Question

In youth's double slit experiment Imax:Imin = 49:9. Calculate the ratio of the intensities of the individuals sources.

Answer 1

Answer :

The ratio of the intensities of the individuals sources = 25 : 4

Explanation :

Given :

In Young's double slit experiment, Iₘₐₓ : Iₘᵢₙ = 49 : 9

To find :

the ratio of the intensities of the individual sources

Solution :

Let the intensities of individual sources be I₁ and I₂

Iₘₐₓ = (√I₁ + √I₂)²

 Iₘᵢₙ = (√I₁ - √I₂)²

$\sf \dfrac{I_{max}}{I_{min}} =\dfrac{49}{9} \\\\ \sf \dfrac{(\sqrt{I_1}+\sqrt{I_2})^2}{(\sqrt{I_1}-\sqrt{I_2})^2} =\dfrac{49}{9} \\\\ \sf \bigg(\dfrac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\bigg)^2 =\dfrac{49}{9} \\\\ \sf \dfrac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}} =\sqrt{\dfrac{49}{9}} \\\\ \sf \dfrac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}} =\sqrt{\dfrac{7^2}{3^2}} \\\\ \sf \dfrac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}} =\dfrac{7}{3}$

3(√I₁ + √I₂) = 7(√I₁ - √I₂)

3√I₁ + 3√I₂ = 7√I₁ - 7√I₂

7√I₁ - 3√I₁ = 3√I₂ + 7√I₂

 4√I₁ = 10√I₂

  $\sf \dfrac{\sqrt{I_1} }{\sqrt{I_2} } =\dfrac{10}{4} \\\\ \sf \dfrac{I_1 }{I_2} =\bigg(\dfrac{10}{4} \bigg)^2 \\\\ \sf \dfrac{I_1 }{I_2} =\dfrac{100}{16} \\\\ \sf \dfrac{I_1 }{I_2} =\dfrac{4 \times 25}{4 \times 4} \\\\ \sf \dfrac{I_1 }{I_2} =\dfrac{25}{4}$

∴ The ratio of the intensities of the individuals sources = 25 : 4