Q129 / 180 · KCET 2022 · Physics
The fringe width for red colour as compared to that for violet colour is approximately
- Correct answerA2 times
- B4 times
- C8 times
- D3 times
Solution
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,
$$
\begin{aligned}
\lambda_{\text {red }} & \approx 2 \lambda_{\text {violet }} \\
\frac{\beta_{\text {red }}}{\beta_{\text {violet }}} & =\frac{\lambda_{\text {red }}}{\lambda_{\text {violet }}}=\frac{2 \lambda_{\text {violet }}}{\lambda_{\text {violet }}}=2 \\
\Rightarrow \quad \beta_{\text {red }} & =2 \beta_{\text {violet }}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.