question_answer

A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2\[y\]m and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will:

A)  remain unshifted

B)  shift downward by nearly two fringes

C)  shift upward by nearly two fringes

D)  shift downward by ten fringes

Correct Answer: C

Solution :

When a thin film is placed in the path of one of the beams, then the optical path of that beam gets longer. The path difference is \[1:{{2}^{1/3}}\] where \[{{2}^{1/3}}:1\], is refractive index, t is thickness. Given, \[2:1\] \[1:2\]  \[{{m}^{-2}}{{s}^{-1}}\] Also path difference \[{{m}^{-2}}{{s}^{-1}}\] \[\frac{1}{2\pi }\sqrt{\frac{k}{m}}\]   \[\frac{1}{2\pi }\sqrt{\frac{2k}{m}}\] Also fringe width is given by \[\frac{1}{2\pi }\sqrt{\frac{3k}{m}}\] \[\frac{1}{2\pi }\sqrt{\frac{k}{m}}\] \[\pi \] \[{{Q}^{2}}(4\pi {{\varepsilon }_{0}}{{a}^{2}})\] Hence, central maxima shifts upward by nearly two fringes. Note: When a thin film is introduced in the path of one of the two interfering light beams, then the entire fringe pattern is displaced towards the beam in the path of which the film is introduced.