Question

Light of wavelength 613.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 83.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.85 cm. What is the width a of the slit in micrometers (μm)?

Answer 1

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

=> 1.85 * 10^(-2) = 5 * 613*10^(-9) * 83.5*10^(-2) / a

=> a = 5 * ( 613*10^(-9) * 83.5) / 1.85

=> a = 138.34 µM (ANSWER).