20: Dispersion And Spectra : Neet-xii-physics

Home Learn Fast select Course

Change Unit Change Subject Last Menu

NEET-XII-Physics

20: dispersion and Spectra

with Solutions - page 2

Note: Please signup/signin free to get personalized experience.

10 minutes can boost your percentage by 10%

Note: Please signup/signin free to get personalized experience.

Qstn #1
A narrow beam of white light goes through a slab having
parallel faces.
(a) The light never splits in different colours.
(b) The emergent beam is white.
(c) The light inside the slab is split into different
colours.
(d) The light inside the slab is white.
digAnsr: b,c
Ans : (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:

Qstn #2
By properly combining two prisms made of different
materials, it is possible to
(a) have dispersion without average deviation
(b) have deviation without dispersion
(c) have both dispersion and average deviation
(d) have neither dispersion nor average deviation.
digAnsr: a,b,c
Ans : (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.
Page No 442:

Qstn #3
In producing a pure spectrum, the incident light is
passed through a narrow slit placed in the focal plane
of an achromatic lens because a narrow slit
(a) produces less diffraction
(b) increases intensity
(c) allows only one colour at a time
(d) allows a more parallel beam when it passes through
the lens.
digAnsr: d
Ans : (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:

Qstn #4
Which of the following quantities related to a lens
depend on the wavelength or wavelengths of the incident
light ?
(a) Power. (b) Focal length.
(c) Chromatic aberration. (d) Radii of curvature.
digAnsr: a,b,c
Ans : (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:

Qstn #5
Which of the following quantities increase when
wavelength is increased ? Consider only the magnitudes.
(a) The power of a converging lens.
(b) The focal length of a converging lens.
(c) The power of a diverging lens.
(d) The focal length of a diverging lens.
digAnsr: b,d
Ans : (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.

#
Section : iv

Qstn #1
A flint glass prism and a crown glass prism are to be combined in such a way that the deviation of the mean ray is zero. The refractive index of flint and crown glasses for the mean ray are 1.620 and 1.518 respectively. If the refracting angle of the flint prism is 6.0°, what would be the refracting angle of the crown prism?
Ans : 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`` °``.
Page No 442:

Qstn #2
A certain material has refractive indices 1.56, 1.60 and 1.68 rfor red, yellow and violet lightespectively.
Ans : 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°

#2-a
Calculate the dispersive power.
Ans : Dispersive power `` \left(\omega \right)`` is given by
`` \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)}``
`` =\frac{0.12}{0.60}=0.2``

#2-b
Find the angular dispersion produced by a thin prism of angle 6° made of this material.
Ans : 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:

Qstn #3
The focal lengths of a convex lens for red, yellow and violet rays are 100 cm, 98 cm and 96 cm respectively. Find the dispersive power of the material of the lens.
Ans : 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:
`` \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]``
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}``
`` \,\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}``
`` \Rightarrow \omega =0.0408``
Thus, the dispersive power of the material of the lens is 0.048.
Page No 442:

Qstn #4
The refractive index of a material changes by 0.014 as the colour of the light changes from red to violet. A rectangular slab of height 2.00 cm made of this material is placed on a newspaper. When viewed normally in yellow light, the letters appear 1.32 cm below the top surface of the slab. Calculate the dispersive power of the material.
Ans : 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\,}}``
`` \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``
`` \,\mathrm{\,Also\,},``
`` \,\mathrm{\,Dispersive\,}\,\mathrm{\,power\,},\omega =\frac{{\mu }_{v}-{\mu }_{r}}{{\mu }_{y}-1}``
`` =\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:

Qstn #5
A thin prism is made of a material having refractive indices 1.61 and 1.65 for red and violet light. The dispersive power of the material is 0.07. It is found that a beam of yellow light passing through the prism suffers a minimum deviation of 4.0° in favourable conditions. Calculate the angle of the prism.
Ans : 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\,}:``
`` \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°``
Thus, the angle of the prism is 7`` °``.
Page No 442:

Qstn #6
The minimum deviations suffered by, yellow and violet beams passing through an equilateral transparent prism are 38.4°, 38.7° and 39.2° respectively. Calculate the dispersive power of the medium.
Ans : 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}``
`` =\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)}``
`` ``
`` \Rightarrow \omega =\frac{(0.8)}{38.7}=0.0206``
So, the dispersive power of the medium is 0.0206.
Page No 442:

Qstn #7
Two prisms of identical geometrical shape are combined with their refracting angles oppositely directed. The materials of the prisms have refractive indices 1.52 and 1.62 for violet light. A violet ray is deviated by 1.0° when passes symmetrically through this combination. What is the angle of the prisms?
Ans : 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
`` A=\frac{\delta }{{\mu }_{2}-{\mu }_{1}}=\frac{1}{(1.62)-(1.52)}=\frac{1}{0.1}``
`` \Rightarrow A=10°``
So, the angle of the prisms is 10^∘.
Page No 442:
Page No 442: