Calculate the image distance for an object of height 12 mm at a distance of 0.20 m from a concave lens of focal length 0.30 m, and state the nature and size of the image.
Solution
Focal length of concave lens f = - 0.30 m
Object distance u = - 0.20 m
Height of object h1 = 12 mm = 0.012 m
Height of image h2 =?
Using the lens formula:
`1/f=1/v-1/u`
`1/-0.30=1/v-1/((-0.20)`
`1/v=1/-0.30-1/0.20`
`1/v=(-2-3)/0.60=-1/0.12`
∴ v = - 0.12 m = - 12 cm.
The negative sign indicates that the image is formed on the left side of the lens. Therefore, the image is virtual.
Magnification is given as:
Magnification =`"image distance"/"object distance"="hight of image"/"hight of object"`
`v/u=h_2/h_1`
`(-0.12)/-0.20=h_2/0.012`
`h_2=(0.12xx0.012)/0.20=0.0072m=+7.2mm`
Here, the positive sign shows that the image is erect. Also, the size of the image is smaller than the size of the object and is diminished.
Therefore, the image formed by the concave lens is virtual, diminished and erect.
