Question

a 3 m hight object is placed 24 cm away from the convex lens of focal length 8 cm find tha position ,hight and nature of the image. If the object ​

Answer 1

Question :

A 3 cm high object is placed 24 cm away from the convex lens of focal length 8 cm, find the position , height and nature of the image.

Given :

• Height of the object = 3 cm

• Object Distance = 24 cm

• Focal length = 8 cm

To find :

• Image distance / position of the Image

• Height of the image

• Nature of the image

Solution :

To find the Image Distance :

We know the lens formula i.e,

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

Where :

• f = Focal length
• v = Image distance
• u = Object Distance

By sign-convention , we get the values as :

• f = + 8 cm
• u = - 24 cm

Using the lens formula and substituting the values in it, we get :

$:\implies \bf{\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}}$

$:\implies \bf{\dfrac{1}{8} = \dfrac{1}{v} - \dfrac{1}{(-24)}}$

$:\implies \bf{\dfrac{1}{8} + \dfrac{1}{(-24)} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{1}{8} - \dfrac{1}{24} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{3 - 1}{24} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{2}{24} = \dfrac{1}{v}}$

$:\implies \bf{\dfrac{1}{12} = \dfrac{1}{v}}$

$:\implies \bf{v = 12}$

$\boxed{\therefore \bf{v = 12}}$

Hence the image distance of the convex lens is 12 cm.

Nature of the image :

To find the nature of the image, first we have to find magnification of the image .

We know the formula for magnification i.e,

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

Where :

• m = Magnification
• v = Image distance
• u = Object Distance

By sign-convention , we get :

• v = + 12 cm
• u = (-24) cm

Now using the above formula and substituting the values in it, we get :

$:\implies\bf{m = - \dfrac{12}{(-24)}}$

$:\implies\bf{m = - \dfrac{1}{(-2)}}$

$:\implies\bf{m = \dfrac{1}{2}}$

$:\implies\bf{m = 0.5}$

$\boxed{\therefore \bf{m = 0.5}}$

Hence the Magnification is 0.5.

Since the Magnification is 0.5 , the image is real , erect and high.

Height of the image :

We know that magnification is :

$\boxed{\bf{m = \dfrac{h_{i}}{h_{o}}}}$

Where :

• m = Magnification
• hi = Height of image
• ho = Height of object

Now using the above formula and substituting the values in it, we get :

$:\implies\bf{0.5 = \dfrac{h_{i}}{3}}$

$:\implies\bf{0.5 \times 3 = h_{i}}$

$:\implies\bf{1.5 = h_{i}}$

$:\implies\bf{h_{i} = 1.5}$

$\boxed{\therefore \bf{h_{i} = 1.5\:cm}}$

Hence the height of image is 1.5 cm.