An object of height 6 cm is placed at a distance of 24 cm from a concave mirror of focal length 30 cm find position height and nature of image
Answers
Answer:
Focal length(f) = 10 cm.
Object distance(u) = 15 cm.
Height of the Object = 6 cm.
Using the Mirror's formula,
1/f = 1/v + 1/u
⇒ 1/-10 = 1/v + 1/-15
⇒ 1/v = 1/-10 + 1/15
∴ 1/v = (-3 + 2)/30
⇒ v = -30 cm.
Since, image distance is negative, therefore, it is real and inverted.
Now, magnification = -v/u
= -(-30)/15
= 2
Since, magnification in greater than 1, therefore Image is magnified.
Also, m = H of image/H of object.
⇒ 2 = H of image/6
∴ Height of image = 12 cm.
Characteristics of Image are ⇒
Real, Inverted and Magnified.
Explanation:
Answer:
The image's position, height, and nature are 120cm, 30cm, and virtual, respectively.
Explanation:
From the mirror formula we have,
(1)
Where,
f=focal length of the mirror
v=image distance from the mirror
u=object distance from the mirror
From the question we have,
Height of the object(h₁)=6cm
Object distance from the mirror(u)=-24cm
The focal length of the concave mirror(f)=-30cm
By placing the required values in equation (1) we get;
(2)
The magnification of the mirror is calculated as,
(3)
By putting the values of "v" and "u" in equation (3) we get;
(4)
Also, the magnification in terms of height is given as,
(5)
h₂=height of the image
h₁=height of the object
By equating equations (4) and (5) we get;
(6)
By placing the value of "h₁" in equation (6) we get;
(7)
Hence, the position, height, and nature of the image are 120cm, 30cm, and virtual respectively.
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