Calculating Minimum Width for 5 Interference Maxima

[msk172]

Homework Statement

Ramon has a coherent light source with wavelength 554 nm. He wishes to send light through a double slit with slit separation of 1.5 mm to a screen 90 cm away. What is the minimum width of the screen if Ramon wants to display five interference maxima?

Homework Equations

λ/s = w/D

λ= wavelength (554nm - converted to 5.54e-4 mm)
s= slit separation (1.5mm - provided by question)
w= width between maxima (use 5w do to question asking for 5 interference maxima)
D= distance to screen (90cm - converted to 900mm provided by question)

The Attempt at a Solution

Given that three of the four variables are known values, this should be simple as plug and solve. For some reason it is not liking the answer I am providing. The left side of the equation is simply 5.54e-4 / 1.5 which one would set equal to 5w / 900. Where am I going wrong here? Thanks in advance!

 

[msk172]

Follow up: Figured it out. Initially thought of the question as asking about 5 maxima per side (thus 10 total). One maxima in the center, two per side, 5 total. w * 4 yields correct value.

 

[nlightner]

Your method for solving this problem is correct. However, it seems like you may have made a calculation error when converting the distance to the screen from centimeters to millimeters. 90 cm should be converted to 900 mm, not 90 mm. Double check your calculations and try again. Additionally, make sure to use the correct units for all values in the equation (mm for wavelength and slit separation, and mm for distance to screen). This should give you the correct minimum width for 5 interference maxima. Keep in mind that the minimum width may be a very small value, so make sure to double check your calculations and units.