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A single slit of width $a$ is illuminated by violet light of wavelength $400 \mathrm{~nm}$ and the width of the diffraction pattern is measured as $y$. When half of the slit width is covered and illuminated by yellow light of wavelength $600 \mathrm{~nm}$, the width of the diffraction pattern is
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Verified Answer
The correct answer is:
$3 y$
Given, In first case,
Wavelength of violet light,
$\lambda_{1}=400 \mathrm{~nm}$
We know that, width of the diffraction pattern is given as
$\begin{aligned}
& \beta=y=\frac{2 D \lambda_{1}}{d} \\
\Rightarrow \quad & y=\frac{2 D \lambda_{1}}{d}=\frac{2 D \times 400}{d}
\end{aligned}$
$y=\frac{800 D}{d}...(i)$
In the second case,
$\begin{aligned} \lambda_{2} &=600 \mathrm{~nm}, d^{\prime}=\frac{d}{2} \\ \therefore \quad y^{\prime} &=\frac{2 D \lambda^{\prime}}{d^{\prime}}=\frac{2 D \times 600}{\frac{d}{2}} \\ &=2400 \frac{D}{d}=3 \times \frac{800 D}{d} \quad \text { [from Eq. (i)] } \\ &=3 y \end{aligned}$
Wavelength of violet light,
$\lambda_{1}=400 \mathrm{~nm}$
We know that, width of the diffraction pattern is given as
$\begin{aligned}
& \beta=y=\frac{2 D \lambda_{1}}{d} \\
\Rightarrow \quad & y=\frac{2 D \lambda_{1}}{d}=\frac{2 D \times 400}{d}
\end{aligned}$
$y=\frac{800 D}{d}...(i)$
In the second case,
$\begin{aligned} \lambda_{2} &=600 \mathrm{~nm}, d^{\prime}=\frac{d}{2} \\ \therefore \quad y^{\prime} &=\frac{2 D \lambda^{\prime}}{d^{\prime}}=\frac{2 D \times 600}{\frac{d}{2}} \\ &=2400 \frac{D}{d}=3 \times \frac{800 D}{d} \quad \text { [from Eq. (i)] } \\ &=3 y \end{aligned}$
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