Question

A single slit forms a diffraction pattern, with the first minimum at an angle of 40.0° from central maximum, when monochromatic light of 590-nm wavelength is used. The same slit, illuminated by a new monochromatic light source, produces a diffraction pattern with the second minimum at a 60.0° angle from the central maximum. What is the wavelength of this new light?

a. 421nm

b. 468 nm

c. 445 nm

d. 398 nm

e. 492 nm

Answer 1

Dsin(Theta) = m*wavelength

D*sin(40) = 1 * 590

D = 590 / sin(40)

D = 917.87 m

Now,

917.87*sin(60) = 2*wavelength

So, wavelength = 917.87*sin(60) / 2
= 398 nm