If an object of height 4cm is placed at distance of 12cm from the concave mirror having focal length 24cm , find the position, nature and the height of the image.
Answers
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Answer--->
We have given,
Height of object (O) = 4cm
Object distance (u) = -12cm ( u is always negative)
Focal length (f) = -24cm ( focal length for concave mirror is negative)
image distance(v) =?
Nature of image =?
Height of image (I) =?
TO FIND (v), (I) AND NATURE OF IMAGE,
We know,
Mirror's formula=> 1/f= 1/v+ 1/u
Putting the values in this formula, We get:---->
1/-24 = 1/v + 1/ -12
-1/24 + 1/12 = 1/ v
1/v = 1/ 24
v = + 24 cm
Hence, image is formed at a distance of 24 cm from the concave mirror.
Positive sign ( + ) before 24 cm indicates that image formed is VIRTUAL and ERECT.
Now, to find height of image ( I ) ----->
We know,
I / O = - v / u
Putting values in the formula ,
I / 4 = -24 / - 12
I / 4 = 2
I = 2 ✖ 4 => 8cm
Hence, height of image ( I ) = 8 cm ,
Image distance ( v ) = + 24 cm
And nature of image is VIRTUAL, ERECT and MAGNIFIED .
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Answer:
If an object of height 4cm is placed at distance of 12cm from the concave mirror having focal length 24cm , find the position, nature and the height of the image.
Explanation:
Given:-
Object height, ho = 4cm
Object distance, u = -12cm
Focal length, f =-24cm
To Find:-
Position , nature & height of image.
Solution:-
According to the Question
Firstly,
We calculate the image distance of the object.
By using mirror formula,
1/v + 1/u = 1/f
where,
v is the Image Position
u is the object distance
f is the focal length
Substitute the value we get,
⇒1/-24 = 1/v + 1/ -12
⇒-1/24 + 1/12 = 1/ v
⇒1/v = 1/ 24
⇒v = + 24 cm
Hence, image is formed at a distance of 24 cm from the concave mirror.
Now, calculating the magnification of the image.
In the formula,
⇒ I / O = - v / u
Substitute the value we get,
By putting values in the formula.
then the values are,
⇒ I / 4 = -24 / - 12
⇒ I / 4 = 2
⇒ I = 2 x 4
⇒ 8cm.
Hence, height of image ( I ) = 8 cm ,
Image distance ( v ) = + 24 cm.
The position, nature and the height of the image is virtual, erect and magnified.
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