Question

Alternate bright and dark fringes are observed on a screen. What is the most probable cause of the problem?
A. Light passing through a converging lens
B. Light passing through a diverging lens
C. Light passing through two very narrow slits
D. Photos emitted from a metal
E. Photos in an emission spectrum

Answer 1

Hint: By looking at the question we get an idea that the question is based on the concept of diffraction. Hence we will answer this question by first defining diffraction, later we will talk about constructive and destructive interference and using the knowledge of all these three we will be able to easily answer this question.

Complete answer:
Diffraction: The phenomenon of diffraction occurs when light bends as it passes through a slit or an opening. When light travelled through one of the slits, a different pattern appeared on the screen. In comparison to interference patterns, the pattern observed featured a centre maximum band that was very wide. On both sides, there were alternate dark and light bands whose intensity was decreasing.

Now let's talk about constructive and destructive interference: constructive interference in constructive interference, two waves (of the same wavelength) interact in such a way that they align, resulting in a new wave that is larger than the original wave.

Destructive interference: When the maxima of two waves are 180 degrees out of phase, destructive interference occurs: one wave's positive displacement is exactly negated by the other wave's negative displacement. The resulting wave has a zero amplitude.

Hence the correct answer is option C.

Note: Another way to answer this question would be by eliminating options in the given question. Look at the options given, we know that the question relates to diffraction hence there can be no use of photons, hence we are left with 3 options (a,b,c), now we know that converging and diverging lenses produce only image and no fringes, hence options a and b are also eliminated. Hence the correct answer would be option C.