Understanding Fraunhofer Diffraction and Central Maximum Width

This problem involves the phenomenon of Fraunhofer diffraction through a single slit. When light passes through a narrow opening (slit), it spreads out. In Fraunhofer diffraction, we observe a pattern of bright and dark fringes on a screen placed far away. The central part of this pattern is a bright fringe called the central maximum.

The angular width of this central maximum is related to the wavelength of the light ($\lambda$) and the width of the slit ($a$). Specifically, the positions of the first dark fringes (minima) on either side of the central maximum are given by the equation:

$a \sin \theta = m \lambda$

where '$m$' is an integer representing the order of the minimum ($m = \pm 1, \pm 2, \dots$). For the first minimum, $m = \pm 1$.

For small angles, which is typical in these experiments, we can approximate $\sin \theta \approx \theta$ (where $\theta$ is in radians). Thus, the angular position of the first minimum is approximately:

$\theta_{min} \approx \frac{\lambda}{a}$

The angular width of the central maximum is the angle between the first minimum on one side and the first minimum on the other side, which is $2 \theta_{min}$. Therefore, the angular width ($\theta_{width}$) is directly proportional to the wavelength ($\lambda$) and inversely proportional to the slit width ($a$):

$\theta_{width} \propto \frac{\lambda}{a}$

Step-by-Step Calculation for the New Wavelength $\lambda_2$

Let's break down the problem into steps using the relationship derived above.

Initial Conditions:

• Initial slit width: $a$
• Initial wavelength: $\lambda_1 = 6000 \text{ Å}$
• Initial angular width of the central maximum: $\theta_1$

Using the proportionality, we can write:

$\theta_1 \propto \frac{\lambda_1}{a}$

Final Conditions:

• The slit width is increased by 20%. The new slit width, $a'$, is: $a' = a + 0.20 a = 1.20 a$
• The new wavelength is $\lambda_2$.
• The new angular width of the central maximum, $\theta_2$, is $\frac{3}{5}$ of the initial value: $\theta_2 = \frac{3}{5} \theta_1$

Using the proportionality for the final conditions:

$\theta_2 \propto \frac{\lambda_2}{a'}$

Relating Initial and Final Conditions:

We can set up a ratio using the proportionality. The constant of proportionality remains the same:

$\frac{\theta_2}{\theta_1} = \frac{\lambda_2 / a'}{\lambda_1 / a}$

Solving for $\lambda_2$:

Substitute the known values into the ratio equation:

$\frac{\frac{3}{5} \theta_1}{\theta_1} = \frac{\lambda_2 / (1.20 a)}{\lambda_1 / a}$

Simplify the equation:

$\frac{3}{5} = \frac{\lambda_2}{1.20 a} \times \frac{a}{\lambda_1}$

The slit width '$a$' cancels out:

$\frac{3}{5} = \frac{\lambda_2}{1.20 \lambda_1}$

Now, rearrange the equation to solve for the new wavelength $\lambda_2$:

$\lambda_2 = \frac{3}{5} \times (1.20 \lambda_1)$

Calculate the value:

$\lambda_2 = 0.6 \times 1.20 \times \lambda_1$

$\lambda_2 = 0.72 \lambda_1$

Substitute the initial wavelength $\lambda_1 = 6000 \text{ Å}$:

$\lambda_2 = 0.72 \times 6000 \text{ Å}$

$\lambda_2 = 4320 \text{ Å}$

Conclusion

By analyzing the relationship between the angular width of the central maximum, the wavelength of light, and the slit width in Fraunhofer diffraction, we calculated the new wavelength $\lambda_2$. An increase in slit width and a change in wavelength resulted in a decreased angular width, leading to the final calculated wavelength of $4320 \text{ Å}$.