Question

Light of wavelength 618.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 62.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.71 cm. What is the width of the slit?

Answer 1

y = mDλ/a

where

a = width of slit

D = distance to screen

λ = wavelength of light

m = order number (the number of the minimum you are dealing with)

a = mDλ/y

y = 0.0171 m

D = 0.625 m

λ = 618 x 10^-9 m

and m = 5

a = 5*0.625*618*10^-9/0.0171 = 1.129*10^-4 m

Answer 2

SOLUTION :

Let the width of slit be a meters.

As per following formula,

Distance between 5th order minimum and central maximum  

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

=> 1.71 * 10^(-2) = 5 * 618 *10^(-9) * 62.5*10^(-2) / a

=> a = 5 * ( 618*10^(-9) * 62.5) / 1.71

=> a =  1.129 * 10(-4) m = 0.1129 mm (ANSWER)