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4 years ago

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You will begin with a relatively standard calculation.Consider a concave spherical mirror with a radius of curvature equal to 60

.0 centimeters. An object 6.00 centimeters tall is placed along the axis of the mirror, 45.0 centimeters from the mirror. You are to find the location and height of the image.a) What is the focal length Image for this mirror?b) Now use the spherical mirror equation to find the imagedistance s'.c) Find the magnification m,using sand s'.d) Finally, use the magnification to find the height of theimage y'.e) Solve the spherical mirror equation for s'.f)What is the magnification m? Use your answer from Part E

1 answer:

Strike441 [17]4 years ago

6 0

a) 30.0 cm

For any mirror, the radius of curvature is twice the focal length:

r = 2f

where

r is the radius of curvature

f is the focal length

For the mirror in this problem, we have

r = 60.0 cm is the radius of curvature

Therefore, solving the equation above for f, we find its focal length:

$f=\frac{r}{2}=\frac{60.0 cm}{2}=30.0 cm$

b) 90 cm

The mirror equation is:

$\frac{1}{s'}=\frac{1}{f}-\frac{1}{s}$

where

s' is the distance of the image from the mirror

f is the focal length

s is the distance of the object from the mirror

For the situation in the problem, we have

f = +30.0 cm is the focal length (positive for a concave mirror)

s = 45.0 cm is the object distance from the mirror

Solving the formula for s', we find

$\frac{1}{s'}=\frac{1}{30.0 cm}-\frac{1}{45.0 cm}=0.011 cm^{-1}$

$s'=\frac{1}{0.011 cm^{-1}}=90 cm$

c) -2

The magnification of the mirror is given by

$M=-\frac{s'}{s}$

where in this problem we have

s' = 90 cm is the image distance

s = 45.0 cm is the object distance

Solving the equation, we find:

$M=-\frac{90 cm}{45 cm}=-2$

So, the magnification is -2.

d) -12.0 cm

The magnification can also be rewritten as

$M=\frac{y'}{y}$

where

y' is the height of the image

y is the heigth of the object

In this problem, we know

y = 6.0 cm is the height of the object

M = -2 is the magnification

Solving the equation for y', we find

$y'=My=(-2)(6.0 cm)=-12.0 cm$

and the negative sign means that the image is inverted.

Part e and f are exactly identical as part b) and c).

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