Question

An object 5cm in length is held 30 cm away from a converging lens of

focal length 10cm. Draw the ray diagram and find the position, size and nature

of the image formed​

Answer 1

Given :

In converging lens,

Height of the object : 5 cm

Object distance : 30 cm

Focal length : 10 cm

To find :

The position, size and the nature of the image.

Solution :

Using lens formula that is,

» The formula which gives the relationship between image distance, object distance and focal length of a lens is known as the lens formula.

The lens formula can be written as :

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

where,

• v denotes image distance
• u denotes object distance
• f denotes focal length

by substituting all the given values in the formula,

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

$\dashrightarrow \sf \dfrac{1}{v} - \dfrac{1}{ - 30}= \dfrac{1}{10}$

$\dashrightarrow \sf \dfrac{1}{v} + \dfrac{1}{30}= \dfrac{1}{10}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{30}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{3 - 1}{30}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{2}{30}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{1}{5}$

$\dashrightarrow \sf v = 15$

Thus, the position of the image is 15 cm.

We know that,

» The ratio of image distance to the object distance is equal to the the ratio of image height to the object height

$\dashrightarrow \sf \dfrac{h'}{h} = \dfrac{v}{u}$

$\dashrightarrow \sf \dfrac{h'}{5} = \dfrac{15}{-30}$

$\dashrightarrow \sf \dfrac{h'}{5} = -\dfrac{1}{2}$

$\dashrightarrow \sf h' = - \dfrac{5}{2}$

$\dashrightarrow \sf h' = - 2.5 cm$

Thus, the size of the image is 2.5 cm

Nature of the image :

• The image is erect and virtual.
• The image is diminished.

Answer 2

Hola ⚘⚘

Answer:

• position of the image = 15cm
• size of the image = 2.5cm
• nature of the image is erect , virtual and diminished.

Explanation:

Given :

In converging lens,

• Height of the object = 5 cm
• distance of the object = 30 cm
• Focal length = 10 cm

Tofind :

• position of the image
• size of the image
• nature of the image

Calculation:

• 1/v - 1/u = 1/f

>> 1/v - 1/-30 = 1/10

>> 1/v + 1/30 = 1/10

>> 1/v = 1/10 - 1/30

>> 1/v = 3-1/30

>> 1/v = 2/3

>> 1/v = 1/5

>> v = 15

• h'/h = v/u

>> h'/5 = 15/-30

>> h'/h = -½

>> h' = - ⁵/₂

>> h' = -2.5cm

Hence

• position of the image = 15cm
• size of the image = 2.5cm
• nature of the image is erect , virtual and diminished.