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GUJCET · Physics · Wave Optics
Light of wavelength \(\lambda\) is incident on slit of width a. The resulting diffraction pattern is observed on a screen placed at distance D. The linear width of central maximum is equal to width of the slit, then \(\mathrm{D}=\) __________ .
- A \(\frac{2 \lambda^{2}}{a}\)
- B \(\frac{a^{2}}{2 \lambda}\)
- C \(\frac{a}{\lambda}\)
- D \(\frac{2 \lambda}{a}\)
Answer & Solution
Correct Answer
(B) \(\frac{a^{2}}{2 \lambda}\)
Step-by-step Solution
Detailed explanation
(B) \(\frac{a^{2}}{2 \lambda}\)
in diffraction fringe width
\(
\begin{aligned}
& & \beta & =\frac{\lambda \mathrm{D}}{a} \\
\text { but } & & \beta & =\frac{a}{2} \\
\therefore & & \frac{a}{2} & =\frac{\lambda \mathrm{D}}{a} \\
\therefore & & \mathrm{D} & =\frac{a^{2}}{2 \lambda}
\end{aligned}
\)
in diffraction fringe width
\(
\begin{aligned}
& & \beta & =\frac{\lambda \mathrm{D}}{a} \\
\text { but } & & \beta & =\frac{a}{2} \\
\therefore & & \frac{a}{2} & =\frac{\lambda \mathrm{D}}{a} \\
\therefore & & \mathrm{D} & =\frac{a^{2}}{2 \lambda}
\end{aligned}
\)
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