Question

3. An object 5cm in length is held
25cm away from a converging lens
of focal length 10cm. Determine
the nature position and size
the image.​

Answer 1

Answer:

Hope you got , enjoy dude .

Answer 2

Given :

In a convex lens,

Object height = 5cm

Object distance = - 25cm

focal length = 10cm

To find :

The nature, position and size of the image.

Solution :

Position of the image :

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,

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

$\dashrightarrow \sf \dfrac{1}{v} - \dfrac{1}{ (- 25)}= \dfrac{1}{10}$

$\dashrightarrow \sf \dfrac{1}{v} + \dfrac{1}{ 25}= \dfrac{1}{10}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{25}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{5 - 2}{50}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{3}{50}$

$\dashrightarrow\sf v= \dfrac{50}{3}$

$\dashrightarrow\sf v = 16.6 \: cm$

thus, the position of image is 16.6 cm.

Magnification of image :

The Magnification produced by a lens is equal to the ratio of image distance to the object distance .i.e.,

$\leadsto\bf m = \dfrac{v}{u }$

$\leadsto\sf m = \dfrac{16.6}{ - 25}$

$\leadsto\sf m = 0.6$

we know that,

$\leadsto\bf m = \dfrac{h'}{h}$

where,

• h' denotes height of the image
• h denotes the height of the object

by substituting the value,

$\leadsto\sf 0.6 = \dfrac{h'}{5}$

$\leadsto\sf h' = 0.6 \times 5$

$\leadsto\sf h' = 3 \: cm$

thus, the size of the image is 3cm

Nature of the image :

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