\frac{{\displaystyle \frac{{\delta}_{v}}{A}-\frac{{\delta}_{r}}{A}}}{{\displaystyle \frac{{\delta}_{y}}{A}}}$

$\Rightarrow \omega =\frac{{\delta}_{v}-{\delta}_{r}}{{\delta}_{y}}=\frac{(39.2)-(38.4)}{(38.7)}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \omega =\frac{(0.8)}{38.7}=0.0206$

So, the dispersive power of the medium is 0.0206.

#### Answer:

Let *A* be the angle of the prisms.

Refractive indices of the prisms for violet light, *μ*_{1} = 1.52 and *μ*_{2} = 1.62

Angle of deviation, *δ* = 1.0°

As the prisms are oppositely directed, the angle of deviation is given by*δ* = (*μ*_{2} − 1)*A* − (*μ*_{1} − 1)*A **δ* = (*μ*_{2} −*μ*_{1} )*A*

$A=\frac{\delta}{{\mu}_{2}-{\mu}_{1}}=\frac{1}{(1.62)-(1.52)}=\frac{1}{0.1}\phantom{\rule{0ex}{0ex}}\Rightarrow A=10\xb0$

So, the angle of the prisms is 10^{âˆ˜}.

#### Page No 442:

#### Question 2:

Let *A* be the angle of the prisms.

Refractive indices of the prisms for violet light, *μ*_{1} = 1.52 and *μ*_{2} = 1.62

Angle of deviation, *δ* = 1.0°

As the prisms are oppositely directed, the angle of deviation is given by*δ* = (*μ*_{2} − 1)*A* − (*μ*_{1} − 1)*A **δ* = (*μ*_{2} −*μ*_{1} )*A*

$A=\frac{\delta}{{\mu}_{2}-{\mu}_{1}}=\frac{1}{(1.62)-(1.52)}=\frac{1}{0.1}\phantom{\rule{0ex}{0ex}}\Rightarrow A=10\xb0$

So, the angle of the prisms is 10^{âˆ˜}.

#### Answer:

(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.

#### Page No 442:

#### 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.

#### Answer:

(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:

#### 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.

#### Answer:

(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:

#### 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.

#### Answer:

(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:

$f\propto \frac{1}{{\lambda}^{2}}(\because {\lambda}_{r}{\lambda}_{v})\phantom{\rule{0ex}{0ex}}$

∴ *f*_{v} < *f*_{r}

#### Page No 442:

#### 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:

$f\propto \frac{1}{{\lambda}^{2}}(\because {\lambda}_{r}{\lambda}_{v})\phantom{\rule{0ex}{0ex}}$

∴ *f*_{v} < *f*_{r}

#### Answer:

(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.

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#### 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.

#### Answer:

(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.

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#### 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.

#### Answer:

(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.

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#### 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.

#### Answer:

(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.

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#### 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.

#### Answer:

(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.

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#### 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.

#### Answer:

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°.

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#### 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°.

#### Answer:

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* = $\frac{0.50\times 6\xb0}{0.60}=5\xb0$

(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°

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#### 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* = $\frac{0.50\times 6\xb0}{0.60}=5\xb0$

(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°

#### Answer:

If *μ'*_{v} and *μ'*_{r}_{}are the refractive indices of material *M*_{1}, then we have:*μ'*_{v} – *μ'*_{r} = 0.014

If *μ*_{v}_{}and *μ*_{r} are the refractive indices of material *M _{2}*, then we have:

*μ*

_{v}–

*μ*

_{r}= 0.024

Now,

Angle of prism for

*M*

_{1}

_{, }A' = 5.3°

Angle of prism for

*M*

_{2}, A = 3.7°

(a) When the prisms are oppositely directed, angular dispersion $\left({\delta}_{1}\right)$ is given by

*δ*

_{1}= (

*μ*

_{v}–

*μ*

_{r})

*A*– (

*μ'*

_{v}–

*μ'*

_{r})

*A*'

On substituting the values, we get:

*δ*

_{1}= 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

*δ*

_{2}= (

*μ*

_{v}–

*μ*

_{r})

*A*+ (

*μ'*

_{v}–

*μ'*

_{r})

*A'*

On substituting the values, we get:

*δ*

_{2}= 0.024 × 3.7° + 0.014 × 5.3°

= 0.163°

So, the angular dispersion is 0.163°.

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