Question

An object 5 cm in length is held at 10 cm away from converging lens and focal length is 5 cm

Answer 1

Answer:

Given, height of object = 5cm

Position of object, u = - 25cm

Focal length of the lens, f = 10 cm

Hence, position of image, v =?

We know that,

1/v - 1/u = 1/f

1/v + 1/25 = 1/10

So, 1/v = 1/10 - 1/25

S0, 1/v = (5 - 2)/50

That is, 1/v = 3/50

So, v= 50/3 = 16.66 cm

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

Now, magnification = v/u

That is, m = 16.66/-25 = -0.66

Also, m= height of image/height of object

Or, -0.66 = height of image / 5 cm

Therefore, height of image = -3.3 cm

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

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

Size of image = - 3.3 cm at the opposite side of lens

Nature of image – Real and inverted....

show how would you connect the three resistors, each of resistance 6 so that the combination has a resistance of 9 , 4

Answer 2

Appropriate Question :

An object of 5cm in length is held 10 cm away from converging ( Convex ) lens of focal length 5 cm . Find the position, size and the nature of the image formed .

Solution:

$\sf{}Height_{(object)} = 5\: cm \\ \\ \sf{}distance_{(object)} = ( - 10) \: cm \\ \\ \sf{}focal \: length \: = 5 \: cm$

Now , By using Lens Formula which is :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf { \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }}$

Here , Object Distance is " u " and Image Distance is " v "

By substituting values;

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \sf { \dfrac{1}{5} = \dfrac{1}{v} - \dfrac{1}{ ( - 10)} }} \\ \\ { \sf { \dfrac{1}{5} = \dfrac{1}{v} + \dfrac{1}{ 10} }} \\ \\ \sf{} \frac{1}{v} = ( - \frac{1}{ 5} ) + \frac{1}{10} \\ \\ \sf{} \frac{1}{v} = \frac{ - 2 + 3}{10} \\ \\ \sf{} \frac{1}{v} = \frac{ - 1}{10} \\ \\ \boxed {\sf {v = - 10 \: cm}}$

Therefore, Distance of image from the pole is equals to -10 cm .

Now to Find the nature of the image formed we have to calculate magnification. Which is calculated by ;

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rightarrow\boxed { \sf{m = \frac{v}{u} }}$

By putting Image Distance and Object Distance ;

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \sf{m = \dfrac{- 10}{ - 10} }} \\ \\ { \sf{ m = \cancel{\dfrac{- 10}{ - 10} }}} \\ \\ \boxed {\sf{ m = -1}}$

From th value of m = 1 , we can conclude that the image formed is Equal in size as it is of object which means Size of Object is equals to size of Image .

Now , from that negative sign of image we can conclude that the image formed is Erect and Virtual .

Since the object is placed at the distance twice of the Focal length, which is Centre of Curvature we can say that image is also formed at Centre of Curvature.

Now , height of the image can be calculated by :

$\boxed{\sf{m = \dfrac{Height_{(image)}}{Height_{(object)}}}}$

By substututing the values we get,

${\sf{-1 = \dfrac{Height_{(image)}}{5}}}$

${\sf{Height_{(image)}= 5 × ( - 1 ) }}$

${\sf{Height_{(image)}= -5\: cm }}$

Therefore, height of the image formed is equals to -5 cm.