Question

In a double-slit interference experiment, the wavelength is \(475 \mathrm{nm}\), the slit separation is \(0.120 \mathrm{mm},\) and the screen is $36.8 \mathrm{cm}$ away from the slits. What is the linear distance between adjacent maxima on the screen? [Hint: Assume the small-angle approximation is justified and then check the validity of your assumption once you know the value of the separation between adjacent maxima.] (tutorial: double slit 1 ).

Short answer

Answer: The linear distance between adjacent maxima on the screen is approximately 1.457 mm.

Step-by-step solution

Convert all units to meters

To have consistent units in the calculations, we need to convert all the given values to meters. We have, \(\lambda = 475 \thinspace nm = 475 \times 10^{-9} \thinspace m\), \(d = 0.120 \thinspace mm = 0.120 \times 10^{-3} \thinspace m\), and \(L = 36.8 \thinspace cm = 36.8 \times 10^{-2} \thinspace m\).

Use the small-angle approximation formula for the angular separation

Under the small-angle approximation, the angular separation between adjacent maxima (\(θ\)) for a double-slit interference experiment can be calculated using the formula: \(θ = \frac{m\lambda}{d}\), where \(m\) is the order of the maxima (e.g., \(1\) for the first maxima, \(2\) for the second maxima, etc.). Let's consider the distance between the first maxima on either side of the central maxima (i.e., \(m = 1\)). This will give us the distance between adjacent maxima. \(θ = \frac{\lambda}{d} = \frac{475 \times 10^{-9} \thinspace m}{0.120 \times 10^{-3} \thinspace m} = 3.958 \times 10^{-3}\) radians

Calculate the linear distance

Now, we need to convert the angular separation (\(θ\)) between adjacent maxima into a linear distance (\(y\)) on the screen. We use the following formula for that: \(y = L\cdot tan(θ) \approx L\cdot θ \thinspace\) (assuming the small-angle approximation) Now, plugging the values, we get: \(y = 36.8 \times 10^{-2} \thinspace m \cdot 3.958 \times 10^{-3}\) radians \(y = 1.457 \times 10^{-3} \thinspace m = 1.457 \thinspace mm\) So, the linear distance between adjacent maxima on the screen is approximately \(1.457 \thinspace mm\).

Check the validity of the small-angle approximation

To confirm that the small-angle approximation is valid, we should check if \(tan(θ)\) is approximately equal to \(θ\). If it is, then our answer is accurate. Let's check that: \(tan(3.958 \times 10^{-3}) \approx 3.958 \times 10^{-3}\) Using a calculator, we find that \(tan(3.958 \times 10^{-3}) \approx 3.959 \times 10^{-3}\), which is very close to \(3.958 \times 10^{-3}\). This means that the small-angle approximation is valid and our calculated linear distance between adjacent maxima is accurate.