Solving Doublet Slit Problem: Find Wavelength of Laser 2

[noellcm]

Homework Statement

A laser with wavelength d/8 is shining light on a double slit with slit separation 0.350 mm. This results in an interference pattern on a screen a distance L away from the slits. We wish to shine a second laser, with a different wavelength, through the same slits.

What is the wavelength lambda_2 of the second laser that would place its second maximum at the same location as the fourth minimum of the first laser, if d = 0.350 mm?

Homework Equations

location of interference maxima: dsin(theta) = m(lambda1)
location of interference minima: dsin(theta) = (m+.5)(lambda2)

The Attempt at a Solution

i know that sin(theta) has to be the same for both of them, so i set these equations equal to sin(theta):
sin(theta) = (m*lambda1)/d
sin(theta) = ((m +.5 )*lambda2)/d

therefore:
(m*lambda1)/d = ((m +.5 )*lambda2)/d

d is given in the problem, so you know d for both sides. you also know that d/8 equals lambda one, so that can be solved. i have tried 2=m for lambda one because it is the second maxima and 3.5=m for lambda two (mastering physics told me to remember that the first minima is zero, not 1). i keep on getting the wrong answer though! please help!

 

[rl.bhat]

What is the wavelength lambda_2 of the second laser that would place its
second maximum at the same location as the fourth minimum of the first laser, if d = 0.350 mm?
So the equation should be
2/d*d/x = 3.5/d*d/8 .Solve for x, and then find the wavelength.

 

[nlightner]

Based on your equations and approach, it seems like you are on the right track. However, there may be some errors in your calculations that are leading to the wrong answer. It would be helpful to show your work and calculations so that I can better assist you in finding the correct solution. Also, make sure to check your units and conversions to ensure they are consistent throughout the problem. Additionally, keep in mind that the integer values for m represent the order of the maxima, not the actual number of maxima. For example, the fourth minimum of the first laser would correspond to m=3.5, not m=4. I hope this helps and good luck with your problem!