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Angular width of the central maxima in the Fraunhoffer diffraction for $\lambda=6000 Å$ is $\theta_0$. When the same slit is illuminated by another monochromatic light, the angular width decreases by $30 \%$. The wavelength of this light is
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Verified Answer
The correct answer is:
$4200 Å$
The angular width of central maxima is given by
$$
2 \theta=\frac{2 \lambda}{\mathrm{a}}
$$
where, $\lambda$ = wavelength of light used
$\mathrm{a}=$ width of the slit
For $\lambda_1=6000 Å, 2 \theta=\theta_0$ (given)
For another light of wavelength $\lambda_2$ (says), the angular wid th decreases by $30 \%$ so,
$$
2 \theta=\left(\frac{100-30}{100}\right) \theta_0=\frac{70}{100} \theta_0=0.7 \theta_0
$$
As slit width is constant, so using Eq. (i) for these values, we get
$$
\begin{gathered}
\frac{\theta_0}{0.7 \theta_0}=\frac{\lambda_1}{\lambda_2} \\
\Rightarrow \lambda_2=\lambda_1 \times 0.7=6000 \times 0.7=4200 Å
\end{gathered}
$$
$$
2 \theta=\frac{2 \lambda}{\mathrm{a}}
$$
where, $\lambda$ = wavelength of light used
$\mathrm{a}=$ width of the slit
For $\lambda_1=6000 Å, 2 \theta=\theta_0$ (given)
For another light of wavelength $\lambda_2$ (says), the angular wid th decreases by $30 \%$ so,
$$
2 \theta=\left(\frac{100-30}{100}\right) \theta_0=\frac{70}{100} \theta_0=0.7 \theta_0
$$
As slit width is constant, so using Eq. (i) for these values, we get
$$
\begin{gathered}
\frac{\theta_0}{0.7 \theta_0}=\frac{\lambda_1}{\lambda_2} \\
\Rightarrow \lambda_2=\lambda_1 \times 0.7=6000 \times 0.7=4200 Å
\end{gathered}
$$
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