**Problem:**Monochromatic light falls on a screen 1.90 m from two slits separated by 2.06 mm. The first- and second-order bright fringes are separated by 0.557 mm. What is the wavelength of the light?**Answer:**\[ \text{Wavelength (\(\lambda\))} = \Box \text{ nm} \]**Explanation:**To solve this problem, we will use the formula for the fringe separation in a double-slit experiment:\[ \Delta y = \frac{\lambda \cdot L}{d} \]Where:- \(\Delta y\) is the separation between fringes (0.557 mm),- \(L\) is the distance from the slits to the screen (1.90 m),- \(d\) is the distance between the slits (2.06 mm),- \(\lambda\) is the wavelength of the light.By rearranging the formula to solve for \(\lambda\):\[ \lambda = \frac{\Delta y \cdot d}{L} \]Remember to convert all measurements to consistent units (e.g., meters) before calculating the wavelength.

Given Distance from screen from slit = D = 1.90 m separation between two slit = d = 2.06 mm…

Answered: nochromatic light falls on a screen…

nochromatic light falls on a screen 1.90 m from two slits separated by 2.06 mm. The first- and second-order bright fringes are separated by 0.557 mm. What is the wavelength of the light? ΠΠ

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Interference of Light

The word "constructive" signifies addition in interference. Constructive interference occurs when the path of the waves that start from the slit differs by 1 entire wavelength or in the intervals of the same. Hence, the waves that are presented on the screen are in phase (crest to crest/trough to trough). The waves interfering constructively forms bright fringes on the screen.

Question

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**Problem:**

Monochromatic light falls on a screen 1.90 m from two slits separated by 2.06 mm. The first- and second-order bright fringes are separated by 0.557 mm. What is the wavelength of the light?

**Answer:**

\[ \text{Wavelength (\(\lambda\))} = \Box \text{ nm} \]

**Explanation:**

To solve this problem, we will use the formula for the fringe separation in a double-slit experiment:

\[ \Delta y = \frac{\lambda \cdot L}{d} \]

Where:
- \(\Delta y\) is the separation between fringes (0.557 mm),
- \(L\) is the distance from the slits to the screen (1.90 m),
- \(d\) is the distance between the slits (2.06 mm),
- \(\lambda\) is the wavelength of the light.

By rearranging the formula to solve for \(\lambda\):

\[ \lambda = \frac{\Delta y \cdot d}{L} \]

Remember to convert all measurements to consistent units (e.g., meters) before calculating the wavelength.

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