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.

Answer 1

Step 1: Given

Object distance = u = -0.2m = -20cm
Focal length = f = -0.3m = -30cm
Image distance = v

Image height = h'

Object height(h) = 12 mm

Step 2: Formula used
$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$

m = $\frac{v}{u}$

Step 3: Finding the image distance
$\frac{1}{v}$ - $\frac{1}{-20}$ = $\frac{1}{-30}$

$\frac{1}{v}$ = $\frac{1}{-20}$ + $\frac{1}{-30}$

$\frac{1}{v}$ = $\frac{-3-2}{60}$

v = $\frac{-60}{5}$ = -12 cm

Step 4: Finding the size of image

m = $\frac{v}{u}$

m = $\frac{-12}{-20}$ = $\frac{3}{5}$

m = $\frac{h^{\prime }}{h}$

$\frac{3}{5}$ = $\frac{h^{\prime }}{12mm}$

h' = $\frac{12\times 3}{5}$

h' = 7.2 mm

Thus, the image is erect and virtual.

Hence, the image distance is 12 cm and the size of the image is 7.2 mm.