Question

If YDSE is performed with a light of wavelength 500 nm, and the 6th bright fringe is found 18 mm from the centered maxima. If the distance between the slits is 5 mm, the distance of the screen from the slits is:

a. 45 cm
b. 90 cm
c. 180 cm
d. 360 cm

Answer 1

The distance of the screen from the slits in a Young's Double-Slit Experiment using the given measurements is 3 meters or 300 cm, not 360 cm. The distance between fringes is used to find the screen distance by rearranging the formula Δy = xλ/d.

Step-by-step explanation:

To find the distance of the screen from the slits in a Young's Double-Slit Experiment (YDSE), we can use the formula for fringe separation. Since the problem states that the sixth bright fringe is 18 mm from the central maximum, this is the y value corresponding to the m=6 fringe. The formula given the small angle approximation where sin θ ≈ θ in radians is:

Δy = xλ/d

Here, Δy is the fringe separation, x is the distance from the slits to the screen, λ is the wavelength of the light, and d is the distance between the slits. We can rearrange the formula to solve for x:

x = (Δy * d) / λ

Plugging in the values Δy = 18 mm / 6 = 3 mm, d = 5 mm, and λ = 500 nm (or 500 x 10-9 m), we get:

x = (3 x 10-3 m * 5 x 10-3 m) / (500 x 10-9 m) = 3 x 10-2 m

Therefore, the distance of the screen from the slits is 3 meters or 300 cm, not 360 cm as in the question's answer choice. It's important to ensure unit consistency when calculating physical quantities.