Vioo vyenye mkunjo - matatizo na suluhisho

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

Inajulikana:

Urefu wa fokasi (f) = sentimita 5

Umbali wa kitu (d_o) = sm 10

Suluhisho :

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 sentimita

Umbali wa picha ni 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 inaonyesha 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.

Inajulikana:

The radius of curvature (r) = sm 20

Urefu wa fokasi (f) = R/2 = 20/2 = 10 cm

Urefu wa kitu (h_o) = sm 5

Suluhisho:

a) urefu wa fokasi (f) = sentimita 10 and the object distance (d_o) = sm 5

Formation of image by concave mirror :

Umbali wa picha (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 sentimita

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

Ukuzaji wa picha (m):

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

Ishara ya kuongeza inaonyesha kwamba picha imesimama wima.

Urefu wa picha (h_i):

m = h_i / h_o

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

The image height is 10 sentimita.

b) Urefu wa fokasi (f) = sentimita 10 and the object distance (d_o) = sm 15

Formation of image by concave mirror :

Umbali wa picha (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 sentimita

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.

Ukuzaji wa picha (m):

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

Ishara ya minus inaonyesha kwamba picha imegeuzwa.

The image is 2 times larger than the object.

Urefu wa picha (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) = sm 20

Formation of image by concave mirror :

Umbali wa picha (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 sentimita

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

Ukuzaji wa picha (m):

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

The negative sign means the image is inverted.

Urefu wa picha (hi):

m =h_i / h_o

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

d) Urefu wa fokasi (f) = sentimita 10 and the object distance (d_o) = sm 30

Umbali wa picha (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 sentimita

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.

Ukuzaji wa picha (m) :

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

Ishara ya minus inaonyesha kwamba picha imegeuzwa.

The image is 0.5 smaller than the object.

Urefu wa picha (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!

Inajulikana:

Ukuzaji wa picha (m= 4

The radius of curvature (r) = sm 20

Urefu wa fokasi (f) = r/2 = 20/2 = 10 cm

Alitaka : Umbali wa kitu (d_o)

Suluhisho:

m = - d_i / d_o

4 = – d_{i /} d_o

- d_i = 4 d_o

d_i = – 4 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)(sekunde 4) = (5)(40)

16 d_o = 200

d_o = 12.5cm

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?

Inajulikana:

Urefu wa kitu (h) = 1 cm

Umbali wa kitu (d)_o) = sm 10

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

Suluhisho:

A. The image distance (d_i)

1/f = 1/d_o + 1/d_i

1/siku_i = 1/f – 1/siku_o = 1/15 – 1/10 = 2/30 – 3/30 = -1/30

d_i = -30/1 = -30 sentimita

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

Urefu wa picha (h_i):

M = h_i / h_o

3 = h_i / 1cm

h_i = 3 (1 cm)

h_i = 3cm

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.

Inajulikana:

Umbali wa kitu (d)_o) = sm 60

Urefu wa fokasi (f) = 20 cm

Inayohitajika: The image magnification (M)

Suluhisho:

The image distance :

1/f = 1/d_o + 1/d_i

1/siku_i = 1/f – 1/siku_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. Kama 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.

Inajulikana:

Umbali wa kitu (d)_o) = sm 6

Umbali wa picha (d)_i) = sm 12

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

Suluhisho:

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/siku_i = 1/f – 1/siku_o = 1/4 – 1/7 = 7/28 – 4/28 = 3/28

d_i = 28/3 = 9.3 sentimita

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

Inajulikana:

Radius of mirror (r) = 8 cm

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

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

Suluhisho:

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.

Jibu sahihi ni 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

Inajulikana:

Radi ya curvature (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

Inayohitajika: Sifa za picha

Suluhisho:

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.

Jibu sahihi ni 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

Inajulikana:

Magnification of image (M) = 5 times

The distance between object and image = 5 meters

Suluhisho:

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) :

Jibu sahihi ni A.

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