Question

Light waves with two different wavelengths, 632 nm and 474 nm, pass simultaneously through a single slit whose width is 8.61 × 10-5 m and strike a screen 1.10 m from the slit. Two diffraction patterns are formed on the screen. What is the distance (in cm) between the common center of the diffraction patterns and the first occurrence of the spot where a dark fringe from one pattern falls on top of a dark fringe from the other pattern?

Answer 1

As we know that;

= a sin theta1= m lamda1

= a sin theta2= n lamda2   

= m lamda1= n lamda2

= since theta1= theta2

= n / m = lamda1/lamda2

= 632 / 474 =1.333

= n = 1.333 m

So,

= m 3, 6, 9

Some other multiple of 3 for n to bean integer

So,

The lowest value of m is m = 3 and n = 4

= sin theta1 = 3 * 6.32 * 10*^-7 / 8.61 * 10^-5

= 0.02202 tan ?

= tan ? = s / D

where,

s= distance torequired minimum

D = distance to screen

= s = 0.02202 * 1.1

= 0.02422 m

= 2.422 cm