Question

(d) (8 marks) An aperture in the form of a narrow slit 0.28 mm wide forms a diffraction pattern on a screen placed 7.0 m away when illuminated with light of wavelength 600 nm. (1) Calculate the width of the central peak of the diffraction pattern (i.e. from minimum to minimum) on the screen. (11) If instead of using one slit the light illuminates two such slits separated by 0.90 mm, what is the separation of the interference maxima formed on the screen? (111) Would you expect any of these interference maxima to be missing? Explain your answer.

Answer 1

2LA =

x= width of central maxima

L= distance between screen and slit

$\lambda$= wavelength of light

a= slit width

2 x 7 x 600 x 10 0.28x 10-3

0.03m Bcm

(b) the separation between interference maxima or fringe width is given by

LA

x= separation between maxima

$\lambda$= wavelength of light

d= separation between slits

7 x 600 x 10 0.9 x 10-3

0.00466m = 4.66mm

(c) yes, there will be a missing maxima

as we know minima for diffraction is given by

asinfn

and for interferance the maxima will be given by

dsin=m .

so for missing they will overlap

nA asin dsin mA

$\frac{a}{d} =\frac{n}{m}$

0.28 0.9 m