Physics, asked by AnandMishra7010, 11 months ago

(a) If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.
(b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light?

Answers

Answered by bestwriters
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(a) Ratio of the maximum and minimum intensity of the fringe is 33.8

Let the intensity of light from one slit = I = I₁

Let the intensity of the light from another slit = 0.5I = I₂

Maximum intensity:

The maximum intensity is given by the formula:

I_{\max }=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}

On substituting the values in above formula, we get,

I_{\max }=(\sqrt{I}+\sqrt{0.5 I})^{2}=2.9 I

Minimum intensity:

The minimum intensity is given by the formula:

I_{\min }=(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}

On substituting the values in above formula, we get,

I_{\min }=(\sqrt{I}-\sqrt{0.5 I})^{2}=0.086 I

Thus, the ratio of intensity is:

\frac{I_{\max }}{I_{\min }}=\frac{2.9}{0.086}=33.8

(b) The kind of fringe is white fringe.

  • The monochromatic light has only one colour whereas the white light has seven colours.
  • So, when the white light undergoes interference then the white fringes are observed at the centre whereas coloured fringes are found in different position of the screen.
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