An object of 5 cm height is placed at a distance of 15 cm from a concave mirror.Find the position, height and nature of its image.The focal length of the mirror is 10 cm
Answers
Given :
▪ Height of object = 5cm
▪ Distance of object = 15cm
▪ Focal length of mirror = 10cm
▪ Type of mirror : concave
To Find :
▪ Distance of image.
▪ Height of image.
▪ Nature of image.
Formula :
✏ Mirror formula :
☆ 1/u + 1/v = 1/f
✏ Lateral magnification :
☆ m = -v / u = h' / h
Calculation :
- u = -15 cm
- f = -10 cm
- h = 5 cm
☢ Distance of image (v) :
→ 1/ u + 1/v = 1/f
→ 1/(-15) + 1/v = 1/(-10)
→ 1/v = -1/10 + 1/15
→ 1/v = (-3+2)/30
→ v = -30 cm
☢ Height of image (h') :
→ -v / u = h' / h
→ -(-30)/(-15) = h' / 5
→ -2 = h' / 5
→ h' = -10 cm
☢ Nature of image :
- Real
- Inveted
- Enlarged
In the above Question , the following information is given -
An object of 5 cm height is placed at a distance of 15 cm from a concave mirror.
The focal length of the mirror is 10 cm .
Here , we have to calculate the position ,height and nature of the image formed .
Solution -
We know that according to the mirror Formulae ,
( 1 / u ) + ( 1 / v ) = ( 1 / f )
Where ,
u is the object distance from the pole .
v is the image distance
f is the focal length .
Now ,
Here , the given mirror is concave .
So ,
u = -15 cm
f = -10 cm
( 1 / -15 ) + ( 1 / v ) = - ( 1 / 10 )
=> ( 1 /v ) = ( 1 / 15 ) - ( 1 / 10 )
=> ( 1 / v ) = ( 2-3 ) / ( 30 )
=> ( 1 / v ) = ( -1 / 30 )
=> v = -30 cm .
Hence , the position of the image is 30 cm behind the mirror .
Now ,
Magnification = ( - v / u )
=> ( h / h 0 ) = ( - v / u ) = -2
Now ,
In the question , h 0 = 5 cm
=> h = - 10 cm .
This means that the image is inverted any has a height of 10 cm .