Question

In a single slit diffraction experiment, the angular position of the first (secondary ) maximum is found to be 5.2°, when the slit width is 0.01mm. If sin 52° =0.0906 , then find the wavelength of the light used.​

Answer 1

Given:  The angular position of the first (secondary ) maximum is found to be 5.2°, the slit width is 0.01 mm

To find: The wavelength of the light used.​

Solution:

• Now we have given angular position of the first (secondary ) maximum is found to be 5.2°, the slit width is 0.01 mm.

• So :

                 a sin x = 3λ/2

• Now :

                 a =  0.01 mm = 10^-5 m , x = 5.2° , sin 52° =0.0906

• So putting all these values in formula, we get:

                 10^-5 ( 0.0906) = 3λ/2

                 λ = 2/3 x 10^-5 ( 0.0906)

                 λ = 6040 Armstrong

Answer:

       So the wavelength of the light used. is 6040 Armstrong.