Double-Slit Interference and Diffraction Gratings AP Physics 2 MCQ – Exam Style Questions etc.
Double-Slit Interference and Diffraction Gratings AP Physics 2 MCQ
Unit 14: Waves , Sound , and Physical Optics
Weightage : 15–18%
Exam Style Practice Questions ,Double-Slit Interference and Diffraction Gratings AP Physics 2 MCQ
Question
Which of the following changes to a double-slit interference experiment would increase the widths of the fringes in the interference pattern that appears on the screen?
(A) Use light of a shorter wavelength.
(B) Move the screen closer to the slits.
(C) Move the slits closer together.
(D) Use light with a lower wave speed.
Answer/Explanation
Ans:C
Relative to the central maximum, the locations of the bright fringes on the screen are given by the expression m\(\left ( \frac{\lambda L}{d} \right ),\) where A is the wavelength, L is the distance to the screen, d is the slit separation, and m is any integer. The width of a fringe is therefore ( m + 1 ) \(\left ( \frac{\lambda L}{d} \right )-m\left ( \frac{\lambda L}{d} \right )=\left ( \frac{\lambda L}{d} \right ).\) The slit spacing will increase if there is a decrease in d.
Question
In a double slit experiment, students are attempting to increase the spacing of the fringes observed on the screen. Which modifications to the set up will result in increased fringe separation? Select two answers.
(A) Doubling the wavelength only
(B) Doubling the wavelength and doubling the slit separation
( C) Doubling the distance to the screen only
(D) Doubling the distance to the screen and doubling the slit separation
Answer/Explanation
Ans:A, C
For the double slit experiment, mλ = dsin(θ) →mλ =dx\L. The spacing of the fringes, x, increases for increasing λ and L and decreases by increasing d.
Question
In Young’s double slit experiment, the second order bright band of one light source overlaps the third order band of another light source. If the first light source has a wavelength of 660 nm, what is the wavelength of the second light source?
A) 1320 nm B) 990 nm C) 440 nm D) 330 nm
Answer/Explanation
Ans:C
Solution: Using m λ = d sin θ, the value of sin θ is the same for both sources since the location of the spot is the same, but the first source is at m=2, and the second source is at m=3. Equating d sin θ for each gives m1 λ1 = m2 λ2 … (2)(660) = 3 (λ2) … λ2 = 440 nm.