Question

Find the position and nature of the image of an object 5 cm high and 10 cm in front of a convex lens of focal length 6 cm

Answer 1

Answer:

The image is formed at a distance of 15 cm on the other side of the lens and the image is real and inverted.

Explanation:

Given that,

Distance of the object u = -5 cm

Focal length f = 6 cm

Height of the object h= 5 cm

Using lens's formula

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

$\dfrac{1}{6}=\dfrac{1}{v}-\dfrac{1}{-10}$

$\dfrac{1}{v}=\dfrac{1}{15}$

$v = 15\ cm$

The image is formed at a distance of 15 cm on the other side of the lens.

The magnification of the lens is

$m =\dfrac{v}{-u}$

$m = \dfrac{15}{-10}$

$m= -1.5$

The height of the image is

$m = \dfrac{h'}{h}=\dfrac{v}{u}$

$\dfrac{h'}{5}=\dfrac{15}{-10}$

$h'=-7.5 cm$

The height of the image is -7.5 cm.

Hence, The image is formed at a distance of 15 cm on the other side of the lens and the image is real and inverted.

Answer 2

Answer:

Explanation:

h1 = 5 cm

u= -10 cm

f = 6 cm

1/v - 1/u = 1/f

1/v – 1/-10 =1 /6

1/v = 1/6 – 1/10 = 2/30 = 1/15

V = 15 cm

Image is formed 15 cm behind the convex lens and it is real and inverted