Answer:
image is real, Inverted and magnified
Focal length of the mirror is 12.7 cm
Explanation:
As we know that object is placed at distance 20 cm in front of mirror and its image formed at 35 cm in front of mirror
So here since object and image both are on same side of mirror so this must be REAL image
now we know that magnification is given as
[tex]M = -\frac{d_i}{d_o}[/tex]
[tex]M = - \frac{35}{20}[/tex]
[tex]M = -1.75 [/tex]
so image is real, Inverted and magnified
now for focal length of the mirror we can use the equation
[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}[/tex]
[tex]\frac{1}{20} + \frac{1}{35} = \frac{1}{f}[/tex]
[tex]f = 12.7 cm[/tex]
The concave mirror that has an object placed 20cm in front of it. and an image is formed at 35 cm in front of the mirror.
The nature of the image formed is
(i) It is Formed beyond C
(ii) It is enlarged
(iii) It is magnified
The magnification is 1.75
The Focal length of the mirror is 12.73 cm
From the question, we need to calculate the focal length of the mirror using mirror formula.
1/f = (1/v)+(1/u)............ Eqaution 1
Where f = focal length, v = object distance, u = image distance.
From the question,
Given: v = 20 cm, u = 35 cm
Substitute these values into equation 1
1/f = (1/20)+(1/35)
1/f = (20+35)/(20×35)
f = (20×35)/(20+35)
f = 700/55
f = 12.73 cm.
Hence the mirror focal length is 12.73 cm
Since the mirror is placed between F and C in a concave mirror
The nature of the image formed is
(i) It is Formed beyond C
(ii) It is enlarged
(iii) It is magnified
Finally,
magnification = image distance/object distance = D/D'
mag = 35/20
mag = 1.75.
Hence the magnification of the image is 1.75
Learn more about concave mirrors here: https://brainly.com/question/21953487
