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In a Young's double slit experiment, a laser light of wave length $560 \mathrm{~nm}$ produces an interference pattern with consecutive bright fringe's separation of $7.2 \mathrm{~mm}$. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of $8.1 \mathrm{~mm}$. The wavelength of second light is
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Verified Answer
The correct answer is:
$630 \mathrm{~nm}$
Wave length of light, $\lambda=560 \mathrm{~nm}=560 \times 10^{-9} \mathrm{~m}$
Fringes separation,
$$
\begin{aligned}
& \beta_1=7.2 \times 10^{-3} \mathrm{~m} \\
& \beta_2=8.1 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$
As we know
$$
\begin{aligned}
& \because \beta=\frac{\lambda \mathrm{D}}{\mathrm{d}} \\
& \frac{\beta_1}{\beta_2}=\frac{\lambda_1}{\lambda_2} \Rightarrow \lambda_2=\frac{\lambda_1 \beta_2}{\beta_1} \\
& =\frac{560 \times 10^{-9} \times 8.1 \times 10^{-3}}{7.2 \times 10^{-3}}=630 \mathrm{~nm}
\end{aligned}
$$
Fringes separation,
$$
\begin{aligned}
& \beta_1=7.2 \times 10^{-3} \mathrm{~m} \\
& \beta_2=8.1 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$
As we know
$$
\begin{aligned}
& \because \beta=\frac{\lambda \mathrm{D}}{\mathrm{d}} \\
& \frac{\beta_1}{\beta_2}=\frac{\lambda_1}{\lambda_2} \Rightarrow \lambda_2=\frac{\lambda_1 \beta_2}{\beta_1} \\
& =\frac{560 \times 10^{-9} \times 8.1 \times 10^{-3}}{7.2 \times 10^{-3}}=630 \mathrm{~nm}
\end{aligned}
$$
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