Espellos cóncavos: problemas e solucións

1. An object is placed 10 cm from a concave mirror. The focal length is 5 cm. Determine (a) The image distancia (b) the magnification of image

Coñecido:

A distancia focal (f) = 5 cm

A distancia do obxecto (do) = 10 centímetros

solución :

Formación da imaxe nun espello cóncavo:

Concave mirror – problems and solutions 1

The image distance :

1/di = 1/f – 1/do = 1/5 – 1/10 = 2/10 – 1/10 = 1/10

di = 10/1 = 10 centímetros

A distancia da imaxe é 10 cm.

The magnification :

m = –di / do = -10/10 = -1

1 means that the image is the same as the object.

The minus sign indica that the image is inverted. If the sign is positive than the image is upright.

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2. A 5-cm-high object is placed in front of a concave mirror with a radius of curvature of 20 cm. Determine the image height if the object distance is 5 cm, 15 cm, 20 cm, 30 cm.

Coñecido:

O radio de curvatura (r) = 20 centímetros

A distancia focal (f) = R/2 = 20/2 = 10 cm

A altura do obxecto (ho) = 5 centímetros

Solución:

a) a distancia focal (f) = 10 cm e a distancia do obxecto (do) = 5 centímetros

Formación da imaxe nun espello cóncavo:

Concave mirror – problems and solutions 2

A distancia da imaxe (di):

1/di = 1/f – 1/do = 1/10 – 1/5 = 1/10 – 2/10 = -1/10

di = -10/1 = -10 centímetros

The minus sign indicates that the image is virtual or the image is behind the mirror.

A ampliación da imaxe (m):

m = -di / do = -(-10)/5 = 10/5 = 2

O signo máis indica que a imaxe está en posición vertical.

A altura da imaxe (hi):

m = hi / ho

hi = ho m = (5 cm)(2) = 10 cm

The image height is 10 cm.

b) A distancia focal (f) = 10 cm e a distancia do obxecto (do) = 15 centímetros

Formación da imaxe nun espello cóncavo:

Concave mirror – problems and solutions 3

A distancia da imaxe (di):

1/di = 1/f – 1/do = 1/10 – 1/15 = 3/30 – 2/30 = 1/30

di = 30/1 = 30 centímetros

The plus sign indicates that the image is real or the image is 30 cm in front of the mirror, on the same side as the object.

A ampliación da imaxe (m):

m = -di / do = -30/15 = -2

O signo menos indica que a imaxe está invertida.

The image is 2 times larger than the object.

A altura da imaxe (hi):

m = hi / ho

hi = ho m = (5 cm)(2) = 10 cm

The image height is 10 cm.

c) The focal length (f) = 10 cm and the object distance (do) = 20 centímetros

Formación da imaxe nun espello cóncavo:

Concave mirror – problems and solutions 4

A distancia da imaxe (di):

1/di = 1/f – 1/do = 1/10 – 1/20 = 2/20 – 1/20 = 1/20

di = 20/1 = 20 centímetros

The positive sign indicates that the image is real or a imaxe is 20 cm in front of the mirror, on the same side as the object.

A ampliación da imaxe (m):

m = -di / do = -20/20 = -1

The negative sign means the image is inverted.

A altura da imaxe (hi):

m = hi / ho

hi = h m = (5 cm)(1) = 5 cm

d) A distancia focal (f) = 10 cm e a distancia do obxecto (do) = 30 centímetros

Concave mirror – problems and solutions 5

A distancia da imaxe (di):

1/di = 1/f – 1/do = 1/10 – 1/30 = 3/30 – 1/30 = 2/30

di = 30/2 = 15 centímetros

The plus sign indicates that the image is real or the image is 15 cm in front of the mirror, on the same side as the object.

A ampliación da imaxe (m) :

m = -di / do = -15/30 = -0.5

O signo menos indica que a imaxe está invertida.

A imaxe é 0.5 máis pequena que o obxecto.

A altura da imaxe (hi):

m = hi / ho

hi = ho m = (5 cm)(0.5) = 2.5 cm

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3. An image an by a concave mirror is 4 times greater than the object. If the radius of curvature 20 cm, determine the object distance in front of the mirror!

Coñecido:

A ampliación da imaxe (m) = 4

O radio de curvatura (r) = 20 centímetros

A distancia focal (f) = r/2 = 20/2 = 10 cm

Querido : A distancia do obxecto (do)

Solución:

m = - di / do

4 = – di / do

- di = 4 do

di = – 4 do

1/f = 1/do + 1/di

1/10 = 1/do + 1/4do

4/40 = 4/4do + 1/4do

4/40 = 5/4do

(4)(4s) = (5)(40)

16 do = 200

do = 12.5 cm

The object distance = 12.5 cm.

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4. A 1-cm high object is placed 10 cm from a concave mirror with the focal length, f = 15 cm. Determine :

A. The image distance ?

B. The image height?

C. The properties of image formed by the concave mirror?

Coñecido:

A altura do obxecto (h) = 1 cm

A distancia ao obxecto (do) = 10 centímetros

A distancia focal do espello cóncavo (f) = 15 cm

Solución:

A. The image distance (di)

1/f = 1/do + 1/di

1/di = 1/f – 1/do = 1/15 – 1/10 = 2/30 – 3/30 = -1/30

di = -30/1 = -30 centímetros

The negative sign indicates that the image is virtual or the image is behind the mirror.

B. The image height (hi)

The magnification of the image (M) :

M = -di/do = hi/ho

M = -(-30)/10 = 30/10 = 3 times

A altura da imaxe (hi):

M = hi / ho

3 = hi / 1 cm

hi = 3 (1 cm)

hi = 3 cm

the image height is 3 cm. The plus sign indicates that the image upward.

C. The properties of the image :

Virtual, upward, larger than object

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5. The magnification of the image, according to the image below.

Coñecido:Concave mirror problems with solutions 1

A distancia ao obxecto (do) = 60 centímetros

A distancia focal (f) = 20 cm

Buscase: A ampliación da imaxe (M)

Solución:

The image distance :

1/f = 1/do + 1/di

1/di = 1/f – 1/do = 1/20 cm – 1/60 cm = 3/60 cm – 1/60 cm = 2/60 cm

di = 60/2 cm = 30 cm

The magnification of the image (M) :

M = di/do = 30 cm / 60 cm = 1/2 times

6. Se o object is placed 6 cm from a concave mirror, the image distance is 12 cm as shown in figure below. Whhat is the image distance if the object is moved from the original position 1 cm away from the mirror.

Coñecido:

A distancia ao obxecto (do) = 6 centímetros

A distancia da imaxe (di) = 12 centímetros

Buscase: if the object distance (do) = 7 cm then the image distance is …

Solución:

1/f = 1/do + 1/di = 1/6 + 1/12 = 2/12 + 1/12 = 3/12

f = 12/3 = 4 cm

The focal length is positive, means that the focal point is real or the rays pass through the point.

The image distance :

1/di = 1/f – 1/do = 1/4 – 1/7 = 7/28 – 4/28 = 3/28

di = 28/3 = 9.3 centímetros

7. A dentist observes and checks the patient’s teeth using a mirror with an 8 cm radius. In order for the hole to be seen clearly by the doctor, what is the distance between the patient’s teeth and the mirror?

A. less than 4 cm in front of a concave mirror

B. less than 4 cm in front of a convex mirror

C. more than 4 cm in front of the concave mirror

D. more than 4 cm in front of the convex mirror

Coñecido:

Radius of mirror (r) = 8 cm

The focal length of mirror (f) = r / 2 = 8 / 2 = 4 cm

Buscase: The distance between the patient’s teeth and the mirror

Solución:

The mirror used is a concave mirror or a convex mirror? In order for the tooth hole to be clearly visible by the doctor, the mirror used should be able to enlarge the image of the tooth and the image must be upright. Convex mirror always produces inverted images and the size of the image is smaller than the size of the object. Conversely a concave mirror can produce an upright image if the object distance (d) is smaller than the focal length (f). If the object distance is greater than the focal length (f) then the concave mirror produces an inverted image.

The focal length (f) of the concave mirror is 4 cm, therefore the patient’s teeth should be less than 4 cm in front of a concave mirror.

A resposta correcta é A.

8. A concave mirror has a radius of curvature of 24 cm. If the object is placed 20 cm in front of the mirror then determine the properties of the image.

A. Real, upright and enlarged

B. Real, inverted and enlarged

C. Virtual, upright and enlarged

D. Virtual, inverted and smaller

Coñecido:

Raio de curvatura (r) = 24 cm

Focal length (f) = R/2 = 24/2 = +12 cm

The focal length of the concave mirror is positive or real because the light passes through the focal point of the mirror.

Object distance (d) = 20 cm

Buscase: Propiedades da imaxe

Solución:

Image is virtual or real? Calculate the image distance (s’):

1/d + 1/d’ = 1/fConcave mirror – problems and solutions 1

1/d’ = 1/f – 1/d

1/d’ = 1/12 – 1/20

1/d’ = 5/60 – 3/60

1/d’ = 2/60

d’ = 60/2

d’ = 30 cm

The image distance signed positive means that the image is real because it is passed by light.

Image enlarged ? Upright or inverted? First calculate the image magnification (M):

M = -d’ / d = -30/20 = -1.5

M > 1 means the image is enlarged, M has a negative sign means an inverted image. So the image properties are real, inverted, enlarged.

A resposta correcta é B.

9. A spherical mirror produces an image has size 5 times greater than the object on a screen, 5 meters away from the object. The mirror is…..

A. concave with the focal length of 25/24 m

B. convex with the focal length of 25/24 m

C. concave with the focal length of 24/25 m

D. convex with the focal length of 24/25 m

Coñecido:

Magnification of image (M) = 5 times

The distance between object and image = 5 meters

Solución:

The size of the image produced by a convex mirror is always smaller than the size of the object, therefore, the mirror is a concave mirror.


Object distance (d) = x

Image distance (d’) = x + 5

Image magnification (M) = 5 times

The formula of image magnification :

Concave mirror – problems and solutions 2

The formula of the focal length (f) :

Concave mirror – problems and solutions 3

A resposta correcta é A.

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