To solve this problem, we can use the mirror equation:
1/f = 1/d_i + 1/d_o
Where:
f = focal length of the mirror
d_i = distance of the image from the mirror
d_o = distance of the object from the mirror
First, let's find the distance of the image from the mirror (d_i). We are given that the focal length (f) is 20 cm and the height of the image is 2.4 cm. Since the mirror is diverging (concave), the image formed is virtual and magnified:
h_i/h_o = -d_i/d_o
Where:
h_i = height of the image
h_o = height of the object (given)
d_i = distance of the image from the mirror (to be found)
d_o = distance of the object from the mirror (given)
Plugging in the values:
2.4 cm / h_o = -d_i / d_o
Since the magnification, m = h_i/h_o, is negative (indicating an inverted image), we can rewrite the equation as:
-2.4 cm / h_o = -d_i / d_o
Now, let's solve for d_i:
-2.4 cm / 2.4 cm = -d_i / d_o
-1 = -d_i / d_o
d_i = d_o
Now, let's use the mirror equation to find the distance of the object from the mirror (d_o), using the given focal length (f = 20 cm):
1/f = 1/d_i + 1/d_o
1/20 cm = 1/d_i + 1/d_o
Since we found that d_i = d_o, we can rewrite the equation as:
1/20 cm = 1/d_i + 1/d_i
1/20 cm = 2/d_i
d_i = 20 cm / 2
d_i = 10 cm
Therefore, the distance of the object from the front of the mirror is 10 cm.
A diverging mirror forms an image of height 2.4cm if the focal length of the mirror is 20cm,find the height and distance of the object from the front of the mirror
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