KPK 11th Class Physics Chapter 9 Physical Optics Short Questions Answers
KPK 11th Class Physics Chapter 9 Physical Optics Short Questions with answers are combined for all 11th class(Intermediate/hssc) Level students.Here You can prepare all Physics Chapter 9 Physical Optics short question in unique way and also attempt quiz related to this chapter. Just Click on Short Question and below Answer automatically shown. After each question you can give like/dislike to tell other students how its useful for each.
Class/Subject: 11th Class Physics
Chapter Name: Physical Optics
Board: All KPK Boards
- Malakand Board 11th Class Physics Chapter 9 Physical Optics short questions Answer
- Mardan Board 11th Class Physics Chapter 9 Physical Optics short questions Answer
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- Swat Board 11th Class Physics Chapter 9 Physical Optics short questions Answer
- Dera Ismail Khan Board 11th Class Physics Chapter 9 Physical Optics short questions Answer
- Kohat Board 11th Class Physics Chapter 9 Physical Optics short questions Answer
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KPK 11th Class Physics Chapter 9 Physical Optics Short Questions Answers
When the soap bubble is about to break, the thickness of the soap film becomes too small i.e. almost equal to zero. So the path difference b/w the two parts of the beam is zero. As one part of the beam is reflected from the denser medium [upper surface] therefore a path difference of ‘s/2’ occurs for ray ‘I’ and the 2nd ray is reflected from a rare medium [Lower surface] so there is no path difference for it. Now the total path difference b/w its ray and 2nd ray become ‘s/2’ due to which destructive interference takes palace and as a result, the soap bubble looks black.
In a Michelson interferometer, a 2nd glass plate is used to introduce the same retardation in beam ‘2’ as introduced in beam ‘1’ by its two passages through its plat ‘A’ so the path difference b/w the two beam of light is zero and as a result construction interference occurs due to which we can observe bright fringes with the help of telescope.
Interference
Diffraction
It is the superposition of two light waves.
It is the bending of light around edges.
Interference fringes may or may not of same width.
Diffraction fringes are not of the same width.
The interference fringes have high intensity.
The diffraction fringes have low intensity.
Points of minimum intensity are perfectly dark and all bright bands are of uniform intensity.
Points of minimum intensity are not perfectly dark and all bright bands are not of uniform intensity.
The Brewster’s law gives us the relation b/w refractive index of two medium and tangent of angle of incidence. i.e. Explain: let an un polarized beam of light is allowed to fall on the surface of medium (1) as shown in the figure at point of incidence, one part of incident beam is reflected and is found to be polarized totally. The other part is refracted and is found to be partially polarized. Let ‘ip‘ represent the angle of incidence or polarizing angle and ‘r’ represents the angle of refraction, then from figure, we have, ip+ 90°+r=180° Now from snell’s law, we have, n1sin ip = n2sin r …………………………(2) Putting eq (1) in eq (2) we get, n1 sin i p= n2sin 90°-ip Eq(3) represents the Brewster’s law this equation shows that for given values of n1, n1, we can calculate the angle of incidence or polarizing angle ‘ip‘.
n2/n1=tan i1
1) Condition for constructive interference: For constructive interference, the path difference b/w the two waves is an integral multiple of the wavelength i.e. P.d = mλ…………………..(1) Where, m 0, 1, 2, 3, ……………………….. And λ= wave length of light coming the sources. For destructice interference, the path difference b/w the two waves is an odd of integral multiple of hold of wavelength i.e. P.d = (m+ 1/2)λ ………………………(2) Where m = 0,1,2,3………………. And λ= wave length of light.
The wave trains from the two separate head lamp are in phase. Their phase difference is not constant. Thus the coherence condition are not satisfied. Also these source are not monochromatic. Due to the these reasons, it is not possible to see the interference where the light beam from the head lamps of a car overlap.
The sound light both are waves. But the wavelength of sound waves is very large as compared to the wavelength of light waves. Hence the sound waves are bent round corners of pole. So they are heard. On the other hand, the wavelength of light waves is not comparable to the dimensions of the pole and therefore, it is not diffracted. Thus the light waves can not be observed around the corners.
In young’s double slit experiment, double slits are used to observe interference wave having very short wave length 10-10 m as compare to visible light. Therefore they do not show diffraction effect by using ordinary diffracting object, diffraction grating etc. As in young,s double slit experiment, an ordinary diffracting objects I,e slits are used, so it is not possible to young,s double slits experiment.
The Huygen,s principle tell us that how the wavefronts propagate in the space from point to point. So we can apply the huygen,s principle to radar waves.
The phenomenon of polarization revealed that light waves are transverse in nature. It is because, polarization effect occurs only in case of transverse waves. The longitudinal waves shows no polarization effect. Therefore, polarization of light has only decided that light waves are transverse in nature.
