Question

An object is placed vertically at a distance of 20 from a convex lens. If the height of the object is 10 cm and the focal length of the lens is 10 cm. The position size & nature of the image is​

Answer 1

Answer:

Answer :

The real, inverted image of same size is formed at a distance of 20 cm from the lens.

Solution :

Given: Height of the object

, <br> focal length (f) = 10 cm, <br> distance of the object (u) = -20 cm <br> To find: Image distance (v), height of the image

, magnification (M) <br> Formulae: i.

<br> ii. Magnification (M)

<br> Calculation: From formula (i), <br>

<br>

<br>

<br>

<br>

<br> From formula (ii), <br>

<br>

<br>

<br>

<br>

<br> The negative sign of the height of the image and the magnification shows that the image is inverted and real. It is below the principal axis and is of the same size as the object.

Answer 2

Given :

In convex lens,

object distance = - 20 cm

object height = 10 cm

Focal length = 10 cm

To find :

The position , size and 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,

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

$\dashrightarrow \sf \dfrac{1}{v} - \dfrac{1}{ - 20}= \dfrac{1}{10}$

$\dashrightarrow \sf \dfrac{1}{v} + \dfrac{1}{20}= \dfrac{1}{10}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{20}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{2 - 1}{20}$

$\dashrightarrow \sf \dfrac{1}{v} = \dfrac{1}{20}$

$\dashrightarrow \sf v = 20 \: cm$

Thus, the position of the image is 20 cm.

We know that,

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

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

$\dashrightarrow \sf \dfrac{h'}{10} = \dfrac{20}{ - 20}$

$\dashrightarrow \sf \dfrac{h'}{10} = - 1$

$\dashrightarrow \sf h'= - 10 \: cm$

Thus, the size of the image is 10 cm

Nature of the image :

• The minus sign shows that this height is in the downwards direction, that is image is formed below the axis. hence the image is real and inverted.