Diffraction — Question 5

Question 5

Immersion changes the wave spreading of a fixed opening

A slit of width $a=0.10\,\mathrm{mm}$ and a screen $L=1.50\,\mathrm m$ away are entirely inside a transparent medium of index 1.33. The source vacuum wavelength is $532\,\mathrm{nm}$. Use $\theta_1\simeq\lambda/a$ and neglect dispersion.

1. Calculate the wavelength in the medium and the central diffraction width at the screen.

2. Find the slit width in air that would give the same central width at the same screen distance.

Question 5 - Solution

Immersion changes the wave spreading of a fixed opening

(a) Method and calculation

$$\boxed{\lambda=\lambda_0/n=400\,\mathrm{nm}}.$$$$\boxed{W\simeq2L\lambda/a=12.0\,\mathrm{mm}}.$$

(b) Interpretation and further calculation

$$\frac{2L\lambda_0}{a_{\mathrm{air}}}=\frac{2L\lambda_0}{na}\Longrightarrow a_{\mathrm{air}}=na.$$$$\boxed{a_{\mathrm{air}}=0.133\,\mathrm{mm}}.$$

Physical check

The frequency is unchanged by immersion; the reduced wavelength narrows the diffraction pattern.