Step 1: Understanding the Concept:
In Fraunhofer diffraction due to a single slit, alternating bright and dark fringes are formed on the screen.
The positions of the secondary maxima (bright fringes) depend on the wavelength of light, the slit width, and the distance to the screen.
Step 2: Key Formula or Approach:
The condition for the $n$-th secondary maximum in a single slit diffraction pattern is given by:
\[ a \sin \theta_n = (2n + 1)\frac{\lambda}{2} \]
where $a$ is the slit width and $\lambda$ is the wavelength.
For small angles, $\sin \theta_n \approx \tan \theta_n = \frac{y_n}{D}$, where $y_n$ is the distance from the center and $D$ is the distance to the screen.
Substituting this, the position $y_n$ of the $n$-th maximum is:
\[ y_n = \frac{(2n + 1)\lambda D}{2a} \]
Step 3: Detailed Explanation:
Given values are:
Distance to screen, $D = 1.3 \text{ m}$
Slit width, $a = 0.65 \text{ mm} = 0.65 \times 10^{-3} \text{ m}$
Position of the second maximum ($n=2$), $y_2 = 2.6 \text{ mm} = 2.6 \times 10^{-3} \text{ m}$
Using the formula for the second maximum ($n=2$):
\[ y_2 = \frac{(2(2) + 1)\lambda D}{2a} = \frac{5\lambda D}{2a} \]
Rearranging to solve for the wavelength $\lambda$:
\[ \lambda = \frac{2a \cdot y_2}{5D} \]
Substitute the given values into the equation:
\[ \lambda = \frac{2 \times (0.65 \times 10^{-3}) \times (2.6 \times 10^{-3})}{5 \times 1.3} \]
\[ \lambda = \frac{1.3 \times 10^{-3} \times 2.6 \times 10^{-3}}{6.5} \]
\[ \lambda = \frac{3.38 \times 10^{-6}}{6.5} \]
\[ \lambda = 0.52 \times 10^{-6} \text{ m} \]
To convert this to Angstroms ($1 \text{ \AA} = 10^{-10} \text{ m}$):
\[ \lambda = 5200 \times 10^{-10} \text{ m} = 5200 \text{ \AA} \]
Step 4: Final Answer:
The wavelength of light used is $5200 \text{ \AA}$.