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White light is passed through a double slit and interference is observed on a screen 1.5m away. The separation between the slits is 0.3mm. The first violet and red fringes are formed 2.0mm and 3.5mm away from the central white fringes. The difference in wavelengths of red and violet light is………nm.

2 Views | Posted 3 months ago

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Vishal Baghel | Contributor-Level 10

3 months ago

For first fringe

$\frac{d y_{1}}{D} = 1 \times λ_{1} [n = 1] λ_{1} \to$ Red wavelength

$\frac{d y_{2}}{D} = 1 \times λ_{2} [n = 1] λ_{2} \to$ violet wavelength

$\frac{d}{D} (y_{1} - y_{2}) = λ_{1} - λ_{2}$

$y_{1} - y_{2} = \frac{D}{d} (λ_{1} - λ_{2})$

= $\frac{0.3 \times 1 0^{- 3}}{1.5} (3.5 - 2) \times 1 0^{- 3}$

= 300 × 10^-9

= 300 nm

For first fringe <math> <mrow> <mfrac> <mrow> <mi>d</mi> <msub> <mrow> <mi>y</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <mi>D</mi> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mo>×</mo> <msub> <mrow> <mi>λ</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mrow> <mo> [</mo> <mrow> <mi>n</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>]</mo> </mrow> <msub> <mrow> <mi>λ</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>→</mo> </mrow> </math> Red wavelength <math> <mrow> <mfrac> <mrow> <mi>d</mi> <msub> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <mi>D</mi> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mo>×</mo> <msub> <mrow> <mi>λ</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mrow> <mo> [</mo> <mrow> <mi>n</mi> <mo>=</mo> <mn>1</mn> </mrow> <mo>]</mo> </mrow> <msub> <mrow> <mi>λ</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>→</mo> </mrow> </math> violet wavelength <math> <mrow> <mfrac> <mrow> <mi>d</mi> </mrow> <mrow> <mi>D</mi> </mrow> </mfrac> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>y</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <msub> <mrow> <mi>λ</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>λ</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </math>           <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1815574570"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <msub> <mrow> <mi>y</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mi>D</mi> </mrow> <mrow> <mi>d</mi> </mrow> </mfrac> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>λ</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>λ</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> </math>        <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1815574571"> </o:OLEObject></xml><! [endif]->= <math> <mrow> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>3</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mo>.</mo> <mn>5</mn> </mrow> </mfrac> <mrow> <mo> (</mo> <mrow> <mn>3</mn> <mo>.</mo> <mn>5</mn> <mo>−</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>3</mn> </mrow> </msup> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1029" DrawAspect="Content" ObjectID="_1815574572"> </o:OLEObject></xml><! [endif]->= 300 × 10-9= 300 nm

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