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Wave Optics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsWave Optics3 Marks
Question
Determine the angular spread between the central maximum and first order maximum of the diffraction pattern due to a single slit of width $0.25 mm$, when light of $6650 A$ is incident on it normally.
✓

Answer

$
\text { Data } \begin{aligned}
\lambda & =6650 Å=6650 \times 10^{-10} m , a =0.25 mm \\
a \sin \theta & =(2 m+1) \frac{\lambda}{2} \\
& =(2 \times 1+1) \frac{\lambda}{2}=\frac{3 \lambda}{2} \quad(\text { for } m=1)
\end{aligned}
$
When $\theta$ is small and expressed in radian, $\sin \theta \simeq \theta$
$
\begin{aligned}
\therefore a \theta & =\frac{3 \lambda}{2} \text {} \\
\therefore \theta & =\frac{3 \lambda}{2 a}=\frac{3 \times 6650 \times 10^{-10}}{2 \times 0.25 \times 10^{-3}} \\
& =3.99 \times 10^{-3} rad
\end{aligned}
$
This is the required angular spread.

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