Question:

A student is performing an experiment to study the phenomenon of interference using a double-slit setup. The slit separation is 0.5 mm, and the slit width is 0.2 mm. The distance from the double slits to a screen is 2 m. The student shines red light with a wavelength of 650 nm onto the double slits. Determine:

a) The separation between the central bright fringe and the first-order bright fringe on the screen. b) The angular position of the second-order dark fringe when viewed from the center of the central bright fringe. c) The number of bright fringes that can be seen on the screen when the distance between the double slits and the screen is changed to 1 m.

Assume the incident wavefront is parallel and the slits are small enough to consider only the central maxima.

Answer:

a) To find the separation between the central bright fringe and the first-order bright fringe, we can use the formula for the path difference:

$\text{{Path Difference}} = d \cdot \sin(\theta)$

where $d$ is the slit separation and $\theta$ is the angle between the incident wavefront and the first-order bright fringe.

For the first bright fringe, $\theta$ is very small, so we can approximate $\sin(\theta) \approx \theta$.

Using the small angle approximation, the path difference becomes:

$\text{{Path Difference}} = d \cdot \theta$

For constructive interference, the path difference must be equal to an integer multiple of the wavelength ($\lambda$):

$d \cdot \theta = m \cdot \lambda$

where $m$ is the order of the bright fringe.

To find the separation between the central bright fringe and the first-order bright fringe, we need to find the angle $\theta$ for $m = 1$.

Rearranging the equation, we have:

$\theta = \frac{{d \cdot \lambda}}{{m}}$

Substituting the given values:

$d = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m}$

$\lambda = 650 \, \text{nm} = 650 \times 10^{-9} \, \text{m}$

$m = 1$

$\theta = \frac{{0.5 \times 10^{-3} \, \text{m} \times 650 \times 10^{-9} \, \text{m}}}{{1}} = 0.325 \times 10^{-3} \, \text{rad}$

The separation between the central bright fringe and the first-order bright fringe is approximately $0.325 \times 10^{-3}$ radians.

b) To find the angular position of the second-order dark fringe, we can use a similar approach.

For a dark fringe, the path difference must be equal to an odd multiple of half the wavelength.

Using the same equation as before, we have:

$\theta = \frac{{d \cdot \lambda}}{{m}}$

Substituting the given values:

$d = 0.5 \times 10^{-3} \, \text{m}$

$\lambda = 650 \times 10^{-9} \, \text{m}$

$m = 2$ (second-order dark fringe)

$\theta = \frac{{0.5 \times 10^{-3} \, \text{m} \times 650 \times 10^{-9} \, \text{m}}}{{2}} = 0.1625 \times 10^{-3} \, \text{rad}$

The angular position of the second-order dark fringe is approximately $0.1625 \times 10^{-3}$ radians.

c) To determine the number of bright fringes seen when the distance between the double slits and the screen changes to 1 m, we can use the formula:

$X = \frac{{m \cdot \lambda \cdot D}}{{d}}$

where $X$ is the number of bright fringes, $m$ is the order of the fringe, $\lambda$ is the wavelength, $D$ is the new distance between the double slits and the screen, and $d$ is the slit separation.

Substituting the given values:

$m = 1$ (since we are considering the central bright fringe)

$\lambda = 650 \times 10^{-9} \, \text{m}$

$D = 1 \, \text{m}$

$d = 0.5 \times 10^{-3} \, \text{m}$

$X = \frac{{1 \times 650 \times 10^{-9} \, \text{m} \times 1 \, \text{m}}}{{0.5 \times 10^{-3} \, \text{m}}} = 1300$

Therefore, when the distance between the double slits and the screen is changed to 1 m, approximately 1300 bright fringes can be seen on the screen.