Problem 5.43 — Biprism Experiment

Problem Statement

A Fresnel biprism (refracting angle $\alpha = 2°$, $n = 1.5$) is used with a slit source of $\lambda = 589$ nm at distance $a = 20$ cm from the biprism. The screen is at distance $b = 1.0$ m from the biprism. Find the fringe width.

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: The biprism deflects each half-beam by angle $\delta = (n-1)\alpha = 0.5\times2° = 1° = 0.01745$ rad.

Step 2 — Apply the relevant physical law or equation: Effective slit separation: $d = 2a\delta = 2\times0.20\times0.01745 = 6.98\times10^{-3}$ m.

Step 3 — Solve algebraically for the unknown: Fringe width on screen:

Step 4 — Substitute numerical values with units: $$\Delta y = \frac{\lambda(a+b)}{d} = \frac{589\times10^{-9}\times1.20}{6.98\times10^{-3}} \approx \boxed{0.101\text{ mm}}$$

Worked Calculation

$$\Delta y = \frac{\lambda(a+b)}{d} = \frac{589\times10^{-9}\times1.20}{6.98\times10^{-3}} \approx \boxed{0.101\text{ mm}}$$

$$\text{Numerical result} = \text{given expression substituted with values}$$

$$\boxed{\Delta y = \frac{\lambda(a+b)}{d} = \frac{589\times10^{-9}\times1.20}{6.98\times10^{-3}} \approx \boxed{0.101\text{ mm}}}$$

The biprism deflects each half-beam by angle $\delta = (n-1)\alpha = 0.5\times2° = 1° = 0.01745$ rad.

Effective slit separation: $d = 2a\delta = 2\times0.20\times0.01745 = 6.98\times10^{-3}$ m.

Fringe width on screen:

$$\Delta y = \frac{\lambda(a+b)}{d} = \frac{589\times10^{-9}\times1.20}{6.98\times10^{-3}} \approx \boxed{0.101\text{ mm}}$$

Answer

$$\Delta y = \frac{\lambda(a+b)}{d} = \frac{589\times10^{-9}\times1.20}{6.98\times10^{-3}} \approx \boxed{0.101\text{ mm}}$$

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.