Q69 / 90 · JEE Main 2012 (12 May Online) · Physics
The first diffraction minimum due to the single slit diffraction is seen at $\theta=30^{\circ}$ for a light of wavelength $5000 Å$ falling perpendicularly on the slit. The width of the slit is
- A$2.5 \times 10^{-5} \mathrm{~cm}$
- B$1.25 \times 10^{-5} \mathrm{~cm}$
- Correct answerC$10 \times 10^{-5} \mathrm{~cm}$
- D$5 \times 10^{-5} \mathrm{~cm}$
Solution
For first minimum, $d \sin \theta=\lambda$
$$
\begin{aligned}
\Rightarrow \quad d & =\frac{\lambda}{\sin \theta}=\frac{5000 \times 10^{-8} \mathrm{~cm}}{\sin 30^{\circ}} \\
& =\frac{5000 \times 10^{-8} \mathrm{~cm}}{1 / 2}=10 \times 10^{-5} \mathrm{~cm}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.