In a diffraction experiment, the screen is placed 2 m away from a narrow slit. The first minimum observed is at 5 mm on either side of the central maximum for wavelength of $5 × 10^{-5} cm$. The width of the slit is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.02 cm
Given:
Distance of screen from slit, $L = 2\ \text{m}$
Distance of first minimum from central maximum, $y = 5\ \text{mm} = 0.005\ \text{m}$
Wavelength of light, $\lambda = 5 \times 10^{-5}\ \text{cm} = 5 \times 10^{-7}\ \text{m}$
For a single slit, position of first minimum:
$y = L \frac{\lambda}{a}$
where $a$ = width of the slit.
Rearranging:
$a = \frac{L \lambda}{y} = \frac{2 \times 5 \times 10^{-7}}{0.005} = \frac{1 \times 10^{-6}}{0.005} = 2 \times 10^{-4}\ \text{m}$
∴ Width of the slit = $2 \times 10^{-4}\ \text{m} = 0.2\ \text{mm}$