Physics quiz question: double slit interference maxima spacing on screen

Physics 205B Quiz 3, fall semester 2010
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 25.31

Light incident on a pair of slits produces an interference pattern on a screen 5.00 m from the slits. The slit separation distance is 0.0120 cm and the spacing between the central and first maximum on the screen is 2.30 cm. The wavelength of the light is:
(A) 479 nm.
(B) 522 nm.
(C) 552 nm.
(D) 690 nm.

Correct answer (highlight to unhide): (C)

Maxima angles (bright fringes) are given by the relationship:

d·sinθ = m·λ.

The angle for the centerline m = 0 maximum is θ = 0°, while the angle θ for the m = 1 bright fringe can be found from:

d·sinθ = (1)λ,

where the given (or assumed to be known) quantities (such as the slit separation distance d = 0.0120 cm = 1.20×10^–4 m), unknown quantities (the first maximum angle θ), and quantities to be explicitly solved for (the wavelength λ) are denoted.

From trigonometry, the relation between the first maximum angle θ and the adjacent leg L = 5.00 m (distance from the slits to the screen) and the opposite leg y = 2.30 cm = 2.30×10^–2 m (measured from the 0° centerline to the first maximum on the screen) is given by:

tanθ = y/L,

such that the first maximum angle can then be solved for:

θ = tan^–1(y/L),

θ = tan^–1((2.30×10^–2 m)/(5.00 m)) = 0.2635587...°,

and so the wavelength λ can then be solved for:

λ = d·sinθ,

λ = (1.20×10^–4 m)·sin(0.2635587...°),

λ = 5.5199...×10^–7 m,

or to three significant figures, the wavelength of this light is 552×10^–9 m = 552 nm.

(Response (A) is L·y/(2·d); response (B) is 2·L·d/y; response (D) is L·d·y/2.)

Student responses
Section 70856
(A) : 3 students
(B) : 1 student
(C) : 7 students
(D) : 0 students

Success level: 63%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.67