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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.

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Most popular questions from this chapter

In Fig. 34-26, stick figure Ostands in front of a spherical mirrorthat is mounted within the boxed region;the central axis through themirror is shown. The four stick figures I1to I4suggest general locationsand orientations for the images that might be produced by themirror. (The figures are onlysketched in; neither their heightsnor their distances from the mirror are drawn to scale.) (a) Whichof the stick figures could not possibly represent images? Of thepossible images, (b) which would be due to a concave mirror, (c)which would be due to a convex mirror, (d) which would be virtual,and (e) which would involve negative magnification?

A short straight object of lengthLlies along the central axis of a spherical mirror, a distance pfrom the mirror. (a) Show that its image in the mirror has alength, L'=L(f/(p-f))2(Hint: Locate the two ends of the object.) (b) Show that the longitudinal magnification is equal tom'=(L'/L) is equal to m2, where m is the lateral magnification.

a real inverted imageof an object is formed by a particular lens (not shown); the object–image separation is, measured along the central axis of the lens. The image is just half the size of the object. (a) What kind of lens must be used to produce this image? (b) How far from the object must the lens be placed? (c) What is the focal length of the lens?

(a) Show that if the object O in Fig. 34-19c is moved from focal point F1toward the observer’s eye, the image moves in from infinity and the angle (and thus the angular magnification mu) increases. (b) If you continue this process, where is the image when mu has its maximum usable value? (You can then still increase, but the image will no longer be clear.) (c) Show that the maximum usable value of ismθ=1+25cmf.(d) Show that in this situation the angular magnification is equal to the lateral magnification.

An object is placed against the center of a converging lens and then moved along the central axis until it is 5.0mfrom the lens. During the motion, the distance between the lens and the image it produces is measured. The procedure is then repeated with a diverging lens. Which of the curves in Fig. 34-28 best gives versus the object distance p for these lenses? (Curve 1 consists of two segments. Curve 3 is straight.)

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