Question

4. Light incident on a pair of slits produces an interference pattern on a screen 2.5 m from the slits. Hint: Is the small-angle approximation justified in the following?

(a) Determine the wavelength of the light if the slit separation is 0.0150 cm and the distance between adjacent bright fringes in the pattern is 0.760 cm.

(b) Determine the distance between adjacent dark fringes if a 560 nm wavelength light source were shined through the same set up.

Answer 1

a)

Given, separation between screen and slit (D) =2.5m

Slit separation (a) =0.0150cm=0.00015m

Separation between two adjacent fringe(∆x)

∆x=0.760cm=0.0076m

From Young double slit experiment, we know

asin = m .......... (1) .

Now near the center of the interference pattern, the angle of deviation is so small, applying small angle approximation

Sin = tan​​​​​​ =

tan​​​​​​m=m= Xm/D

m is the order of fringe, so

From equation (1)

a×Xm/D =m

Xm=mD/a

Now, distance between adjacent fringe is

∆X=Xm+1-Xm

=D/a

so, wavelength () =∆X×a/D ........... (2)

= 0.0076×0.00015/2.5=456×10^-9m=456nm

b)

Given, wavelength () =560nm=560×10^-9m

So, putting in the equation (2)

Distance between adjacent dark fringe =Distance between adjacent bright fringe=(∆X) =D/a

Putting the values

∆X=560×10^-9×2.5/0.00015=0.93×10^-2m