Question:
The direction of the second order diffraction minima in Fraunhofer diffraction at a single slit (\(a=\) slit width) is given by
The direction of the second order diffraction minima in Fraunhofer diffraction at a single slit (\(a=\) slit width) is given by
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For single-slit Fraunhofer diffraction,
\[
a\sin\theta=n\lambda
\]
gives the positions of minima.
\[
n=1 \Rightarrow \text{First minimum},
\qquad
n=2 \Rightarrow \text{Second minimum},
\qquad
n=3 \Rightarrow \text{Third minimum}.
\]
Updated On: Jul 9, 2026
- \(\sin\theta=\dfrac{\lambda}{2a}\)
- \(\sin\theta=\dfrac{2\lambda}{a}\)
- \(\sin\theta=2\lambda a\)
- \(\sin\theta=\dfrac{a}{2\lambda}\) \bigskip
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The Correct Option is B
Solution and Explanation
Concept:
For Fraunhofer diffraction at a single slit of width \(a\), the minima occur at
\[
a\sin\theta=n\lambda,
\]
where
\[
n=1,2,3,\ldots
\]
is the order of the minimum.
Step 1: Write the condition for the second-order minimum. For the second-order minimum, \[ n=2. \] Therefore, \[ a\sin\theta=2\lambda. \]
Step 2: Find the direction of the minimum. \[ \sin\theta = \frac{2\lambda}{a}. \]
Step 3: Write the final answer. \[ \boxed{\sin\theta=\frac{2\lambda}{a}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Write the condition for the second-order minimum. For the second-order minimum, \[ n=2. \] Therefore, \[ a\sin\theta=2\lambda. \]
Step 2: Find the direction of the minimum. \[ \sin\theta = \frac{2\lambda}{a}. \]
Step 3: Write the final answer. \[ \boxed{\sin\theta=\frac{2\lambda}{a}} \] \[ \boxed{\text{Answer = (B)}} \]
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