Question

A screen is placed 1.40 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.60 cm wide-that is, the two first-order diffraction minima are separated by 1.60 cm. What is the distance between the two second-order minima?

Answer 1

Given

R = 1.4 m distance from slit to screen

2y1 = 1.6 cm , y1 = 0.8 cm

and the condition is Tan theta1 = y/R = 0.8*10^-2/1.4 = 0.0057142857 ==> theta = 0.3274 degrees

and the condition for minimum is d sin theta = m*lambda

  

so d sin theta1 = 1*lambda

d sin theta 2 = 2*lambda

dividing these sin theta2 = 2* sin theta1

sin theta2 = 2* sin(0.3274)

sin theta2 = 0.0114283537484

theta2 = 0.655 degrees

now for second order minima  

tan theta2 = y2/R

y2 = R*tan theta2

y2 = 1.4*tan(0.655) m

Y2 = 0.0160053664859 M

SO the distance between two second order minima is 2y1= 2*0.0160053664859m = 0.0320107329718 m m = 32 cm