Cracking the SAT Physics Subject Test
CHAPTER 15 DRILL
- B From the equation λf = c, we find that
- D Since the fringe is bright, the waves must interfere constructively. This
implies that the difference in path lengths must be a whole number times the
wavelength, eliminating (B), (C), and (E). The central maximum is
equidistant from the two slits, so ∆ℓ = 0 there. At the first bright fringe above
the central maximum, we have ∆ℓ = λ.
- C First, eliminate (A) and (B): The index of refraction is never smaller than 1.
Refer to the following diagram: