Question

Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 94.5 cm from the slit. The distance on the screen between the second order minimum and the central maximum is 1.21 cm. What is the width of the slit, in mm?

Answer 1

In single slit diffraction

path difference = a*sintheta

for minimum diffraction path difference = m*lambda

a*sintheta = m*lambda

for small angle sintheta = tnatheta = y/R

a*y/R = m*lambda

a = m*lambda*R/y

given for m = 2 ,

y = 1.21 cm = 0.0121 m

slit width a = ?

R = distacne between slit and screen = 94.5 cm 0.945 m

wavelength = lambda = 608*10^-9 m

a = 2*608*10^-9*0.945/0.0121

a = 9.496*10^-5 m

a = 0.09496 mm   = 0.095 mm<<<-----ANSWER

Answer 2

SOLUTION :

Let the width of slit be a meters.

As per following formula,

Distance between 2nd order minimum and central maximum  

=  Order of minimum * Wave length * Distance of screen from the slit / Width of slit

=> 1.21 * 10^(-2) = 2 * 608*10^(-9) * 94.5*10^(-2) / a

=> a = 2 * ( 608*10^(-9) * 94.5) / 1.21

=> a = 94.97 µM (ANSWER).