Finding the distance between two fringes in a double slit experiment

[prettykitty]

Homework Statement

Blue light of wavelength 400 nm passes through two slits which are a distance
d = 1×10−4 m apart. This produces a double slit pattern on a screen L = 12 m away. (The
screen is parallel to the plane of the two slits.) If the central bright fringe is denoted the
“m=0 fringe”, what is the distance between the m = 15 and m = 14 bright fringe on the
screen? (All angles are sufficiently small that tan θ ≈ sin θ ≈ θ.)

Homework Equations

dsin(θ) = mλ
y = L tan(θ)

The Attempt at a Solution

arcsin[(15*400e-9)/1e-4] = 3.44°
y= 12 * tan (3.44) = .72 m

arcsin[(14*400e-9)/1e-4] = 3.21°
y= 12 * tan (3.21) = .67 m

y2-y1 = .72-.67 = .05 m

That was my attempt. First to find out how far the 14th and 15th bright fringe were from the center and then to find the distance between them.
The answer given by the instructor is: .113 m or 11.3 cm.
But why? I still can't find anything wrong with my train of thought. Thanks!

 

[Ibix]

The problem specifies small angles, so $\sin\theta\simeq \tan\theta\simeq y/L$ and hence $y=mL\lambda/d$. The distance between the 14th and 15th fringes is, therefore, $\Delta y=(15-14)L\lambda/d$. Plugging in your numbers gives me 4.8cm.

In other words, I agree with you.

 

[davenn]

The problem specifies small angles, so $\sin\theta\simeq \tan\theta\simeq y/L$ and hence $y=mL\lambda/d$. The distance between the 14th and 15th fringes is, therefore, $\Delta y=(15-14)L\lambda/d$. Plugging in your numbers gives me 4.8cm.

In other words, I agree with you.

you realized you grabbed this thread and the other one out of 3 year old archives ??

 

[Ibix]

They were in the Open Practice Problems forum when I answered them, so yes I did. I used to be an optics guy, and was enjoying a spot of nostalgia answering them.