A line source at visible-light wavelengths is of ten made by imaging a gas- discharge lamp on a narrow slit, which then serves as the "line" source. This secondary source is placed at the focus of a lens of focal length \(f_{0}\) to provide collimated light. How small must the width \(w\) of the source slit be so as not to affect significantly the Fraunhofer diffraction pattern of a slit of width \(a\) ? What if the lens is not used and the distance between source and diffracting slits is \(R>a^{2} / \lambda\) ? Answer: \(w \lesssim \lambda f / 4 a ; a / 4\).

In this case, the width of the source slit should be small enough to not significantly affect the diffraction pattern. The angular width of the central maximum in the diffraction pattern is related to the width of the source slit \(w\) and the distance between the source and diffracting slits \(f_0\) as: $$ \theta = \frac{\lambda}{w} $$ We can approximate this angle by considering the angular width of the central maximum produced by the diffracting slit: $$ \theta \approx \frac{\lambda}{a} $$ Since we don't want the source slit's width to affect the diffraction pattern, we can equate these two angles: $$ \frac{\lambda}{w} \lesssim \frac{\lambda}{a} $$ Solving for \(w\): $$ w \lesssim \frac{a\lambda}{\lambda} $$ Finally, we need the width to be much smaller than this value. A reasonable approximation would be to divide the result by 4: $$ w \lesssim \frac{\lambda f_0}{4a} $$

In this case, we can use the condition given in the problem, \(R > a^2/\lambda\), to estimate the width \(w\) of the source slit. Since the distance between the slits is not changing significantly, we can still use the previously mentioned approximation for the angular width of the central maximum: $$ \theta \approx \frac{\lambda}{a} $$ Now we'll use the condition \(R > a^2/\lambda\): $$ R \approx \frac{a^2}{\lambda} $$ Substituting this value of \(R\) into the angular width equation, we get: $$ \theta \approx \frac{\lambda}{a} \approx \frac{w}{R} $$ Solving for \(w\): $$ w \approx \frac{a\lambda}{a} $$ As before, we need the width to be much smaller than this value. A reasonable approximation would be to divide the result by 4: $$ w \approx \frac{a}{4} $$ To summarize, for the given scenarios: Case 1 (with a lens): \( w \lesssim \frac{\lambda f_0}{4a} \) Case 2 (without a lens, \(R>a^2/\lambda\)): \( w \approx \frac{a}{4} \)