/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Light of wavelength \(5000 \math... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Light of wavelength \(5000 \mathrm{~A}\) falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

Short Answer

Expert verified
The wavelength is 5000 Å and frequency is \(6 \times 10^{14} \text{ Hz}\). The angle of incidence is 45°.

Step by step solution

01

Understand the problem

Light with a wavelength of 5000 Ã… (angstroms) hits a reflecting surface. We need to find the wavelength and frequency of the light after reflection, and also find the angle of incidence where the reflected ray is normal to the incident ray.
02

Wavelength of Reflected Light

When light reflects off a surface, it does not change its wavelength. Therefore, the wavelength of the reflected light remains the same as that of the incident light, i.e., 5000 Ã….
03

Frequency of Reflected Light

The frequency of light remains constant regardless of the medium or reflection. Using the speed of light \( c \) and the given wavelength \( \lambda \), we find the frequency \( f \) with the equation:\[c = f \lambda\]Given \( \lambda = 5000 \text{ Ã…} = 5000 \times 10^{-10} \text{ m} \) and \( c = 3 \times 10^8 \text{ m/s} \), solve for \( f \):\[ f = \frac{c}{\lambda} = \frac{3 \times 10^8}{5000 \times 10^{-10}} = 6 \times 10^{14} \text{ Hz}\]
04

Angle of Incidence for Perpendicular Reflection

For the reflected ray to be perpendicular to the incident ray, the angle between the incident ray, normal, and reflected ray must be a right angle (90°). This means that the angle of reflection \( r \) is 90° minus the angle of incidence \( i \):\[ r = 90^\circ - i\]Since angle of incidence is equal to angle of reflection in a plane mirror:\[ i = 90^\circ - i\]Solving gives \( i = 45^\circ \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wavelength of Light
When discussing light, one of the fundamental properties to consider is its wavelength. Wavelength refers to the distance between consecutive peaks (or troughs) of a wave. In the context of light, this distance is often measured in angstroms (Ã…), where 1 Ã… equals 10^{-10} meters. Light with a wavelength of 5000 Ã… is typical for visible light, which ranges approximately from 3800 Ã… to 7500 Ã….

One vital point to understand about wavelength is that it does not change when light reflects off a surface. This means if light with a certain wavelength, like our 5000 Ã…, strikes a mirror, the light maintains this wavelength upon reflection. Reflection ensures that both the incident and the reflected light share identical wavelengths, a concept crucial for maintaining color consistency in mirrors.
Frequency of Light
Frequency is another core concept when working with light. It represents how many wave peaks pass a given point each second. Measured in hertz (Hz), frequency stays constant regardless of reflection or the medium that light travels through.

To calculate the frequency of light, you can use the speed of light, denoted as \( c = 3 \times 10^8 \text{ m/s} \), and the wavelength. The relationship between these quantities is given by the equation \( c = f \lambda \). Substituting known values, such as the wavelength \( \lambda = 5000 \times 10^{-10} \text{ m} \), allows us to solve for the frequency \( f \). Through this calculation, the frequency of light is determined to be \( 6 \times 10^{14} \text{ Hz} \).

Frequency remains unaltered because, during reflection, light's speed and wavelength stay the same, maintaining the uniformity of this parameter.
Angle of Incidence
Understanding the angle of incidence is essential when analyzing light reflection. This angle is the measure between the incoming light ray and the normal—a line perpendicular to the reflecting surface. The law of reflection states that the angle of incidence \( i \) equals the angle of reflection \( r \).

A special case occurs when the reflected ray is perpendicular to the incident ray. Mathematically, this means that their angle forms a right angle (90°). Here, the angle of reflection can be expressed as \( r = 90^\circ - i \). Given that \( i = r \), it follows that \( i = 45^\circ \) when setting up the rays to be perpendicular.

In practical terms, this implies that, when you want the reflected ray to travel directly sideways in relation to the incident light, you should place the light source at a 45° angle relative to the normal of the mirror.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a double-slit experiment the angular width of a fringe is found to be \(0.2^{\circ}\) on a screen placed \(1 \mathrm{~m}\) away. The wavelength of light used is \(600 \mathrm{~nm}\). What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be \(4 / 3\).

Two towers on top of two hills are \(40 \mathrm{~km}\) apart. The line joining them passes \(50 \mathrm{~m}\) above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?

In deriving the single slit diffraction pattern. it was stated that the intensity is zero at angles of \(n \lambda / a\). Justify this by suitably dividing the slit to bring out the cancellation.

Answer the following questions: (a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? (b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? (c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why? (d) Two students are separated by a \(7 \mathrm{~m}\) partition wall in a room \(10 \mathrm{~m}\) high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily. (e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?

In a Young's double-slit experiment, the slits are separated by \(0.28 \mathrm{~mm}\) and the screen is placed \(1.4 \mathrm{~m}\) away. The distance between the central bright fringe and the fourth bright fringe is measured to be \(1.2 \mathrm{~cm} .\) Determine the wavelength of light used in the experiment.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.