Question

in ydse the distance between the slits is 1.2mm and screen is 0.9m from the slits. if the wavelength of light used is 800nm, calculate fringe width. what is the percentage of change in fringe width if entire apparatus immersed in water of R.I=1.33.

Answer 1

Answer: Separation is given by

y=  

d

nλD

​  

 

where

d=3mm=3×10  

−3

 

D=2m

λ  

1

​  

=480nm=480×10  

−9

m

λ  

2

​  

=600nm=600×10  

−9

m

n  

1

​  

=n  

2

​  

=5

y  

2

​  

−y  

1

​  

=?

So,

y  

1

​  

=  

d

nλ  

1

​  

D

​  

 

y  

1

​  

=  

3×10  

−3

 

5×480×10  

−9

×2

​  

 

y  

1

​  

=1.6×10  

−3

m

Also,  

y  

2

​  

=  

d

nλ  

2

​  

D

​  

 

y  

2

​  

=  

3×10  

−3

 

5×600×10  

−9

×2

​  

 

y  

2

​  

=2×10  

−3

m

As y  

2

​  

>y  

1

​  

 

y  

2

​  

−y  

1

​  

=2×10  

−3

−1.6×10  

−3

 

             =4×10  

−

4m

Therefore the separation on the screen between the fifth order bright fringes of the two interference patterns is 4×10  

−4

m

Explanation: