KCET · Physics · Wave Optics
In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by \(5 \times 10^{-2} \mathrm{~m}\) towards the slits, the change in fringe width is \(3 \times 10^{-5} \mathrm{~m}\). If separation between the slits is \(10^{-3} \mathrm{~m}\), the wavelength of light used is
- A \(6000 Å\)
- B \(5000 Å\)
- C \(3000 Å\)
- D \(4500 Å\)
Answer & Solution
Correct Answer
(A) \(6000 Å\)
Step-by-step Solution
Detailed explanation
\[
\begin{aligned}
\beta &=\frac{\lambda \mathrm{D}}{\mathrm{d}} \\
\Rightarrow & \frac{\beta}{\beta} & \propto \mathrm{D} \\
\Rightarrow & \frac{\beta_{1}}{\beta_{2}} &=\frac{\mathrm{D}_{1}}{\mathrm{D}_{2}} \\
\Rightarrow \quad \frac{\beta_{1}-\beta_{2}}{\beta_{2}} &=\frac{\mathrm{D}_{1}-\mathrm{D}_{2}}{\mathrm{D}_{2}} \\
\Rightarrow \quad \frac{\Delta \beta}{\Delta \mathrm{D}} &=\frac{\beta_{2}}{\mathrm{D}_{2}}=\frac{\lambda_{2}}{\mathrm{~d}_{2}} \\
\Rightarrow \quad \lambda_{2} &=\frac{3 \times 10^{-5}}{5 \times 10^{-2}} \times 10^{-3} \\
&=6 \times 10^{-7} \mathrm{~m}=6000 Å
\end{aligned}
\]
\begin{aligned}
\beta &=\frac{\lambda \mathrm{D}}{\mathrm{d}} \\
\Rightarrow & \frac{\beta}{\beta} & \propto \mathrm{D} \\
\Rightarrow & \frac{\beta_{1}}{\beta_{2}} &=\frac{\mathrm{D}_{1}}{\mathrm{D}_{2}} \\
\Rightarrow \quad \frac{\beta_{1}-\beta_{2}}{\beta_{2}} &=\frac{\mathrm{D}_{1}-\mathrm{D}_{2}}{\mathrm{D}_{2}} \\
\Rightarrow \quad \frac{\Delta \beta}{\Delta \mathrm{D}} &=\frac{\beta_{2}}{\mathrm{D}_{2}}=\frac{\lambda_{2}}{\mathrm{~d}_{2}} \\
\Rightarrow \quad \lambda_{2} &=\frac{3 \times 10^{-5}}{5 \times 10^{-2}} \times 10^{-3} \\
&=6 \times 10^{-7} \mathrm{~m}=6000 Å
\end{aligned}
\]
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