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You have a concave spherical mirror with a 10.9-cm radius of curvature. You place an object on the mirror\'s axis, 18.5 cm in front of the mirror. How far is the object's image from the mirror? If it can be determined, is the image real or virtual? If it can be determined, is the image upright or inverted with respect to the object?

Respuesta :

skyluke89

1) The distance of the image is 7.7 cm

2) The image is real

3) The image is inverted

Explanation:

1)

We can solve the problem by using the mirror equation:

[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]

where

f is the focal length of the mirror

p is the distance of the object from the mirror

q is the distance of the image from the mirror

In this problem, we have:

- The radius of curvature of the mirror is

R = 10.9 cm

The focal length is half the radius of curvature, so:

[tex]f=\frac{10.9}{2}=5.45 cm[/tex] (positive for a concave mirror)

- The distance of the object from the mirror is

p = 18.5 cm

Therefore, by applying the equation, we find the distance of the image:

[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{5.45}-\frac{1}{18.5}=0.129 \rightarrow q=\frac{1}{0.129}=7.7 cm[/tex]

2)

The sign of q tells us whether the image is real or virtual:

  • If the sign of q is positive, it means that the image is on the same side of the object with respect to the mirror, so the image is real
  • If the sign of q is negative, it means that the image is on the opposite side to the object with respect to the mirror, so the image is virtual

In part 1), we found that the value of q is positive, therefore the image is real.

3)

To find the orientation of the image, we have to use the magnification equation:

[tex]M=\frac{y'}{y}=-\frac{q}{p}[/tex]

where

M is the magnification

y' is the size of the image

y is the size of the object

We have two cases:

  • If the sign of M is positive, the image is upright
  • If the sign of M is negative, the image is inverted

In this case, we have

p = 18.5 cm

q = 7.7 cm

Substituting, we find

[tex]M=-\frac{7.7}{18.5}=-0.42[/tex]

Therefore, the image is inverted.

#LearnwithBrainly

batolisis

A) The distance of the object's image from the mirror is : 7.7 cm

B) The image is real

C) The image is inverted

Given data :

Radius of curvature of mirror = 10.9 cm

Distance of object from mirror = 18.5 cm

Determine the distance of the object's image from the mirror

Applying the mirror equation

[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex] --- ( 1 )

f = focal length , p = object distance , q = image distance.

f = 10.9 / 2 = 5.45 cm ,  p = 18.5 ,  q = ?

insert values into equation ( 1 )

1 / 5.45 = 1/18.5  + 1 / q

therefore image distance ( q ) = 7.7 cm

Determine if the image is real or virtual

The image is real because the value of the image distance ( q ) is a positive value, which denotes that the image is on the same side with the object.

Determine if the image is upright or inverted

To determine if the image is upright or inverted we will apply the equation below

M = [tex]- \frac{q}{p}[/tex]  

   = - 7.7 / 18.5

   = -0.42

Given that the magnification value is a negative value the image is inverted.

Hence we can conclude that A) The distance of the object's image from the mirror is : 7.7 cm.  B)The image is real. C) The image is inverted

Learn more about concave spherical mirror : https://brainly.com/question/9895618

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