Young's Interference Experiment — Question 8

Section 35-3 · Physics pages 5766–5767

Question 8 A focal-plane fringe pattern is independent of the later screen-distance formula Two narrow slits are followed immediately by a thin lens with focal length f = 0.30 m. A detector lies in its rear focal plane. For small angles y = fθ. The slit spacing is d = 0.15 mm and wavelength λ = 600 nm. lens P(y) d f Schematic; labels, not drawing lengths, specify the data. (a) Derive and evaluate the fringe spacing in the focal plane. (b) Find the slit spacing required to fit exactly eight bright-fringe intervals across a 6.0 mm detector width with the same lens and wavelength.
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Question 8 - Solution A focal-plane fringe pattern is independent of the later screen-distance formula lens P(y) d f (a) Method and calculation dθm = mλ, ym = fθm =⇒∆y = fλ/d . ∆y = 1.20 mm . (b) Interpretation and further calculation 8∆y = 6.0 mm =⇒∆y = 0.750 mm. d = fλ ∆y = 0.240 mm . Physical check The relevant scale is the lens focal length because the detector is measuring outgoing angle, not free-space propagation from the slit plane.

Question 8

A focal-plane fringe pattern is independent of the later screen-distance formula

Two narrow slits are followed immediately by a thin lens with focal length f=0.30mf=0.30\,\mathrm m. A detector lies in its rear focal plane. For small angles y=fθy=f\theta. The slit spacing is d=0.15mmd=0.15\,\mathrm{mm} and wavelength λ=600nm\lambda=600\,\mathrm{nm}.

Diagram shown in the original SVG worksheet.

  1. Derive and evaluate the fringe spacing in the focal plane.

  2. Find the slit spacing required to fit exactly eight bright-fringe intervals across a 6.0mm6.0\,\mathrm{mm} detector width with the same lens and wavelength.

Question 8 - Solution

A focal-plane fringe pattern is independent of the later screen-distance formula

Diagram shown in the original SVG worksheet.

(a) Method and calculation

dθm=mλ,ym=fθm⇒Δy=fλ/d.d\theta_m=m\lambda,\quad y_m=f\theta_m\Longrightarrow\boxed{\Delta y=f\lambda/d}.Δy=1.20mm.\boxed{\Delta y=1.20\,\mathrm{mm}}.

(b) Interpretation and further calculation

8Δy=6.0mm⇒Δy=0.750mm.8\Delta y=6.0\,\mathrm{mm}\Longrightarrow\Delta y=0.750\,\mathrm{mm}.d=fλΔy=0.240mm.\boxed{d=\frac{f\lambda}{\Delta y}=0.240\,\mathrm{mm}}.

Physical check

The relevant scale is the lens focal length because the detector is measuring outgoing angle, not free-space propagation from the slit plane.