# A beam of monochromatic light is incident on a single slit of width 0.5 mm. A diffraction pattern...

## Question:

A beam of monochromatic light is incident on a single slit of width 0.5 mm. A diffraction pattern forms on a wall 1.5 m beyond the slit. The distance between the positions of zero intensity on both sides of the central maximum is 2.5 mm. Calculate the wavelength of the light.

a) 417 nm

b) 520 nm

c) 625 nm

d) 832 nm

## Single Slit Diffraction:

In single slit diffraction of light, the central maxima is the distance between the first minima on both sides of center point on the screen. The width of central maxima is more compared to the width of central maxima in double slit experiment.

## Answer and Explanation:

**Given Data**

- slit width, {eq}a\ = 0.5\ mm\ = 5\times 10^{-4}\ m{/eq}

- distance of screen from the slit,
*L*= 1.5 m - width of central maxima, {eq}w\ = 2.5\ mm\ = 2.5\times 10^{-3}\ m{/eq}

**Finding the wavelength ({eq}\lambda{/eq}) of light**

The width of central maxima is written as:

- {eq}w\ = \dfrac{2\times \lambda\times L}{a} {/eq}

- {eq}2.5\times 10^{-3}\ = \dfrac{2\times \lambda\times 1.5}{5\times 10^{-4}} {/eq}

- {eq}\lambda\ = 417\ nm{/eq}

**Correct option is (a)**

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from UExcel Physics: Study Guide & Test Prep

Chapter 15 / Lesson 9