Double Slit Problem: Solving Unknown Wavelength

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SUMMARY

The double slit problem involves determining the unknown wavelength of light that overlaps with a known wavelength of 600nm. The solution reveals that the unknown wavelength is 450nm, derived from the relationship between the fringe orders of the two wavelengths. The equation used is dsin(theta) = m(lambda), where the angles for both wavelengths are constant due to their overlapping fringes. The correct application of the fringe order relationship, 3*600nm=4*x, leads to the conclusion that the unknown wavelength is 450nm.

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  • Understanding of wave interference principles
  • Familiarity with the double slit experiment
  • Knowledge of fringe order calculations
  • Proficiency in using the equation dsin(theta) = m(lambda)
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Homework Statement



A double slit is illuminated simultaneously with orange light of wavelength 600nm and light of an unknown wavelength. the m=4 bright fringe of the unknown wavelength overlaps the m=3 bright orange fringe. What is the unknown wavelength?

Homework Equations



dsin(theta) = m(lambda)

The Attempt at a Solution



I tried finding d for orange light by suing the above formula. I substitued m=3
sin 90 and wavelength as 600 x 10^-9
and got d= 0.0000018m
and then I used this d to find unknown wavelength and used the same above equation but this time substitutes m=4 and sin120 and got wavelength= 0.000000389m or (389nm)

but my answer is wrong ... please tell me where I'm making mistake.
 
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Why did you use sin90 in the first case and sin 120 in the second? We don't know what angle the rays make with the axis.
Since its given that 3rd fringe of the orange light is equal to the 4th fringe of the unknown wavelength, here the angle both rays (of different wavelengths) make with the axis is the same.
Therefore, here, dsin(theta) is a constant.
Thus, 3*600nm=4*x, giving the unknown wavelength to be 450nm.
 
chaoseverlasting said:
Why did you use sin90 in the first case and sin 120 in the second? We don't know what angle the rays make with the axis.
Since its given that 3rd fringe of the orange light is equal to the 4th fringe of the unknown wavelength, here the angle both rays (of different wavelengths) make with the axis is the same.
Therefore, here, dsin(theta) is a constant.
Thus, 3*600nm=4*x, giving the unknown wavelength to be 450nm.

Yeah, I realized later on that angles are wrong ... I used
(delta)r = m(wavelength)
first I found the delta r quantity for orange light by substituting m=3 and wavelength = 600*10^-9m

and I got Delta r = 0.0000018 m and then I used the same equation but different value of m =4 (given in the question)
I got the same wavelength 450nm.

But thanks for the help.
 

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