Question

The Fraunhofer diffraction pattern of a single slit is formed at the focal plane of a lens of focal length 1m. The width of the slit is 0.3 mm. If the third minimum is formed at a distance of 5 mm from the central maximum then the wavelength of light will be:

• A. 4500 Å
• B. 3000 Å
• C. 5000 Å
• D. 4000 Å

Answer 1

Answer: (C)

5000 Å

Answer 2

The position of the minima in a single-slit Fraunhofer diffraction pattern is given by $a \sin \theta_n = n \lambda$, where $a$ is the slit width, $\theta_n$ is the angle of diffraction for the $n$-th minimum, $\lambda$ is the wavelength, and $n = 1, 2, 3, \ldots$.

When the diffraction pattern is formed at the focal plane of a lens, the distance $y_n$ of the $n$-th minimum from the central maximum is related to the angle $\theta_n$ by $\tan \theta_n = \frac{y_n}{f}$, where $f$ is the focal length. For small angles, $\sin \theta_n \approx \tan \theta_n \approx \frac{y_n}{f}$.

Therefore, the formula becomes: $a \frac{y_n}{f} = n \lambda$

Given: Slit width, $a = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m}$ Focal length, $f = 1 \, \text{m}$ For the third minimum, $n=3$. Distance of the third minimum from the central maximum, $y_3 = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m}$.

Substitute these values into the equation: $(0.3 \times 10^{-3} \, \text{m}) \times \frac{5 \times 10^{-3} \, \text{m}}{1 \, \text{m}} = 3 \times \lambda$

Solving for $\lambda$: $\lambda = \frac{(0.3 \times 10^{-3}) \times (5 \times 10^{-3})}{3}$ $\lambda = \frac{1.5 \times 10^{-6}}{3}$ $\lambda = 0.5 \times 10^{-6} \, \text{m}$ $\lambda = 5 \times 10^{-7} \, \text{m}$

To convert this wavelength to Angstroms (Å), we use the conversion $1 \, \text{m} = 10^9 \, \text{Å}$: $\lambda = 5 \times 10^{-7} \times 10^9 \, \text{Å}$ $\lambda = 5000 \, \text{Å}$