Gratings: Dispersion and Resolving Power MCQ Quiz in मल्याळम - Objective Question with Answer for Gratings: Dispersion and Resolving Power - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT

Dispersion:

• Visible light, also known as white light, consists of a collection of component colours. 
• Upon passage through the prism, the white light is separated into its component colours - red, orange, yellow, green, blue and violet.
• The separation of visible light into its different colours is known as dispersion.

EXPLANATION:

• The cause of dispersion of light is that white light consists of seven different colours, and each colour has a different angle of deviation.
• Therefore, on passing through the prism different colours deviate through different angles.
• Hence the seven colours of white light separate and form a spectrum.
• Out of seven colours, the red colour deviates the least, and hence the red colour is present at the top of the spectrum. On the other hand, the violet colour deviates most that's why violet colour is present at the lower end of the spectrum.
• This results in the separation of colours, producing a band of seven colours. Hence, option (4) is correct.

CONCEPT:

Dispersion of light:

• Dispersion of light is the phenomenon of splitting of a beam of white light into its constituent colours on passing through a prism.
• The band of seven colours so obtained is called the (visible) spectrum.
• The dispersion of white light occurs because the colours of white light travel at different speeds through the glass prism.

EXPLANATION:

• When the dispersion of white light takes place through a prism, the maximum deviation is suffered by violet colour and minimum deviation suffered red colour. Therefore option 4 is correct.

CONCEPT:

Dispersive power (ω):

• It is the ability of the prism materials to cause dispersion.
• It is defined as the ratio of the angular dispersion to the mean deviation.

\(\omega = \frac{Angular \, dispersion}{Mean\, devaition}\)

\(ω = \frac{{{\delta _v} - {\delta _r}}}{{{\delta _y}}} = \frac{{{μ _v} - {μ _r}}}{{{u_y} - 1}}\)

EXPLANATION:

• From above it is clear that, dispersive power is independent of the angle of the prism and depends only on the material of the prism. Therefore option 2 is correct.

• The phenomenon of splitting of a beam of white light into its constituent seven colors is known as dispersion
• Dispersion of light can be seen in prism, rainbow and similar kind of optical device
• As shown below when the dispersion of white light takes place through a prism, it splits white light into constituent seven color

• Whereas reflection can be defined as it is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated
• Light travels in a straight-line path. When a small opaque object is kept in its path, the light tends to bend around the object and not walk in the same straight line. This is known as the diffraction of light.
• Polarization of light is also defined as the process or phenomenon in which the waves of light or other electromagnetic radiation are restricted to a certain direction of vibration, usually specified in terms of the electric field vector. 

CONCEPT:

Dispersive power (ω):

• It is the ability of the prism materials to cause dispersion.
• It is defined as the ratio of the angular dispersion to the mean deviation.

\(\omega = \frac{Angular \, dispersion}{Mean\, devaition}\)

\(ω = \frac{{{\delta _v} - {\delta _r}}}{{{\delta _y}}} = \frac{{{μ _v} - {μ _r}}}{{{u_y} - 1}}\)

EXPLANATION:

• From above it is clear that, dispersive power is independent of the angle of the prism and depends only on the material of the prism. Therefore option 1 is correct.

CONCEPT:

• Refractive Index: The Refractive index measures the bending of a ray of light when passing from one medium into another medium.
  • Every medium has a different refractive index value.
  • The refractive index of a medium is given by n = c / v

where c is the speed of light and v is the speed of light in the medium.

• Our earth consists of mainly 6 layers.​​

1. ​Troposphere 2. Stratosphere 3. Mesosphere 4. Thermosphere 5. Ionosphere 6. Exosphere

EXPLANATION:

• A star appears to twinkle due to the atmospheric refraction of starlight.
• Our earth consists of many layers.​​
• The starlight when enters the earth’s atmosphere, undergoes refraction through different layers of atmosphere before it reaches the earth.

• The atmospheric refraction occurs in a medium due to gradually changing the refractive index of the different mediums.
• This makes a star to twinkle for our eyes.
• So the correct answer will be option 1.

CONCEPT:

Dispersive power (ω):

• It is the ability of the prism materials to cause dispersion.
• It is defined as the ratio of the angular dispersion to the mean deviation.

\(\omega = \frac{Angular \, dispersion}{Mean\, devaition}\)

\(ω = \frac{{{\delta _v} - {\delta _r}}}{{{\delta _y}}} = \frac{{{μ _v} - {μ _r}}}{{{u_y} - 1}}\)

CALCULATION:

Given - The ration of focal length, f1 : f2 = 2 : 3 

• Condition for achromatism is

\(\Rightarrow \frac{\omega_1}{f_1}+ \frac{\omega_2}{f_2}=0\)

\(\Rightarrow \frac{\omega_1}{\omega_2}= \frac{f_1}{f_2}=\frac{2}{3}\)

Concept:

Prism and Refraction:

• A prism is an optical element with flat, polished surfaces that refract light.
• The refractive index (μ" id="MathJax-Element-314-Frame" role="presentation" style="position: relative;" tabindex="0">μμ $\mu$ ) of a material is a measure of how much the speed of light is reduced inside the material.
• Angle of minimum deviation (δm" id="MathJax-Element-315-Frame" role="presentation" style="position: relative;" tabindex="0">δmδm $\delta_m$ ) is the smallest angle through which light is deviated by a prism.
• Formula for the angle of minimum deviation for a prism:

\( \delta_m = 2 \sin^{-1}\left( \frac{\mu \sin \frac{A}{2}}{\mu'} \right) - A \)

• Here, μ" id="MathJax-Element-316-Frame" role="presentation" style="position: relative;" tabindex="0">μμ $\mu$ is the refractive index of the prism material.
• μ′" id="MathJax-Element-317-Frame" role="presentation" style="position: relative;" tabindex="0">μ′μ′ $\mu'$ is the refractive index of the surrounding medium.
• A is the prism's refracting angle.
• δm" id="MathJax-Element-318-Frame" role="presentation" style="position: relative;" tabindex="0">δmδm $\delta_m$ is the angle of minimum deviation.

Calculation:

Given,

Refractive index of glass, μ=32" id="MathJax-Element-319-Frame" role="presentation" style="position: relative;" tabindex="0">μ=32μ=32 $\mu = \frac{3}{2}$

Refractive index of water, μ′=43" id="MathJax-Element-320-Frame" role="presentation" style="position: relative;" tabindex="0">μ′=43μ′=43 $\mu' = \frac{4}{3}$

Refracting angle of prism, A=60∘=π3" id="MathJax-Element-321-Frame" role="presentation" style="position: relative;" tabindex="0">A=60∘=π3A=60∘=π3 $A = 60^\circ = \frac{\pi}{3}$

The angle of minimum deviation for the prism is,

\( \delta_m = 2 \sin^{-1}\left( \frac{\mu \sin \frac{A}{2}}{\mu'} \right) - A \)

⇒ δm=2sin−1⁡(32sin⁡π643)−π3" id="MathJax-Element-322-Frame" role="presentation" style="position: relative;" tabindex="0">δm=2sin−1(32sinπ643)−π3δm=2sin−1⁡(32sin⁡π643)−π3 $\delta_m = 2 \sin^{-1}\left( \frac{\frac{3}{2} \sin \frac{\pi}{6}}{\frac{4}{3}} \right) - \frac{\pi}{3}$

⇒ δm=2sin−1⁡(32×1243)−π3" id="MathJax-Element-323-Frame" role="presentation" style="position: relative;" tabindex="0">δm=2sin−1(32×1243)−π3δm=2sin−1⁡(32×1243)−π3 $\delta_m = 2 \sin^{-1}\left( \frac{\frac{3}{2} \times \frac{1}{2}}{\frac{4}{3}} \right) - \frac{\pi}{3}$

⇒ δm=2sin−1⁡(34×12)−π3" id="MathJax-Element-324-Frame" role="presentation" style="position: relative;" tabindex="0">δm=2sin−1(34×12)−π3δm=2sin−1⁡(34×12)−π3 $\delta_m = 2 \sin^{-1}\left( \frac{3}{4} \times \frac{1}{2} \right) - \frac{\pi}{3}$

⇒ δm=2sin−1⁡(38)−π3" id="MathJax-Element-325-Frame" role="presentation" style="position: relative;" tabindex="0">δm=2sin−1(38)−π3δm=2sin−1⁡(38)−π3 $\delta_m = 2 \sin^{-1}\left( \frac{3}{8} \right) - \frac{\pi}{3}$

⇒ δm=2sin−1⁡(916)−π3" id="MathJax-Element-326-Frame" role="presentation" style="position: relative;" tabindex="0">δm=2sin−1(916)−π3δm=2sin−1⁡(916)−π3 $\delta_m = 2 \sin^{-1}\left( \frac{9}{16} \right) - \frac{\pi}{3}$

∴ The correct answer is option 3: 2sin−1⁡(916)−π3" id="MathJax-Element-327-Frame" role="presentation" style="position: relative;" tabindex="0">2sin−1(916)−π32sin−1⁡(916)−π

Concept:

Resolving power:

• When two objects or their images are very close together, they may appear as one and it may be impossible for the eye to see them separately. 
• If we want to see as separate objects we make use of optical instruments such as telescopes, microscopes, prisms and grating etc.
• The separation of such close objects is known as Resolution.
• The ability of an optical instrument to form distinctly separate
  images of two objects very close together is called its resolving power. 
• where the geometrical position between two near objects is to be “resolved” known as geometrical resolution.
• A small difference of Wavelengths of light in a given source is “resolved” known as spectral resolution.

Case-1 If an optical instrument “just” resolves two spectral lines of wavelength λ and λ+dλ, then,

Resolving power of the instrument = λ/dλ 

Case-2 Ifdθ is the smallest angular separation that can just be resolved by the telescope, then dθ=1.22λ/d. Therefore,

Resolving power of telescope = \(\frac{1}{d\theta } = \frac{d}{1.22λ}\)

Explanation:

From the above discussion, we can conclude that, 

• In the case of spectral resolution, we can use an optical instrument like grating and prism and

Resolving power of these optical instruments = λ/dλ  

Additional Information

Rayleigh's criterion:

• According to Rayleigh's criterion, the two images of such point objects lying close to each other are said to be resolved if the central maximum of one falls on the first secondary minimum of the other.
• In other words, when the central bright image of one falls on the dark ring of the other, two said to be resolved.

CONCEPT:

• Primary colours are those colours which when mixed in specific proportions can form any colour. 
• The spectral colours i.e. blue, red and green are the primary colours. 
• Secondary colours such as yellow, magenta etc. can be produced by mixing two primary colours in the right proportion. 
• Any two colours which when added produce white light are known as complementary colours. 

EXPLANATION:

• From the above diagram, it is clear that when a red light from one source and green light from another source is falling on the white screen, then the screen will appear as yellow. Therefore option 2 is correct.