A small transparent slab containing material of µ = 1.5 is placed along AS₂(figure). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

6 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

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

Explanation- as the refractive index of the class $μ$ , the path difference will be calculated as $? x$ =2dsin $θ$ +( $μ - 1$ )L

For principal maxima ,(path difference is zero)

2dsin $θ$ ₀+( $μ - 1$ )L=0

Sin $θ$ ₀= - $\frac{L (μ - 1)}{2 d}$ = $\frac{- L (0.5)}{2 d}$

Sin $θ$ ₀=-1/16

OP=Dtan $θ$ ₀= Dsin $θ$ ₀=-D/16

For pat $?$ h difference $? \frac{λ}{2}$

2dsin $θ$ ₁+0.5L= $? \frac{λ}{2}$

Sin $θ$ ₁= $\frac{? \frac{λ}{2} - 0.5 L}{2 d}$ = $\frac{? \frac{λ}{2} - \frac{d}{8}}{2 d}$

= $\frac{\frac{λ}{2} - \frac{λ}{8}}{2 λ}$ = $?$ 1/4 -1/16

So two possible values $\frac{1}{4} - \frac{1}{16} = \frac{3}{16}$ and- $\frac{1}{4} - \frac{1}{16}$ = $- \frac{5}{16}$

This is a Long Answer Type Questions as classified in NCERT ExemplarExplanation- as the refractive index of the class <math> <mi>μ</mi> </math> , the path difference will be calculated as <math> <mo>?</mo> <mi>x</mi> </math> =2dsin <math> <mi>θ</mi> </math> +( <math> <mi>μ</mi> <mo>-</mo> <mn>1</mn> </math> )LFor principal maxima ,(path difference is zero)2dsin <math> <mi>θ</mi> </math> 0+( <math> <mi>μ</mi> <mo>-</mo> <mn>1</mn> </math> )L=0Sin <math> <mi>θ</mi> </math> 0= - <math> <mfrac> <mrow> <mrow> <mi>L</mi> <mo>(</mo> <mi>μ</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>d</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mo>-</mo> <mi>L</mi> <mo>(</mo> <mn>0.5</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>d</mi> </mrow> </mrow> </mfrac> </math> Sin <math> <mi>θ</mi> </math> 0=-1/16OP=Dtan <math> <mi>θ</mi> </math> 0= Dsin <math> <mi>θ</mi> </math> 0=-D/16For pat <math> <mo>?</mo> </math> h difference <math> <mo>?</mo> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> 2dsin <math> <mi>θ</mi> </math> 1+0.5L= <math> <mo>?</mo> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> Sin <math> <mi>θ</mi> </math> 1= <math> <mfrac> <mrow> <mrow> <mo>?</mo> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mn>0.5</mn> <mi>L</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>d</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mo>?</mo> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <mrow> <mi>d</mi> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>d</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>λ</mi> </mrow> </mrow> </mfrac> </math> = <math> <mo>?</mo> </math> 1/4 -1/16So two possible values <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </mfrac> </math>  and- <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </mfrac> </math> = <math> <mo>-</mo> <mfrac> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </mfrac> </math>

0 Upvotes 0 Downvotes

A small transparent slab containing material of µ = 1.5 is placed along AS₂(figure). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

A small transparent slab containing material of µ = 1.5 is placed along AS2(figure). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

A small transparent slab containing material of µ = 1.5 is placed along AS₂(figure). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?