CBSE 12th Standard Physics Subject Wave Optics HOT Questions 3 Mark Questions With Solution 2021
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CBSE 12th Standard Physics Subject Wave Optics HOT Questions 3 Mark Questions With Solution 2021
12th Standard CBSE
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Reg.No. :
Physics
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A jar of height is filled with a transparent liquid of refractive index \(\mu \)(figure). At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the centre, the dot is invisible.
The problem is based on the principle of total inter reflection and area of visibility. -
Show that for a material with refractive index \(\mu \ge \sqrt { 2 } \), light incident at any angle shall be guided along a length perpendicular to the incident face.
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(i) Light passes through two polaroids \({ P }_{ 1 }and{ P }_{ 2 }\) with pass axis of \({ P }_{ 2 }\) with pass axis of \({ P }_{ 2 }\) making in angle \(\theta \) with the pass axis of \({ P }_{ 1 }\) . For what value of \(\theta \) is the intensity of emergent light zero?
(ii) A third polaroid is placed between \({ P }_{ 1 }and{ P }_{ 2 }\) with its pass axis making an angle \(\beta \) with the pass axis is \({ P }_{ 1 }\) . Find the value of \(\beta \) for which the intensity of light from \({ P }_{ 2 }\) is \(\frac { { I }_{ 0 } }{ 8 } \) , where \({ I }_{ 0 }\) is the intensity of light on the polaroid \({ P }_{ 1 }\) . -
Explain with the help of ray diagram, the working of an astronomical telescope. The magnifying power of a telescope in its normal adjustment is 20. If the length of the telescope is 105 cm in this adjustment, find the focal length of the two lenses.
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In an experiment with two coherent sources, the amplitude of the intensity variation is found to be 5% of the average intensity. What will be the relative intensities of the light waves of interfering source.
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CBSE 12th Standard Physics Subject Wave Optics HOT Questions 3 Mark Questions With Solution 2021 Answer Keys
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Let d be the diameter of the disc. The sopt shall be invisible, if the incident rays OA and OB suffer total internal reflection.
Let i be the angle of incidence
Using relationship between refractive index and critical angle, then
\(\sin { i=\frac { 1 }{ \mu } }\)
Using geometry and trigonometry,
Now,
\(\frac { d/2 }{ h } =\tan { i } \Rightarrow \frac { d }{ 2 } =h\tan { i } =h[\sqrt { { \mu }^{ 2 }-1{ ] }^{ -1 } }\)
\( [\because \ From \ the \ figure,\ tani=\frac { 1 }{ \sqrt { { \mu }^{ 2 } } -1 } ]\)
\( d=\frac { 2h }{ \sqrt { { \mu }^{ 2 } } -1 } \)
This is required expression of d. -
Any ray entering at an angle i shall be guided along AC, if the ray makes an angle \(\phi \) with the face AC greater than the critical angle as the principle of total internal reflection,\(\phi \)+r = 90o, therefore, sin \(\phi \) = cos r.
\(\Rightarrow \ sin\phi \ge \frac { 1 }{ \mu } \ \Rightarrow \ cos\ r\ge \frac { 1 }{ \mu } \)
or \(1-{ cos }^{ 2 }r\le 1-\frac { 1 }{ { \mu }^{ 2 } } \Rightarrow { sin }^{ 2 }r\le 1-\frac { 1 }{ { \mu }^{ 2 } } \left[ \therefore 1-{ cos }^{ 2 }r={ sin }^{ 2 }r \right] \) -
(i) By Malus law, intensity of emergent light from \({ P }_{ 2 }\) is \(I={ I }_{ \circ }cos^{ 2 }\theta \) , where \(\theta \) is the angle between \({ P }_{ 1 }and{ P }_{ 2 }\)|
When \(\theta ={ 90 }^{ \circ }\)
\(\\ I={ I }_{ \circ }\times 0\)
Intensity of emergent light, I = 0
(ii) Intensity of light from \({ P }_{ 3 }\)
\(=\left( \frac { { I }_{ \circ } }{ 2 } cos^{ 2 }\beta \right) \left[ cos^{ 2 }\left( { 90 }^{ \circ }-\beta \right) \right] \)
\(=\frac { { I }_{ \circ } }{ 2 } cos^{ 2 }\beta sin^{ 2 }\beta =\frac { { I }_{ \circ } }{ 8 } sin^{ 2 }2\beta \)
\(As,\ \frac { { I }_{ \circ } }{ 8 } sin^{ 2 }2\beta =\frac { { I }_{ \circ } }{ 8 } (given)\)
\(so, \ \left( sin2\beta \right) ^{ 2 }=1 \ \Rightarrow 2\beta ={ 90 }^{ \circ } \ \Rightarrow \beta ={ 45 }^{ \circ }\) -
In normal adjustments, m = \(|\frac { { f }_{ 0 } }{ { f }_{ e } } |=20\\ \)
Also, length of telescope, f0 + fe = 105
20f0 + f e = 105
21f e = 105
f e = 5 cm
f0 = 20 fe = 20 x 5 = 100 cm -
As we know that, \(I={ I }_{ av }\pm 0.05{ I }_{ av }\)
\(\Rightarrow \) \({ I }_{ max }={ I }_{ av }\left( 1+0.05 \right) =1.05{ I }_{ av }\)
\({ I }_{ min }={ I }_{ av }\left( 1-0.05 \right) =0.95{ I }_{ av }\)
Dividing Eqs. (i) and (ii), we get
\(\Rightarrow \) \(\frac { { I }_{ max } }{ { I }_{ min } } =\frac { 1.05 }{ 0.95 } \)
\(\Rightarrow \) \(\left( \frac { r+1 }{ r-1 } \right) ^{ 2 }=\frac { 105 }{ 95 } \)
\(\Rightarrow \) \({ r }^{ 2 }+1+2r)95=105({ r }^{ 2 }+1-2r)\)
\(\Rightarrow \) \({ 10r }^{ 2 }-400r+10=0\Rightarrow r=40\)
\(\frac { { I }_{ 1 } }{ { I }_{ 2 } } ={ r }^{ 2 }=({ 40) }^{ 2 }=1600:1\)
