Chapter 19.2, Problem 20PP

(a)

To determine

The filter that will produce the wider band.

Answer to Problem 20PP

The red filter produces the wider band.

Explanation of Solution

Introduction:

In single slit experiment, when a light ray from a monochromatic light source falls on a single slit placed at a distance, the light gets diffracted by the edges of the single slit and falls on the screen. The brightest band appears at the middle of the screen.

The expression for the width of bright band $\left(2x_1\right)$ in single-slit diffraction is

  $2x_1=\frac{2\lambda L}{w}$

Here, $\lambda$ is the wavelength, $L$ is the distance to the screen, and $w$ is the width of the slit.

When a monochromatic blue light passes through a single minute opening, whose size is larger than the wavelength of the light, the light gets diffracted by both edges of the slit and a series of dark and bright bands appears on a distant screen

The width of the central band is directly proportional to the wavelength of the light emitted by the monochromatic light.

As the wavelength of the red light is more than the wavelength of the blue light, the width of the central band corresponding to red filter is more than that of the blue filter.

Conclusion:

Thus, the red filter produced the wider band.

(b)

To determine

The width of the central bright band for blue-violet filter and red filter.

Answer to Problem 20PP

The width of the central bright band for blue-violet filter is $18\text{ mm}$ .

The width of the central bright band for red filter is $25\text{ cm}$ .

Explanation of Solution

Given:

The width of the slit is $w=0.050\text{ mm}$ .

The wavelength of the blue-violet filter is $\lambda =441\text{ nm}$ .

The wavelength of the red filter is $\lambda =622\text{ nm}$ .

The distance is $L=1.0\text{ cm}$ .

Formula used:

The expression for the width of bright band $\left(2x_1\right)$ in single-slit diffraction is

  $2x_1=\frac{2\lambda L}{w}$

Here, $\lambda$ is the wavelength, $L$ is the distance to the screen, and $w$ is the width of the slit.

Calculation:

The width of the central bright band for blue-violet filter is

  $\begin{array}{l}2x_1=\frac{2\lambda L}{w}\\ 2x_1=\frac{2\times \left(441\text{ nm}\times \frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)\times \left(1.0\text{ m}\right)}{\left(0.050\text{ mm}\times \frac{1\text{ m}}{1,000\text{ mm}}\right)}\\ 2x_1=0.018\text{ m}\times \frac{1,000\text{ mm}}{1\text{ m}}\\ 2x_1=18\text{ mm}\end{array}$

The width of the central bright band for red filter is,

  $\begin{array}{l}2x_1=\frac{2\lambda L}{w}\\ 2x_1=\frac{2\times \left(\text{622 nm}\times \frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)\times \left(1.0\text{ m}\right)}{\left(0.050\text{ mm}\times \frac{1\text{ m}}{1,000\text{ mm}}\right)}\\ 2x_1=0.025\text{ m}\times \frac{1,000\text{ mm}}{1\text{ m}}\\ 2x_1=25\text{ mm}\end{array}$

Conclusion:

Thus, the width of the central bright band for blue-violet filter is $18\text{ mm}$ and the width of the central bright band for red filter is $\text{25 mm}$ .