# 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. - Science

#### Question

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.

Is there an error in this question or solution?

#### APPEARS IN

Lakhmir Singh Solution for Physics for Class 10 (2018 (Latest))
Chapter 5: Refraction of Light
| Q 8 | Page 256