GOAL Calculate properties of a convex mirror. PROBLEM An object 3.00 cm high is

Question

GOAL Calculate properties of a convex mirror.

PROBLEM An object 3.00 cm high is placed 20.0 cm from a convex mirror with a focal length of magnitude 8.00 cm. Find (a) the position of the image, (b) the magnification of the mirror, and (c) the height of the image.

STRATEGY This problem again requires only substitution into the mirror and magnification equations. Multiplying the object height by the magnification gives the image height.

SOLUTION

(A) Find the position of the image.

Because the mirror is convex, its focal length is negative. Substitute into the mirror equation.

Solve for q.

q = -5.71 cm

(B) Find the magnification of the mirror.

Substitute into the magnification equation.

(C) Find the height of the image.

Multiply the object height by the magnification.

h' = hM = (3.00 cm)(0.286) = 0.858 cm

LEARN MORE

REMARKS The negative value of q indicates the image is virtual, or behind the mirror. The image is upright because M is positive.

QUESTION Is the image that a convex mirror produces real or virtual?

The image is always virtual.It depends on the focal length of the mirror.     It depends on the distance to the object.The image is always real.

PRACTICE IT

Use the worked example above to help you solve this problem. An object 3.15 cm high is placed 19.1 cm from a convex mirror with a focal length of 7.50 cm.

(a) Find the position of the image.
cm

(b) Find the magnification of the mirror.


(c) Find the height of the image.
cm

EXERCISE HINTS:  GETTING STARTED  |  I'M STUCK!

Suppose the object is moved so it is 3.75 cm from the same mirror. Repeat parts (a) through (c).

(a) q =  cm

(b) M =  

(c) h' =  cm

The image is  ---Select--- upright and virtual upright and real inverted and virtual inverted and real .

Explanation / Answer

practice question:

part a:

given that,

object distance=p=19.1 cm

focal length=f=-7.5 cm

then if image distance is q,

(1/p)+(1/q)=1/f

==>(1/19.1)+(1/q)=(-1/7.5)

==>q=1/((-1/7.5)-(1/19.1))=-5.3853 cm

so image position is -5.3853 cm

part b:

magnitification=-q/p=0.28195

part c:

height of image=height of object*magnification

=3.15*0.28195

=0.88816 cm

part d:

new object distance p=3.75 cm

then q=1/((-1/7.5)-(1/3.75))=-2.5 cm

part e:

M=-q/p

=0.67

part f:

h'=M*h

=0.67*3.15

=2.1 cm

the image is upright (as magnification is positive) and virtual (as image distance is -ve)

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