Light is incident on two slits that are separated by 0.2 mm. The figure shows the resulting interference pattern observed on a screen 1.0 m from the slits. Determine the wavelength of light used in this experiment.
A)0.05 nm
B)0.50 nm
C)50 nm
D)500 nm
E)5000 nm

Question

Light is incident on two slits that are separated by 0.2 mm. The figure shows the resulting interference pattern observed on a screen 1.0 m from the slits. Determine the wavelength of light used in this experiment. 
A)0.05 nm 
B)0.50 nm 
C)50 nm 
D)500 nm 
E)5000 nm

Quizplus · Accepted Answer

Using the formula for the position of minima in a double-slit experiment, we have:

d*sin(theta) = m*lambda

where d is the distance between the two slits (0.2 mm), theta is the angle between the incident wave and the line connecting the midpoint between the two slits and the point on the screen (we can assume small angles and use tan(theta) = y/L, where y is the distance from the central maximum to the m-th minimum on the screen, and L is the distance from the slits to the screen), m is the order of the minimum, and lambda is the wavelength of the light.

For the first minimum (m=1), we have:

0.2*10^(-3)*sin(theta) = lambda

tan(theta) = y/L = 0.28/1.0 = 0.28

theta = arctan(0.28) = 15.6 degrees

sin(theta) = 0.27

lambda = 0.2*10^(-3)/0.27 = 740 nm

Therefore, the closest choice is D) 500 nm.

Light Is Incident on Two Slits That Are Separated by 0.2