A concave shaving mirror has a radius of curvature of 35cm. It is positioned so that the (upright) image of a man’s face is 2.5 times the size of the face. How far is the mirror from the face?

Short Answer

Expert verified

Distance of mirror from the face is p=10.5cm.

Step by step solution

01

Step 1: Given

The radius of curvature isr=35.0cm

Magnification is m=2.50

02

Determining the concept

Use the concept of magnification and focal length of the mirror. Find the expression for image distance from the formula for magnification and substitute it in the mirror formula. Substitute the given values to find the distance of the mirror from the face.

Formulae are as follows:

m=-ip

1f=1p+1i

where, m is the magnification, p is the pole, fis the focal length.

03

Determining the distance of the mirror from the face

The magnification is given by,

m=-ipi=-mp

Plugging this in the mirror equation,

1f=1p+1i1f=1p-1mp

Use the equation of focal length.

For spherical mirrors,f=r2

Therefore,

2r=1p-1mp

Solve it for p:

2r=1p1-1mp=r21-1m

Plugging the values,

p=35.021-12.50p=10.5cm

Therefore, the distance of the mirror from the face is p=10.5cm.

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

An object is placed against the center of a thin lens and then moved away from it along the central axis as the image distance is measured. Figure 34-41 gives i versus object distance p out to ps=60cm. What is the image distancewhen p=100cm?

58 through 67 61 59 Lenses with given radii. An object Ostands in front of a thin lens, on the central axis. For this situation, each problem in Table 34-7 gives object distance O, index of refraction n of the lens, radius of the nearer lens surface, and radius of the farther lens surface. (All distances are in centimeters.) Find (a) the image distance and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real or virtual , (d) inverted from the object Oor non-inverted , and (e) on the same side of the lens as object or on the opposite side.

Figure 34-37 gives the lateral magnification mof an object versus the object distanc pfrom a spherical mirror as the object is moved along the mirror’s central axis through a range of values p. The horizontal scale is set by Ps=10.0mm. What is the magnification of the object when the object is 21cm from the mirror?

32 through 38 37, 38 33, 35 Spherical refracting surfaces. An object Ostands on the central axis of a spherical refracting surface. For this situation, each problem in Table 34-5 refers to the index of refraction n1where the object is located, (a) the index of refraction n2on the other side of the refracting surface, (b) the object distancep, (c) the radius of curvature rof the surface, and (d) the image distance i. (All distances are in centimeters.) Fill in the missing information, including whether the image is (e) real (R)or virtual (V)and (f) on the same side of the surface as the object Oor on the opposite side.

A peanut is placed 40cmin front of a two-lens system: lens 1 (nearer the peanut) has focal length f1 =20cm, lens 2 has f2=-15cm and the lens separation is d=10cm. For the image produced by lens 2, what are (a) the image distance i2(including sign), (b) the image orientation (inverted relative to the peanut or not inverted), and (c) the image type (real or virtual)? (d) What is the net lateral magnification?

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