The diagram shows an experiment which has been set up to demonstrate two-source interference. Microwaves of wavelength λ pass through two slits S1 and S2.

Question 25

The diagram shows an experiment which has been set up to demonstrate two-source interference. Microwaves of wavelength λ pass through two slits S1 and S2.

The detector is moved from point O in the direction of the arrow. The signal detected decreases until the detector reaches point X, and then starts to increase again as the detector moves beyond X.

Which equation correctly determines the position of X?

A OX = λ

B OX = λ / 2

C S2X – S1X = λ

D S2X – S1X = λ / 2

Reference: Past Exam Paper – June 2014 Paper 12 Q27

Solution:

Answer: D.

A minima is found at X since the signal keeps on decreasing until it reaches X (so at X, the signal has the lowest amplitude). X is the first minima.

S₂X and S₁X are distances.

The condition for destructive interference is that the path difference should be nλ / 2.

For the first minima (n = 1),

Path difference = λ / 2

S₂X – S₁X = λ / 2

This distance (λ/2) is equal to the path difference between S₂X and S₁X which is (S₂X – S₁X). [D is correct]

The distance between the central maxima (O) and the first maxima is called the fringe separation x. It can be obtained from

x = λD / a                    (double-slit experiment)

The distance between the central maxima (O) and the first minima (at X) is half the fringe separation (= x/2            and x is given by the above formula).

So, OX is NOT equal to λ / 2.