/*! 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 8 A man uses a concave mirror for ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A man uses a concave mirror for shaving. He keeps his face at a distance of \(25 \mathrm{~cm}\) from the mirror and gets an image which is \(1 \cdot 4\) times enlarged. Find the focal length of the mirror.

Short Answer

Expert verified
The focal length of the concave mirror is approximately \(-14.6 \text{ cm}\).

Step by step solution

01

Understand the Mirror Equation and Magnification

When dealing with mirrors, we use the mirror equation \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance. For magnification \( m \), which is the image height divided by object height, we have the relationship \( m = \frac{v}{u} \). Here, magnification \( m = 1.4 \), and the object distance \( u = -25 \text{ cm} \).
02

Calculate Image Distance Using Magnification

The formula for magnification is \( m = \frac{v}{u} \). Given \( m = 1.4 \) and \( u = -25 \text{ cm} \), we rearrange to find \( v \): \( v = m \times u = 1.4 \times (-25) = -35 \text{ cm} \). The negative sign indicates that the image is on the same side as the object, which is typical for a real image formed by a concave mirror.
03

Use Mirror Formula to Find Focal Length

Substitute \( v = -35 \text{ cm} \) and \( u = -25 \text{ cm} \) into the mirror equation: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]. This becomes \[ \frac{1}{f} = \frac{1}{-35} + \frac{1}{-25} \].
04

Calculate the Values for Each Term

Calculate \( \frac{1}{-35} = -0.02857 \) and \( \frac{1}{-25} = -0.04 \). Add these values: \( \frac{1}{f} = -0.02857 + (-0.04) = -0.06857 \).
05

Solve for Focal Length

Take the reciprocal of \( -0.06857 \) to find \( f \): \( f = \frac{1}{-0.06857} \approx -14.6 \text{ cm} \). The negative sign indicates a concave mirror.

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.

Mirror Equation
When discussing concave mirrors, the mirror equation is a fundamental concept that helps us understand how mirrors form images. It is given by the formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \). Here, \( f \) represents the focal length of the mirror, \( v \) is the distance from the mirror to the image, and \( u \) is the distance from the object to the mirror.
In practice, this equation allows us to relate the distances associated with image and object positions to find unknown values. By using the signs convention, where in concave mirrors the object distance is negative because it is measured against the direction of the incoming light, we can accurately solve real-world problems like determining the focal length.
Focal Length
The focal length of a mirror is a measure of how strongly it converges or diverges light. In a concave mirror, this length determines the radius at which parallel rays of light are focused to a point. The focal length \( f \) of a concave mirror is half the radius of curvature.
Knowing the focal length helps us predict how an image will form. In the exercise provided, by rearranging the mirror equation as \( f = \frac{1}{\frac{1}{v} + \frac{1}{u}} \), the calculated focal length is approximately -14.6 cm. The negative sign is crucial as it signifies that the mirror is concave, focusing the rays inwards.
Magnification
Magnification in mirrors refers to how much larger or smaller an image appears compared to the actual object. It is described by the formula \( m = \frac{v}{u} \), where \( m \) is the magnification, and \( v \) and \( u \) are the image and object distances respectively.
In the given situation, the magnification is 1.4, meaning the image appears 1.4 times larger than the object. Understanding this concept helps to deduce not only the size but also the orientation of the image - in this case, an enlarged upright image since it appears on the same side of the mirror as the object.
Real Image
Real images are formed when light rays actually converge at a point. For concave mirrors, real images are produced on the same side as the object when the object is placed beyond the focal point.
Real images are usually inverted, however, since our case involves a magnified direct image, the position must be carefully analyzed with measurements. The calculated image distance \( v = -35 \) cm indicates the image forms in front of the mirror, true to concave mirror behavior. This characteristic allows us to use the mirror for practical applications like shaving, where an enlarged and clear view is essential.

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

A hemispherical portion of the surface of a solid glass sphere \((\mu=1 \cdot 5)\) of radius \(r\) is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance \(3 r\) from the centre of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. Locate the final image formed.

One end of a cylindrical glass rod \((\mu=1 \cdot 5)\) of radius \(1 \cdot 0 \mathrm{~cm}\) is rounded in the shape of a hemisphere. The rod is immersed in water \((\mu=4 / 3)\) and an object is placed in the water along the axis of the \(\operatorname{rod}\) at a distance of \(8 \cdot 0 \mathrm{~cm}\) from the rounded edge. Locate the image of the object.

An object \(P\) is focussed by a microscope \(M .\) A glass slab of thickness \(2 \cdot 1 \mathrm{~cm}\) is introduced between \(P\) and \(M .\) If the refractive index of the slab is \(1 \cdot 5\), by what distance should the microscope be shifted to focus the object again?

A converging lens of focal length \(15 \mathrm{~cm}\) and a converging mirror of focal length \(10 \mathrm{~cm}\) are placed \(50 \mathrm{~cm}\) apart. If a pin of length \(2 \cdot 0 \mathrm{~cm}\) is placed \(30 \mathrm{~cm}\) from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?

A paperweight \((\mu=1 \cdot 5)\) in the form of a hemisphere of radius \(3 \cdot 0 \mathrm{~cm}\) is used to hold down a printed page. An observer looks at the page vertically through the paperweight. At what height above the page will the printed letters near the centre appear to the observer?

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.