Problem Statement
A slit of width $a = 0.3$ mm is illuminated by a mercury lamp filtered to $\lambda = 546$ nm. A lens of focal length $f = 60$ cm focuses the Fraunhofer pattern. Find the full width of the central maximum and the positions of the first secondary maxima.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: First minima at: $y_1 = f\lambda/a = 0.60\times546\times10^{-9}/0.3\times10^{-3} = 0.60\times1.82\times10^{-3} = 1.09\times10^{-3}$ m.
Step 2 — Apply the relevant physical law or equation: Full width of central max: $w = 2y_1 = \boxed{2.18\text{ mm}}$.
Step 3 — Solve algebraically for the unknown: First secondary maxima at $\alpha \approx 3\pi/2$: $y_{sec} = 1.5y_1 = \boxed{1.64\text{ mm}}$ from centre.
Worked Calculation
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{y_{sec} = 1.5y_1 = \boxed{1.64\text{ mm}}}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
First minima at: $y_1 = f\lambda/a = 0.60\times546\times10^{-9}/0.3\times10^{-3} = 0.60\times1.82\times10^{-3} = 1.09\times10^{-3}$ m.
Full width of central max: $w = 2y_1 = \boxed{2.18\text{ mm}}$.
First secondary maxima at $\alpha \approx 3\pi/2$: $y_{sec} = 1.5y_1 = \boxed{1.64\text{ mm}}$ from centre.
Answer
$$\boxed{y_{sec} = 1.5y_1 = \boxed{1.64\text{ mm}}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
Leave a Reply