KCET · Physics · Wave Optics
The fringe width for red colour as compared to that for violet colour is approximately
- A 2 times
- B 4 times
- C 8 times
- D 3 times
Answer & Solution
Correct Answer
(A) 2 times
Step-by-step Solution
Detailed explanation
We know that, fringe width,
\[
\beta=\frac{D \lambda}{d}
\]
where, \(D=\) distance between the screen and source,
\(d=\) distance between the slits
and \(\lambda=\) wavelength of light used.
\(\Rightarrow \quad \beta \propto \lambda\)
As we know that,
\[
\begin{aligned}
\lambda_{\text {red }} & \approx 2 \lambda_{\text {violet }} \\
\frac{\beta_{\text {red }}}{\beta_{\text {violet }}} & =\frac{\lambda_{\text {red }}}{\lambda_{\text {violet }}}=\frac{2 \lambda_{\text {violet }}}{\lambda_{\text {violet }}}=2 \\
\Rightarrow \quad \beta_{\text {red }} & =2 \beta_{\text {violet }}
\end{aligned}
\]
\[
\beta=\frac{D \lambda}{d}
\]
where, \(D=\) distance between the screen and source,
\(d=\) distance between the slits
and \(\lambda=\) wavelength of light used.
\(\Rightarrow \quad \beta \propto \lambda\)
As we know that,
\[
\begin{aligned}
\lambda_{\text {red }} & \approx 2 \lambda_{\text {violet }} \\
\frac{\beta_{\text {red }}}{\beta_{\text {violet }}} & =\frac{\lambda_{\text {red }}}{\lambda_{\text {violet }}}=\frac{2 \lambda_{\text {violet }}}{\lambda_{\text {violet }}}=2 \\
\Rightarrow \quad \beta_{\text {red }} & =2 \beta_{\text {violet }}
\end{aligned}
\]
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