Question:
The fringe width obtained in a given Young’s double slit experimental set up for red light, blue light and green light are, respectively, $\beta_R$, $\beta_B$ and $\beta_G$. Then
The fringe width obtained in a given Young’s double slit experimental set up for red light, blue light and green light are, respectively, $\beta_R$, $\beta_B$ and $\beta_G$. Then
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In YDSE:
- Larger wavelength $\Rightarrow$ larger fringe width
- Red light gives widest fringes
Updated On: Apr 30, 2026
- $\beta_B > \beta_R$
- $\beta_R > \beta_G$
- $\beta_G > \beta_R$
- $\beta_B > \beta_R$
- $\beta_R = \beta_B = \beta_G$
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The Correct Option is B
Solution and Explanation
Concept:
Fringe width in Young’s double slit experiment:
\[
\beta = \frac{\lambda D}{d}
\]
Thus, $\beta \propto \lambda$.
Step 1: Compare wavelengths.
\[ \lambda_R > \lambda_G > \lambda_B \]
Step 2: Relate fringe width.
\[ \beta_R > \beta_G > \beta_B \]
Step 1: Compare wavelengths.
\[ \lambda_R > \lambda_G > \lambda_B \]
Step 2: Relate fringe width.
\[ \beta_R > \beta_G > \beta_B \]
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