Question

$\tt\huge\bold\red{Qu}\pink{e}\green{s} \blue{ti}\orange{o}\red{n}$

~~An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.​

Answer 1

$\bf\pink{♤Answer}$

GIVEN THAT:

The height of object = 5cm

Position of object, u = – 25cm

The focal length of the lens, f = 10 cm

TO FIND:

The position of the image, v =?

Size of the image

Nature of the image

FORMULA:

We know that

1/v – 1/u = 1/f

Substituting the known values in the above equation we get,

1/v + 1/25 = 1/10

=> 1/v = 1/10 – 1/25

=> 1/v = (5 – 2)/50

Hence, 1/v = 3/50

So, v= 50/3 = 16.66 cm

Therefore, the distance of the image is 16.66 cm on the opposite side of the lens.

Now, we know that

Magnification = v/u

Hence, m = 16.66/-25 = -0.66

Also, we know that

m= height of image/height of the object

Or, -0.66 = height of image / 5 cm

Hence, height of image = -3.3 cm

The negative sign of the height of the image depicts that an inverted image is formed.

So, the position of image = At 16.66 cm on the opposite side of the lens

So, the position of image = At 16.66 cm on the opposite side of the lensSo, the position of image = At 16.66 cm on the opposite side of the lensSize of image = – 3.3 cm at the opposite side of the lens

So, the position of image = At 16.66 cm on the opposite side of the lensSo, the position of image = At 16.66 cm on the opposite side of the lensSize of image = – 3.3 cm at the opposite side of the lensSo, the position of image = At 16.66 cm on the opposite side of the lensSize of image = – 3.3 cm at the opposite side of the lensNature of image – Real and inverted

Answer 2

Given :-

• Length of the object = 5cm
• Position of the object = - 25 cm
• Focal length = 10cm

To Find :-

• Position of the image
• Size of the image
• Nature of the image

Formula used :-

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

Substituting the values we get :-

$= > \frac{1}{v} - ( - \frac{1}{ 25} ) = \frac{1}{10} \\ = > \frac{1}{v} + \frac{1}{25} = \frac{1}{10} \: \: \: \: \: \: \: \\ = > \frac{1}{v} = \frac{1}{10} - \frac{1}{25} \: \: \: \: \: \: \: \\ = > \frac{1}{v} = \frac{5 - 2}{50} = \frac{3}{50} \\ = > v = \frac{50}{3} = 16.66cm$

We know that :-

Magnification = v/u

$= > Magnification = \frac{16.66}{ - 25} = - 0.66$

Also, we know that :-

• $= > m = \frac{height \: of \: the \: image}{height \: of \: object}$

$= > - 0.66 = \frac{x}{5} \: \: \: \: \: \: \\ = > x = 5 \times - 0.66 \\ = > x = - 3.3cm \: \: \:$

Here, the negative sign indicates that the image is inverted.

Therefore :-

• Position of the image = 16.66 cm on the opposite side of the lens.
• Size of the image = - 3.3 cm
• Nature of the image = Real and inverted.

NOTE :- For the ray diagram kindly refer the attachment. :)

Hope the answer helps you mate :)