Question

Monochromatic light of wavelength 687 nm is incident on a narrow slit. On a screen 1.65 m away, the distance between the second diffraction minimum and the central maximum is 2.09 cm. (a) Calculate the angle of diffraction θ of the second minimum. (b) Find the width of the slit. Please explain with your solution

Answer 1

Answer:

a ) 1.267 radian

b ) 1.084 10⁻³ mm

Explanation:

Distance of screen D = 1.65 m

Width of slit d = ?

Wave length of light   λ  = 687 nm.

Distance of second minimum fro centre y = 2.09 cm

Angle of diffraction = y / D

=  2.09 /1.65  

= 1.267. radian

Angle of diffraction of second minimum

= 2 λ / d

so 2 λ / d = 1.267

d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm

=1084.45 nm = 1.084 x 10⁻³ mm.