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Wave Optics Physics Ncert Solutions Class 12th Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Ten

Figure shown a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I 0 �� 0 is the intensity of the principal maxima when no polariser is present, calculte in the present case, the intensity of the principal maxima as well as the first minima.

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Vishal Baghel | Contributor-Level 10

4 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- when polariser is not used

A=A_perp+A

letA₁= asinwt and A₂=asin(wt+ $\emptyset$ )

now superposition principle for perpendicular polariser

A_R= asinwt+ asin(wt+ $\emptyset$ )

A_R=a(2cos $\emptyset / 2$ sin(wt+ $\emptyset$ ))

A_R=2acos $\emptyset / 2$ sin(wt+ $\emptyset$ )

This eqn is also same for parallel polariser

A_R=2acos $\emptyset / 2$ sin(wt+ $\emptyset$ )

And we know that intensity is directly proportional to square of amplitude

(A_R)²= (A_perp)²+(A)²

So resultant intensity is

I=4(a)²cos^{2
$\emptyset / 2 \frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t + \emptyset)$}dt + 4(a)²cos^{2
$\emptyset / 2 \frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t + \emptyset)$}dt

I= 8(a)²cos^{2
$\emptyset / 2$}(1/2) &nb

...more

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- when polariser is not used

A=A_perp+A

letA₁= asinwt and A₂=asin(wt+ $\emptyset$ )

now superposition principle for perpendicular polariser

A_R= asinwt+ asin(wt+ $\emptyset$ )

A_R=a(2cos $\emptyset / 2$ sin(wt+ $\emptyset$ ))

A_R=2acos $\emptyset / 2$ sin(wt+ $\emptyset$ )

This eqn is also same for parallel polariser

A_R=2acos $\emptyset / 2$ sin(wt+ $\emptyset$ )

And we know that intensity is directly proportional to square of amplitude

(A_R)²= (A_perp)²+(A)²

So resultant intensity is

I=4(a)²cos^{2
$\emptyset / 2 \frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t + \emptyset)$}dt + 4(a)²cos^{2
$\emptyset / 2 \frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t + \emptyset)$}dt

I= 8(a)²cos^{2
$\emptyset / 2$}(1/2) (after integrating sin value we got 1/2)

I= 4(a)²cos^{2
$\emptyset / 2$}

So I= 4(a)²

But when there is a polariser

Perpendicular polariser is blocked

I=2(a)²cos^{2
$\emptyset / 2 \frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t + \emptyset)$}dt + (a)^{2
$\frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t)$}

I=2(a)²cos^{2
$\emptyset / 2 +$}1/2(a)²

When $\emptyset = 0$

I_max=2(a)²+1/2(a)²

I_max=5/2(a)²

I_max=5/2(I/4)=5I/8

At first minima it becomes I/8

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This is a Long Answer Type Questions as classified in NCERT ExemplarExplanation- when polariser is not usedA=Aperp+AletA1= asinwt and A2=asin(wt+ <math> <mi>∅</mi> </math> )now superposition principle for perpendicular polariserAR= asinwt+ asin(wt+ <math> <mi>∅</mi> </math> )AR=a(2cos <math> <mi>∅</mi> <mo>/</mo> <mn>2</mn> </math> sin(wt+ <math> <mi>∅</mi> </math> ))AR=2acos <math> <mi>∅</mi> <mo>/</mo> <mn>2</mn> </math>  sin(wt+ <math> <mi>∅</mi> </math> )This eqn is also same for parallel polariserAR=2acos <math> <mi>∅</mi> <mo>/</mo> <mn>2</mn> </math>  sin(wt+ <math> <mi>∅</mi> </math> )And we know that intensity is directly proportional to square of amplitude(AR)2= (Aperp)2+(A)2So resultant intensity isI=4(a)2cos2 <math> <mi>∅</mi> <mo>/</mo> <mn>2</mn> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> </mfrac> <msubsup> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mi>T</mi> </mrow> </msubsup> <mrow> <mrow> <msup> <mrow> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mo>(</mo> <mi>w</mi> <mi>t</mi> <mo>+</mo> <mi>∅</mi> <mo>)</mo> </math> dt + 4(a)2cos2 <math> <mi>∅</mi> <mo>/</mo> <mn>2</mn> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> </mfrac> <msubsup> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mi>T</mi> </mrow> </msubsup> <mrow> <mrow> <msup> <mrow> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mo>(</mo> <mi>w</mi> <mi>t</mi> <mo>+</mo> <mi>∅</mi> <mo>)</mo> </math> dtI= 8(a)2cos2 <math> <mi>∅</mi> <mo>/</mo> <mn>2</mn> </math> (1/2)                                     (after integrating sin value we got 1/2)I= 4(a)2cos2 <math> <mi>∅</mi> <mo>/</mo> <mn>2</mn> </math> So I= 4(a)2But when there is a polariserPerpendicular polariser is blockedI=2(a)2cos2 <math> <mi>∅</mi> <mo>/</mo> <mn>2</mn> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> </mfrac> <msubsup> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mi>T</mi> </mrow> </msubsup> <mrow> <mrow> <msup> <mrow> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mo>(</mo> <mi>w</mi> <mi>t</mi> <mo>+</mo> <mi>∅</mi> <mo>)</mo> </math> dt + (a)2 <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> </mfrac> <msubsup> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mi>T</mi> </mrow> </msubsup> <mrow> <mrow> <msup> <mrow> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mo>(</mo> <mi>w</mi> <mi>t</mi> <mo>)</mo> </math> I=2(a)2cos2 <math> <mi>∅</mi> <mo>/</mo> <mn>2</mn> <mo>+</mo> </math> 1/2(a)2When <math> <mi>∅</mi> <mo>=</mo> <mn>0</mn> </math> Imax=2(a)2+1/2(a)2Imax=5/2(a)2Imax=5/2(I/4)=5I/8At first minima it becomes I/8

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