/*! 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} Q.43 High-power lasers are used to cu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

High-power lasers are used to cut and weld materials by focusing the laser beam to a very small spot. This is like using a magnifying lens to focus the sun light to a small spot that can burn things. As an engineer you have designed a laser cutting device in which the material to be cut is placed 5.0cmbehind the lens. you have selected a high-power laser with a wavelength of

your calculation indicates that the laser must be focused to a 5.0-μm-diameterspot in order to have sufficient power to make the cut. what is the minimum diameter of lens you must install?

Short Answer

Expert verified

The minimum diameter of the lens is2.58cm.

Step by step solution

01

Given information

We have given that:

High power laser wavelength λ=1.06x10-6m

Laser cutting devicef=5cm→5.0x10-2m

sufficient power to cut laser Ӭ=5μm→5.0x10-6m

we need to find the diameter of lens to install.

02

Simplification

Let us use this formula,

Ӭ=2.44λfD

Here, fis the focal length, λis wavelength and Ӭ is the value for laser must be focused.

Now multiplying both sides by D,

localid="1649142735572" D×Ӭ=D×2.44λfD

localid="1649142750541" D×Ӭ=2.44λf

Dividing both the sides by Ó¬,

D×ӬӬ=2.44λfӬ

D=2.44λfӬ

Let us substitute values in the equation,

D=2.44(1.06x10-6m)(5.0x10-2m)5.0x10-6m

D=0.0258m→2.58cm.

Here,Dis the diameter of the lens.

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!

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

Two converging lenses with focal lengths of 40 cm and 20 cm are 10 cm apart. A 2.0-cm-tall object is 15 cm in front of the 40-cm-focal-length lens. a). Use ray tracing to find the position and height of the image. Do this accurately using a ruler or paper with a grid, then make measurements on your diagram.

b). Calculate the image position and height. Compare with your ray-tracing answers in part a.

Suppose you wanted special glasses designed to let you see underwater without a face mask. Should the glasses use a converging or diverging lens? Explain.

A 2.0-cm-tall object is 20cmto the left of a lens with a focal length of10cmA second lens with a focal length of 15cmis30cmto the right of the first lens.

a. Use ray tracing to find the position and height of the image. Do this accurately using a ruler or paper with a grid, then make measurements on your diagram.

b. Calculate the image position and height. Compare with your ray-tracing answers in part a.

The rays leaving the two-component optical system of FIGUREP35.27produce two distinct images of the1.0cm-tall object. what are the position (relative to the lens), orientation, and height of each image?

A narrow beam of white light is incident on a sheet of quartz. The beam disperses in the quartz, with red light (l≈400nm)traveling at an angle of 26.3°with respect to the normal and violet light (l≈400nm) traveling at 25.7° . The index of refraction of quartz for red light is 1.45. What is the index of refraction of quartz for violet light?

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.