Do sound waves undergo reflection, rarefaction and polarization phenomena

1. Do sound waves undergo reflection, rarefaction and polarization phenomena?
Ans: Sound wave undergoes reflection and refraction but it does not follow polarization. Since sound wave is longitudinal wave, the vibration of particle occurs to and fro in the direction of the propagation of the wave. The vibrations of such type cannot be limited or controlled by any barriers and so polarization is not possible in them
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2. Is velocity of sound more in damp air or in dry air? Explain.
Ans: The velocity of air is more in damp air than in dry air. In damp air, the density of air molecule is less than that of dry air due of the presence of humidity. Also, the velocity of sound is inversely proportional to the square root of density as per the relation given below.
$$\text{v} \propto \frac{1}{\sqrt{\rho}}$$

Hence, the velocity of sound is more in damp air than in dry air.
3. Although the density of solid is high, the velocity of sound is greater in air. Why ?
Ans: We know that the velocity of sound in solid is given by $\text{v} \propto \frac{\text{Y}}{\sqrt{\rho}}$ where Y is the Young’s modulus and ρ is the density of the solid. Though density of solid is high, the velocity of sound is greater in solid because the value of Young’s modulus is very high in case s of solid.

4. Sound is heard better on wet days than in dry day. Why?
Ans: On a wet day, the air contains large number of water vapour. So, the density of air molecule is less than that in dry days. As per the relation, $\text{v} \propto \frac{1}{\sqrt{\rho}}$, the velocity of sound is inversely proportional to density. So, sound is heard better on wet days than in dry days.

5. A person walking on a railway line hears two sounds for the same explosion occurring far away, why?
Ans: The sound of explosion travels through metal and air. The velocity of propagation of sound in metal is high due to high value of Young’s modulus while the velocity of sound in air is less in compared to that of solid. So, listener hears two sounds.

6. Do sound wave need medium to travel from one medium to another medium in space? What properties of the medium are relevant ?
Ans: Yes, since sound wave is a longitudinal wave, it needs medium for its propagation in space. Some of relevant properties of the medium are as follows.

• The medium should have property of elasticity.
• The medium should possess the property of inertia.
• The medium should have minimum friction.

7. Why explosions on other planet are not heard in earth?
Ans: We know that sound waves require material medium for its propagation. Since there is a vacant space between earth and other planets, sound produced in other planets cannot be heard in earth.

8. Why sound made at a distance can be heard distinctly at night than at day time?
Ans: It is based on the variation of velocity of sound with temperature. The speed of sound and temperature have direct relation given by $\text{v} \propto \text {T}$. So, if temperature is high, the velocity of sound is high and if less, the velocity is also less.
At night, the ground is cooler than the air above  as air high above the air is hotter than those at near ground. So sound waves travelling in slanted manner will travel faster whereas those travelling near the ground will travel slowly. This makes the sound waves become more near each other near the sound but farther apart high above the ground. In addition the waves which travelled up will head back to earth after bending caused due to the difference in velocities.
So, an observer at a distance will get sound that

Laboratory preparation of Hydrogen Gas

Hydrogen gas is prepared in laboratory by the action of dilute HCl upon granulated zinc.

Zn+dil⋅H2SO4⟶ZnSO4(aq)+H2↑

Note: This is a redox reaction since Zn is oxidized and H⁺ ions is reduced here.

Procedure:
The granulated zinc pieces are taken in a Woulfe’s bottle and the apparatus is fitted as shown in figure. The two necks of woulfe’s bottle is tighten up with cork so that no air enters from outside. It is connected with the delivery tube to a water trough containing water. Dilute sulphuric acid is poured slowly from the thistle funnel. Some crystals of copper sulphate are added to increase the rate of reaction. The hydrogen gas thus formed is collected in a gas jar through delivery tube by the downward displacement of water.

Laboratory preparation of Hydogen gas

Purification: The hydrogen gas thus obtained may contain different impurities like H₂S, NO₂, CO₂, moisture etc. So, these impurities are removed by using suitable process to get pure hydrogen.
Very pure hydrogen gas can be prepared in lab by the following reaction. This is a short and sweet method.

Mg+dil⋅H2SO4⟶MgSO4+H2↑

Laboratory Preparation of Oxygen

Oxygen is prepared in lab generally in two ways either by the application of heat or no application of heat.
Using heat:
Oxygen in lab is prepared by heating the mixture of powdered potassium chlorate and manganese dioxide in the ratio 4:1 in a hard glass test tube. The oxygen gas is observed in a gas jar through the downward displacement of water. The reaction involved is given below:

2KClO3−→−−−−−MnO2200−300∘C2KCl+3O2↑

Preparation of oxygen on application of heat

Without heat:
The dry sodium peroxide is taken in a conical flask and the apparatus is fitted as shown in figure. The two necks of woulfe’s bottle is tighten up with cork so that no air enters from outside. It is connected with the delivery tube to a water trough containing water. Water is poured slowly from the thistle funnel. Here sodium peroxide reacts with water at ordinary temperature to give hydrogen gas. The hydrogen gas thus formed is collected in a gas jar through delivery tube by the downward displacement of water. The reaction takes place as below:

2Na2O2+2H2O⟶4NaOH+O2↑

Note: This reaction is very quick and hydrogen gas is collected in a gas jar within a fraction of seconds.

Preparation of oxygen without the application of heat

Diferentiate between plane wavefront and spherical wavefront.

1.  What is wavefront and wavelet?
Ans: The locus of vibrating particles in medium travelling in some phase at that instant is called wavefront. The new spherical wave originating from wavefront of the wave which travels with the speed of wave at that medium is called wavelet.

2. Diferentiate between plane wavefront and spherical wavefront.
Ans: If the source of light is point source, then the locus formed at  any instant is called the spherical wave front. If the source of light is in infinite distance, then the wavefront formed is called plane wavefront.

3. How can we change spherical wavefront to plane wavefront and vice versa?
Ans: We can interchange from one form to another form by using convex lens. When a spherical wavefront is kept at the focus of the lens, it changes into plane wavefront. Likewise if convex lens is kept in the path of plane wavefront, spherical wavefront is formed.


4. What are the advantages of Foucault’s method over Fizeau’s method of measure speed of light?
Ans: The advantages of  Foucault’s method over Fizeau’s method are:

• Foucault’s method can be applied in any optical medium while Fizeau’s method can be applied only in light or vacuum.
• Experiment setup for Foucault’s method can be conducted in a small laboratory while that of Fizeau’s method needs large distance of around 10-12 km.

5. Which parameter of light doesn’t changes on refraction?
Ans: When light undergo refraction, velocity and wavelength of light changes but its frequency remains constant. So, frequency is the fundamental property of light.

6. Explain Huygen’s principle.
Ans:  Huygen’s principle states that:

• Each light source sends out disturbance in all directions that travels in medium called ether medium.
• Light is a longitudinal wave.
• Different colors of light is due to the difference in wavelength.

7. What suggest the dual nature of light?
Ans: The phenomenon like reflection, refraction suggest that light behave like particle while phenomenon like interference, diffraction, polarization suggest that light can behave like wave. So, light is believed to have dual nature.

What is polarizer and analyzer?

What is polarizer and analyzer?
Ans: A polarizer is a substance which is used in cutting the vibration of light radiation in one plane to produce polarized light while analyzer is a substance that detects whether the light is polarized or not.


 What happens to the intensity of light, when it is polarized?
Ans: When the light is polarized, the vibration of electric vector is allowed in only one plane. The vibration to other plane is restricted. So, the intensity of the light gets decreased.

Can sound waves be polarized? Explain.,, What is polaroid? Mention the uses of polaroid.,,What is polarizing angle? Does it depend upon the wavelength of light ?

1. Can sound waves be polarized? Explain.
Ans: No, sound waves cannot be polarized. Sound waves are longitudinal waves. In longitudinal waves, the vibrations of particles are oriented in the direction of motion of the waves. This means sound waves are already polarized.  So, sound waves cannot be polarized further
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2. What is polaroid? Mention the uses of polaroid.
Ans: A polaroid is a device used to produce plane polarized light. Some of the uses of polaroid are as follows:

• It is used for the manufacture of polarized sunglasses.
• It is used in trains and aeroplanes for  the manufacture of polarized window.
• It is used to record and reduce the dimensional moving pictures.
• It is used in LCD screens to display images, letters or numbers in it.

3. What is polarizing angle? Does it depend upon the wavelength of light ?
Ans: The angle of incidence in which complete polarization occurs is called polarizing angle.
From the Canchy’s relation, the refractive index of transparent medium is given by:  $$ \mu = \text{A} +  \frac{\text{B}}{ \lambda^2 }$$
The inverse of this refractive index is the polarizing angle i.e., θ_p = tan^-1(μ). So, polarizing angle depends upon the wavelength of light.

What is the diffreence between Frensel’s diffraction and Fraunhofer diffraction ?,,What happens to the Fraunhofer diffraction of light, when it is placed in water?,, Colored spectrum is seen when we see through muslin cloth. Why?

What is the diffreence between Frensel’s diffraction and  Fraunhofer diffraction ?
 
Ans: During diffraction, If the source of light and screen are at finite distance from the obstacle, then it is called Frensel diffraction whereas if the source of light and screen are kept at infinite distance from the aperture or obstacle, it is called Fraunhofer’s diffraction.
Light does not need to be converged in Frensel’s diffraction. So, no lens is used to converge the beam light. But, lenses and mirrors are used to converge and modification of diffracted light in Fraunhofer’s diffraction.

. What happens to the Fraunhofer diffraction of light, when it is placed in water?
Ans: The wavelength and decreases is less in water than in air. So, wavelength of light is less in water than in air. In diffraction, the width of central maximum has direct relation to the wavelength of the light used i.e., β∝ λ. So, when the apparatus is placed in water, the width of central bright maximum decreases.
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 Colored spectrum is seen when we see through muslin cloth. Why?
Ans: Muslin cloth is made up of very fine thread which acts like thin slits. When white light is passed through this cloth, the light get diffracted and interference of these diffracted beams give rise to a colored spectrum.

What happens to the single slit experiment when the width of the slit is less than wavelength of the waveWhy is diffraction of sound waves more evident (common) than that of light waves?,,We cannot observe the diffraction pattern in a wide slit illuminated by monochromatic light. Why,

What happens to the single slit experiment when the width of the slit is less than wavelength of the wave? 
 
Ans: In single slit experiment, the first central maximum occurs at:

sinθ=λd

Here, λ is the wavelength of light and d is the width of the slit. When the width of slit is less than the wavelength of the wave, the value of λ/d increases. So, diffraction pattern becomes wider.
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 Why is diffraction of sound waves more evident (common) than that of light waves?
Ans

: We know that the wavelength of sound waves greater that of light waves. For a sustained diffraction, the size of the obstacles should be in the order of wavelength of the waves. The wavelength of sound is comparable to the size of the obstacles around which they bend. But, the wavelength of light is in the order of 10^-6 m and thus cannot be compared with the size of obstacles. So, diffraction of light cannot be observed easily.

. We cannot observe the diffraction pattern in a wide slit illuminated by monochromatic light. Why ?

Ans: We know that the size of central maximum is given by:

2y=2λDa

Where D is the distance between the slits and a is the slit width. So, as the slit width increases, the size of central maximum reduces and diffraction pattern cannot be seen. Even the variation of intensity of other maxima and minima become so small to distinguish. Hence, we cannot observe the diffraction pattern in a wide slit illuminated by monochromatic light.

Diffraction grating is better than two slits set up for measuring the wavelength of a monochromatic light. Explain

Diffraction grating is better than two slits set up  for measuring the wavelength of a monochromatic light. Explain. 
Ans: In diffraction grating, the spectrum of the light obtained is very sharp and clear than that of two slit. This helps for precise measurement of wavelength of the monochromatic light.  So, diffraction grating is better than two slits set up for measuring the wavelength of monochromatic light.

What is diffraction of light ? Differentiate between diffraction and interference

Diffraction of light is the phenomenon of bending of light round the corners and spreading into the regions of the geometrical shadow.

Diffraction	Interference
The phenomenon of interaction of light coming from different parts of the same wave front is called diffraction.	The phenomenon of non uniform distribution of light energy (wave) due to the superposition of coherent sources of light is called interference.
In diffraction, the widths of fringes are not equal.	In interference, the width of fringes are equal.
Bands are very less in number.	Bands are very large in number.
Dark fringes in diffraction are not completely dark.	Dark fringes in interference are perfectly dark.

Mirror Formula for Concave Mirror

Let us consider a ray of light AB strikes on the surface of a concave mirror at point B and is reflected back and passes through the point C. Let DB be the normal of the ray. Since angle of incidence is equal to the angle of reflection, we can say that AB = BC. Let α, β and γ be the angle made by the incident ray normal and the reflected ray with the reflecting surface. Now, again a ray of light passes through the principle axis P and strikes at the pole and returns in the same path. Then image is formed at the point I . Also, BE be the perpendicular line drawn from point B to the principle axis. Now,
By using basic geometrical laws,
α + i = β………..(i)
β + i = γ ………. (ii)
On solving equation (i) and (ii),
β – α = γ –β
2 β = γ + β…………(iii)
Since the value of alpha, beta and gamma is very small, we can write Tanα =α,  Tanβ =β and Tanγ= γ.
Now, in triangle BAE,

Tanα=α=BE AE

In triangle BDE,

Tanβ=β=BE DE

Similarly, in triangle BCE,

Tanγ=γ=BE EC

Since, the aperture of a mirror is very small, so the point E lies nearly to the point P, We can write:

α=BE AP

β=BE DP

γ=BE CP

Now, putting the value of α, β and γ in equation (iii) we get:

2BEDP=BECP=BEAP

or,2DP=1CP+1AP

or,2R=1u+1v

1f=1u+1v……(iii)

Equation (iii) is the required mirror formula.

Relation Between Radius of Curvature and Focal Length

Let us consider a ray of light AB strikes on a surface of a concave mirror. Here PD is the radius of curvature and PC is the focal length. Assuming that the aperture of a mirror is very small, we can conclude that PC is nearly equal to PB. Here, angle ABD is an incident angle. BD is a normal and angle DBC is a reflected ray.
Now, using simple geometrical laws in the above figure:
∠ ABD = ∠ CBD  (∵ i = r)
∠ ABD = ∠ BDP( ∵ Being alternate angle)
Since ∆ BDC is an isosceles triangle, CB is equal to CD.
Now, PD = BC + CD
or, PD = PC + CD
or, PD = PC + PC
or, PD = 2PC
∴ R = 2F
This is the required relation between radius of curvature and focal length.