KCET · Physics · Wave Optics
In Young's double slit experiment, fringes of width \(\beta\) are produced on a screen kept at a distance of \(1 \mathrm{~m}\) from the slit. When the screen is moved away by \(5 \times 10^{-2} \mathrm{~m}\), fringe width changes by \(3 \times 10^{-5} \mathrm{~m}\). The separation between the slits is \(1 \times 10^{-3} \mathrm{~m}\). The wavelength of the light used is
- A \(400 \mathrm{~nm}\)
- B \(500 \mathrm{~nm}\)
- C \(600 \mathrm{~nm}\)
- D \(700 \mathrm{~nm}\)
Answer & Solution
Correct Answer
(C) \(600 \mathrm{~nm}\)
Step-by-step Solution
Detailed explanation
In YDSE, we have
\[
\begin{aligned}
&\text { Wavelength } \begin{aligned}
\lambda &=\frac{\Delta \beta \mathrm{d}}{\Delta \mathrm{D}}=\frac{3 \times 10^{-5} \times 1 \times 10^{-3}}{5 \times 10^{-2}} \\
&=6 \times 10^{-7} \mathrm{~m}
\end{aligned} \\
&\text { or } \lambda=600 \mathrm{~nm}
\end{aligned}
\]
\[
\begin{aligned}
&\text { Wavelength } \begin{aligned}
\lambda &=\frac{\Delta \beta \mathrm{d}}{\Delta \mathrm{D}}=\frac{3 \times 10^{-5} \times 1 \times 10^{-3}}{5 \times 10^{-2}} \\
&=6 \times 10^{-7} \mathrm{~m}
\end{aligned} \\
&\text { or } \lambda=600 \mathrm{~nm}
\end{aligned}
\]
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