Question

Object 5 cm in length is held 25cm away from the converging lens of focal length 10 centimetre a diagram and find the position size nature of the image formed.
And also please suggest me some tips how to draw ray diagram

Answer 1

Answer:  

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are  

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

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

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

$\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}$

Thus,

$\frac{h^{\prime}}{h}=\frac{v}{u}$

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}$

$\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}$

$\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}$

$v=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}$

Then using the magnification relation, we can get the image height as follows

$\frac{h^{\prime}}{5}=-\frac{16.66}{25}$

So, the image height will be

$h^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}$

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

Answer 2

Answer:

Postition of image = At a distance 16.66 cm away from lens on right side.

Size of image = 3.3 cm on right side of lens

Nature of image = inverted and real.

Explanation:

Given,

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm.

To find:   The position(v), size and the nature of the image formed.

Solution:

From sign convention,

Distance of image(u) = - 25 cm

Height of object (H_o) = + 5cm

Focal length of the lens ( f) = + 10 cm

Applying Len's formula,

1/v - 1/u = 1/f

1/v - 1/(-25) = 1/10

1/v = 1/10 - 1/25

1/v = (5 - 2)/50

1/v = 3/50

or, v= 50/3 = 16.66 cm

From Magnification formula,

m = - v / u = -16.66/25 = -0.66

Also, magnification can be given as:

m = height of image / height of object

or, m = H_i / H_o

or, -0.66 = H_i / 5

or, H_i = -0.66 * 5 = -3.3 cm

Since, Height of image is negative we can say that image formed will be inverted and real. Ray diagram is attached.

Therefore, Postition of image = At a distance 16.66 cm away from lens on right side.

Size of image = 3.3 cm on right side of lens

Nature of image = inverted and real.