Question

a red laser from the physics lab is marked as producing 632.8-nm light. when light from this laser falls on two closely spaced slits, an interference pattern formed on a wall several meters away has bright red fringes spaced 5.00 mm apart near the center of the pattern. when the laser is replaced by a small laser pointer, the fringes are 5.13 mm apart. part a what is the wavelength of light produced by the pointer? express your answer to three significant figures and include the appropriate units.

Answer 1

We can use the formula for the spacing between fringes in a double-slit interference pattern:

dsin(theta) = mlambda

where d is the distance between the slits, theta is the angle between the incident light and the normal to the screen, m is the order of the fringe, and lambda is the wavelength of the light.

Since the same screen is used for both the red laser and the pointer, we can assume that the angle theta is the same in both cases. Therefore, we can write:

dsin(theta) = mlambda_red (for the red laser)

dsin(theta) = mlambda_p (for the pointer)

Dividing these two equations, we get:

(lambda_red / lambda_p) = (m_p / m_red)

where m_p and m_red are the orders of the fringes for the pointer and the red laser, respectively.

We are given that the spacing between fringes for the red laser is 5.00 mm and for the pointer is 5.13 mm. Since the fringes are evenly spaced, we can assume that we are looking at the central maximum, where m_red = m_p = 0. Therefore:

(lambda_red / lambda_p) = 0/0 = 1

Solving for lambda_p, we get:

lambda_p = lambda_red = 632.8 nm

Therefore, the wavelength of light produced by the pointer is also 632.8 nm.

Step-by-step explanation: