Question

Monochromatic light is incident on (and perpendicular to) two slits separated by 0.200 mm, which causes an interference pattern on a screen 613 cm away. The light has a wavelength of 656.3 nm. (a) What is the fraction of the maximum intensity at a distance of 0.600 cm from the central maximum of the interference pattern

Answer 1

Answer:

I = 0.636*Imax

Explanation:

(a) To find the fraction of the maximum intensity at a distance y from the central maximum you use the following formula:

$I=I_{max}cos^2(\frac{\pi d}{\lambda L}y)$   (1)

I: intensity of light

Imax: maximum intensity of light

d: separation between slits = 0.200mm = 0.200 *10^-3 m

L: distance from the screen = 613cm = 0.613 m

y: distance to the central peak of the interference pattern

λ: wavelength of light = 656.3 nm = 656.3 *10^-9 m

You replace the values of all variables in the equation (1):

$I=I_{max}cos^2(\frac{\pi (0.200*10^{-3}m)}{(656.3*10^{-9}m)(0.613m)}0.600m)\\\\I=I_{max}cos^2(937.06)=0.636I_{max}$

Hence, the fraction of the maximum intensity is I = 0.636*Imax