Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For specular reflection of a light ray, the angle of incidence a) must be equal to the angle of reflection. b) is always less than the angle of reflection. c) is always greater than the angle of reflection. d) is equal to \(90^{\circ}\) - the angle of reflection. e) may be greater than, less than, or equal to the angle of reflection.

Short Answer

Expert verified
Answer: (a) The angle of incidence must be equal to the angle of reflection.

Step by step solution

01

Understand the concept of specular reflection

Specular reflection occurs when a parallel beam of light rays hits a smooth surface, and it reflects these rays such that they remain parallel after reflection. In this type of reflection, the angle between the incident ray and the reflected ray is important.
02

Recall the law of reflection

The law of reflection states that for a smooth surface, the angle of incidence (the angle between the incident light ray and the surface normal) is equal to the angle of reflection (the angle between the reflected light ray and the surface normal). Mathematically, this can be written as: \(angle~of~incidence = angle~of~reflection\)
03

Compare the given statements with the law of reflection

Now, we need to observe the given statements and compare them with the law of reflection: a) must be equal to the angle of reflection. — This statement is true according to the law of reflection. b) is always less than the angle of reflection. — This statement is false, as it contradicts the law of reflection. c) is always greater than the angle of reflection. — This statement is false, as it contradicts the law of reflection. d) is equal to \(90^{\circ}\) - the angle of reflection. — This statement is false because according to the law of reflection, the angle of incidence is equal to the angle of reflection, not their difference. e) may be greater than, less than, or equal to the angle of reflection. — This statement is incorrect since, according to the law of reflection, the angle of incidence must always be equal to the angle of reflection.
04

Conclusion

Based on our analysis, we can conclude that the correct answer is (a) the angle of incidence must be equal to the angle of reflection for a specular reflection of a light ray.

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 Vaia!

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

Answer as true or false with an explanation for the following: The wavelength of He-Ne laser light in water is less than its wavelength in the air. (The refractive index of water is \(1.33 .\)

An object is located at a distance of \(100 . \mathrm{cm}\) from a concave mirror of focal length \(20.0 \mathrm{~cm}\). Another concave mirror of focal length \(5.00 \mathrm{~cm}\) is located \(20.0 \mathrm{~cm}\) in front of the first concave mirror. The reflecting sides of the two mirrors face each other. What is the location of the final image formed by the two mirrors and the total magnification by the combination?

Reflection and refraction, like all classical features of light and other electromagnetic waves, are governed by the Maxwell equations. The Maxwell equations are time-reversal invariant, which means that any solution of the equations reversed in time is also a solution. a) Suppose some configuration of electric charge density \(\rho,\) current density \(\vec{j},\) electric field \(\vec{E},\) and magnetic field \(\vec{B}\) is a solution of the Maxwell equations. What is the corresponding time-reversed solution? b) How, then, do "one-way mirrors" work?

A light ray of wavelength 700 . \(\mathrm{nm}\) traveling in air \(\left(n_{1}=1.00\right)\) is incident on a boundary with a liquid \(\left(n_{2}=1.63\right) .\) a) What is the frequency of the refracted ray? b) What is the speed of the refracted ray? c) What is the wavelength of the refracted ray?

One of the factors that cause a diamond to sparkle is its relatively small critical angle. Compare the critical angle of diamond in air compared to that of diamond in water.

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.

Sign-up for free