Question:

In a Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light In a Young’s double-slit experiment, using mono chromatic light of wave length λ, the intensity of light at a point on the screen is I0, where the path difference between the interfering waves is λ. The path difference between the interfering waves at a point where the intensity is \(\frac{I_o}{2}\) , will be:at a point on the screen is I0​, where the path difference between the interfering waves is λ. The path difference at a point where the intensity is 2I0​​ will be:

Updated On: Nov 6, 2024
  • \(\frac{\lambda}{4}\)
  • \(\frac{\lambda}{2}\)
  • λ
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The Correct Option is B

Solution and Explanation

\(\textbf{Solution:} \text{ The intensity } I \text{ in a Young's double-slit experiment is related to the path difference } \Delta x \text{ between the waves by the formula:}\)

\[I = I_0 \cos^2 \left( \frac{\Delta x}{2 \lambda} \right)\]

\(\text{When the intensity is } \frac{I_0}{2}, \text{ we have:}\)
\(\frac{I_0}{2} = I_0 \cos^2 \left( \frac{\Delta x}{2 \lambda} \right)\)

\(\text{This simplifies to:}\)
\(\cos^2 \left( \frac{\Delta x}{2 \lambda} \right) = \frac{1}{2}\)

\(\text{Thus, } \Delta x = \frac{\lambda}{2}.\)

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