Specula concava – problemata et solutiones

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

Notum:

Longitudo focalis (f) = 5 cm

Distantia obiecti (d_o) = 10 cm

Solutio :

Formation of image by concave mirror :

The image distance :

1/d_i = 1/f – 1/d_o = 1/5 – 1/10 = 2/10 – 1/10 = 1/10

d_i = 10/1 = 10 cm

Distantia imaginis est 10 cm.

The magnification :

m = –d_i / d_o = -10/10 = -1

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

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

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.

Notum:

The radius of curvature (r) = 20 cm

Longitudo focalis (f) = R/2 = 20/2 = 10 cm

Altitudo obiecti (h_o) = 5 cm

solution:

a) longitudo focalis (f) = 10 cm and the object distance (d_o) = 5 cm

Formation of image by concave mirror :

Distantia imaginis (d_i):

1/d_i = 1/f – 1/d_o = 1/10 – 1/5 = 1/10 – 2/10 = -1/10

d_i = -10/1 = -10 cm

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

Amplificatio imaginis (m):

m = -d_i / d_o = -(-10)/5 = 10/5 = 2

Signum plus indicat imaginem esse rectam.

Altitudo imaginis (h_i):

m = h_i / h_o

h_i = h_o m = (5 cm)(2) = 10 cm

The image height is 10 cm.

b) Longitudo focalis (f) = 10 cm and the object distance (d_o) = 15 cm

Formation of image by concave mirror :

Distantia imaginis (d_i):

1/d_i = 1/f – 1/d_o = 1/10 – 1/15 = 3/30 – 2/30 = 1/30

d_i = 30/1 = 30 cm

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.

Amplificatio imaginis (m):

m = -d_i / d_o = -30/15 = -2

Signum minus indicat imaginem inversam esse.

The image is 2 times larger than the object.

Altitudo imaginis (h_i):

m = h_i / h_o

h_i = h_o m = (5 cm)(2) = 10 cm

The image height is 10 cm.

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

Formation of image by concave mirror :

Distantia imaginis (d_i):

1/d_i = 1/f – 1/d_o = 1/10 – 1/20 = 2/20 – 1/20 = 1/20

d_i = 20/1 = 20 cm

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

Amplificatio imaginis (m):

m = -d_i / d_o = -20/20 = -1

The negative sign means the image is inverted.

Altitudo imaginis (hi):

m = h_i / h_o

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

d) Longitudo focalis (f) = 10 cm and the object distance (d_o) = 30 cm

Distantia imaginis (d_i):

1/d_i = 1/f – 1/d_o = 1/10 – 1/30 = 3/30 – 1/30 = 2/30

d_i = 30/2 = 15 cm

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.

Amplificatio imaginis (m) :

m = -d_i / d_o = -15/30 = -0.5

Signum minus indicat imaginem inversam esse.

The image is 0.5 smaller than the object.

Altitudo imaginis (h_i):

m = h_i / h_o

h_i = h_o m = (5 cm)(0.5) = 2.5 cm

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!

Notum:

Amplificatio imaginis (m) II =

The radius of curvature (r) = 20 cm

Longitudo focalis (f) = r/2 = 20/2 = 10 cm

voluit : Distantia obiecti (d_o)

solution:

m = - d_i / d_o

4 = - d_{i /} d_o

- d_i = 4 d_o

d_i = - I d_o

1/f = 1/d_o + 1/d_i

1/10 = 1/d_o +1/4d_o

4 / = 40 4 / 4d_o +1/4d_o

4 / = 40 5 / 4d_o

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

16 d_o = 200

d_o = X cm

The object distance = 12.5 cm.

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?

Notum:

Altitudo obiecti (h) = 1 cm

Distantia obiecti (d)_o) = 10 cm

The focal length of the concave mirror (f) = 15 cm

solution:

A. The image distance (d_i)

1/f = 1/d_o + 1/d_i

VIII / d_i = 1/f – 1/d_o = 1/15 – 1/10 = 2/30 – 3/30 = -1/30

d_i = -30/1 = -30 cm

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

B. The image height (h_i)

The magnification of the image (M) :

M = -d_i/d_o = h_i/h_o

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

Altitudo imaginis (h_i):

M = h_i / h_o

3 = h_i / 1cm

h_i = 3 (1 cm)

h_i = X 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

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

Notum:

Distantia obiecti (d)_o) = 60 cm

Longitudo focalis (f) = 20 cm

Quaesitum: The image magnification (M)

solution:

The image distance :

1/f = 1/d_o + 1/d_i

VIII / d_i = 1/f – 1/d_o = 1/20 cm – 1/60 cm = 3/60 cm – 1/60 cm = 2/60 cm

d_i = 60/2 cm = 30 cm

The magnification of the image (M) :

M = d_i/d_o = 30 cm / 60 cm = 1/2 times

6. Si 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.

Notum:

Distantia obiecti (d)_o) = 6 cm

Distantia imaginis (d_i) = 12 cm

Quaesitum: if the object distance (d_o) = 7 cm then the image distance is …

solution:

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 :

VIII / d_i = 1/f – 1/d_o = 1/4 – 1/7 = 7/28 – 4/28 = 3/28

d_i = 28/3 = 9.3 cm

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

Notum:

Radius of mirror (r) = 8 cm

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

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

solution:

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.

Responsum rectum est 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

Notum:

Radius of 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

Quaesitum: Proprietates imaginis

solution:

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

1/d + 1/d’ = 1/f

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.

Responsum rectum est 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

Notum:

Magnification of image (M) = 5 times

The distance between object and image = 5 meters

solution:

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 :

The formula of the focal length (f) :

Responsum rectum est A.

[wpdm_package id='858′]

1. Problemata et solutiones speculi concavi
2. Problemata et solutiones speculi convexi
3. Problemata et solutiones lentium divergentium
4. Problemata et solutiones lentium convergentium
5. Instrumenta optica, problemata et solutiones oculi humani.
6. Problemata et solutiones lentium contactuum instrumentorum opticorum
7. Specilla instrumentorum opticorum
8. Problemata et solutiones vitrorum magnificantium instrumentorum opticorum
9. Microscopium instrumentorum opticum – problemata et solutiones
10. Problemata et solutiones telescoporum instrumentorum opticorum