【Daily Question】Theoretical Foundation 2020.06.04
Thread Content
Multiple-choice question: In a double-slit interference experiment using monochromatic light with wavelength λ in air, the distance between adjacent bright fringes is 1.33 mm. When the experimental setup is placed in water (with a refractive index of n = 1.33), the distance between adjacent bright fringes becomes: ( ) A. 1.33 mm B. 2.66 mm C. 1 mm D. 2 mm Answer: C Hint: Extra rewards will be given to those who can explain the solution process; the answer can be found in the response!
C. When the light waves emitted by the monochromatic light source S reach the two slits S1 and S2, which are at equal distances from S, the light emitted from S1 and S2 becomes two coherent light sources (with the same frequency, the same direction of light vibration, and a constant phase difference at the point of intersection). These light waves overlap in space, resulting in interference phenomena. If a screen is placed in front of the slits, a series of alternating bright and dark fringes will appear on the screen; this phenomenon is known as Young’s double-slit interference. The position of the bright and dark fringes is determined by the path difference; if the path difference is zero, bright fringes are formed, while if the path difference is (2k+1)/2λ, dark fringes are formed. The distance between adjacent bright (or dark) fringes, where (d represents the distance between the two light sources, and D represents the distance from the light sources to the screen). In a medium with a refractive index of n, the distance between adjacent bright (or dark) fringes is; therefore, the distance between adjacent bright fringes in water is: 1.33 mm / 1.33 = 1 mm.