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A parallel beam of monochromatic light is incident on a narrow rectangular slit of width 1mm. When the diffraction pattern in seen on a screen placed at a distance of 2m, the width of principal maxima is found to be 2.5mm. The wavelength of light is

Option 1 -

6250 Å

Option 2 -

6200 Å

Option 3 -

5890 Å

Option 4 -

6000 Å

2 Views | Posted 2 months ago

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Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

Here the width of principal maxima is 2.5 mm

Then its half width is = $\frac{β}{2}$

$= \frac{2.5}{2} = 1.25 \times 1 0^{- 3} m$

Diffraction angle q = $\frac{(\frac{β}{2})}{D} = \frac{1.25 \times 1 0^{- 3}}{2}$ …(i)

$∴ a θ = λ ∴ θ = \frac{λ}{a}$ …(ii)

From (i) and (ii)

$λ = θ a$

$\frac{1.25 \times 1 0^{- 3}}{2} \times 1 0^{- 3}$

$∴ λ = 6250$ = 6.25 × 10^-7 m

Here the width of principal maxima is 2.5 mmThen its half width is =  <math> <mrow> <mfrac> <mrow> <mi>β</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>   <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mo>.</mo> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mo>.</mo> <mn>2</mn> <mn>5</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> <mi>m</mi> </mrow> </math>             Diffraction angle q = <math> <mrow> <mfrac> <mrow> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mi>β</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mi>D</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mo>.</mo> <mn>2</mn> <mn>5</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>       …(i)  <math> <mrow> <mo>∴</mo> <mi>a</mi> <mi>θ</mi> <mo>=</mo> <mi>λ</mi> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mo>∴</mo> <mi>θ</mi> <mo>=</mo> <mfrac> <mrow> <mi>λ</mi> </mrow> <mrow> <mi>a</mi> </mrow> </mfrac> </mrow> </math>   …(ii)From (i) and (ii)  <math> <mrow> <mi>λ</mi> <mo>=</mo> <mi>θ</mi> <mi>a</mi> </mrow> </math>           <math> <mrow> <mfrac> <mrow> <mn>1</mn> <mo>.</mo> <mn>2</mn> <mn>5</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> </mrow> </math>             <math> <mrow> <mo>∴</mo> <mi>λ</mi> <mo>=</mo> <mn>6</mn> <mn>2</mn> <mn>5</mn> <mn>0</mn> </mrow> </math> = 6.25 × 10-7 m

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A parallel beam of monochromatic light is incident on a narrow rectangular slit of width 1mm. When the diffraction pattern in seen on a screen placed at a distance of 2m, the width of principal maxima is found to be 2.5mm. The wavelength of light is

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