Target Exam

CUET

Subject

Physics

Chapter

Wave Optics

Question:

The first diffraction minima due to a single slit diffraction is at θ = 30° for a light of wavelength 5000 Å. The width of the slit is

Options:

$5 × 10^{-5}\, cm$

$1 × 10^{-4}\, cm$

$2.5 × 10^{-5}\, cm$

$1.25 × 10^{-5}\, cm$

Correct Answer:

$1 × 10^{-4}\, cm$

Explanation:

The correct answer is Option (2) → $1 × 10^{-4}\, cm$

Given:

Wavelength of light: $\lambda = 5000\,\text{Å} = 5000 \times 10^{-10}\,\text{m} = 5 \times 10^{-7}\,\text{m}$

Diffraction minimum angle: $\theta = 30^\circ$

For a single slit, the condition for the first diffraction minimum is:

$a \sin \theta = \lambda$, where $a$ is the slit width

Rearranging: $a = \frac{\lambda}{\sin \theta}$

Substitute values: $a = \frac{5 \times 10^{-7}}{\sin 30^\circ} = \frac{5 \times 10^{-7}}{0.5} = 1 \times 10^{-6}\,\text{m}$

Convert to cm: $a = 1 \times 10^{-6}\,\text{m} = 1 \times 10^{-4}\,\text{cm}$

Answer: $a = 1 \times 10^{-4}\,\text{cm}$