Q. 165.0( 1 Vote )

A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe-width? (b) At what distance from the center will the first maximum be located?

Answer :

Given, separation between the slits

The distance between the screen and the slit,

Wavelength of light,

refractive index of mica,

refractive index of polystyrene,

(a)The fringe width is given by

(b) The optical path difference is

Now we know that

Therefore, number of fringe shifts is

This means that there are 25 full fringe shifts from the center with 0.42w shift on one side and (1-0.42) w=0.58w on the other side

Hence, maximum on one side =

And on the other side,

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Standing Wave44 mins
Sound Wave43 mins
Energy & Power of a Wave60 mins
Transmitted & Reflected Waves39 mins
String WaveFREE Class
Energy in SHM42 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses