Question

A convex lens forms an image 16.0 cm long of an object 4.0 cm long kept at a distance of 6 cm from the lens. The object and the image are on the same side of the lens.
(a) What is nature of the image?
(b) Find (i) the position of the image, and (ii) the focal length of the lens.

Answer 1

Explanation:

Given:-

Image height (I) = 16.0 cm (positive)

Object height(0) = 4.0 cm (positive)

Object distance (u) = 6 cm (negative)

(i) From relation:

⟹ m = I/O = v/u

Where:-

m = magnification

I = image height

O = object height

u = object distance

v = image distance

Since:-

⟹ m = I/0 = v/u = 160/40 = v/-6

or v = -24 cm

Thus, the image is at a distance 24 cm in front of the lens.

(ii) From relation:

⟹ 1/v - 1/u = 1/f

Where:-

v = image distance

u = object distance

f = focal length

Since:-

⟹ 1/f = 1/-24 - 1/(-6)

⟹ 1/f = -1/24 + 1/6 = 1/8

or f = 8 cm

Thus, the focal length of the lens is 8 cm.

Nature of the image:-

Since, the image is magnified and on the same side of the lens as the object, so the image is virtual.

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Answer 2

Information given to us:

• A convex lens forms an image 16.0 cm long
• Height of object is 4.0 cm
• Distance from lens = 4.0 cm

What we have to calculate:

• Nature of image
• Position of image
• focal length of lens

(a) What is nature of the image?

Ans. Nature of the image is virtual

(b) Find :

________________

(i) The position of the image:-

$\red\bigstar{} \underline{\sf{Using \: Formula: - }}$

• $\underline{ \boxed{ \sf{linear \: magnification \: = \: \dfrac{length \: of \: image(I)}{length \: of \: object(0)} \: = \: \dfrac{v}{u} }}}$

$\red\bigstar{} \underline{\sf{Substituting \: the \: values: - }}$

In the formula where,

• I is length of image
• O is length of object
• v is distance of image
• v is distance of object

Remember that distance from the lens would be negative.

$: \implies \: \sf{ \dfrac{160}{40} \: = \: \dfrac{v}{ - 6} }$

$: \implies \: \sf{ \dfrac{ \cancel{160}}{ \cancel{40}} \: = \: \dfrac{v}{ - 6} }$

$: \implies \: \sf{ \dfrac{ 16}{ 4} \: = \: \dfrac{v}{ - 6} }$

$: \implies \: \sf{ \dfrac{ \cancel{16}}{ \cancel{4}} \: = \: \dfrac{v}{ - 6} }$

$: \implies \: \sf{ 4\: = \: \dfrac{v}{ - 6} }$

$: \implies \: \sf{ v \: = \: - 6 \times 4}$

$: \implies \: \boxed{\bf{ v \: = \: - 24}}$

$\underline{ \therefore \: \sf{Position \: of \: image \: is \: at \: - 24cm}}$

_____________________

(ii) The focal length of the lens:-

______________

$\red\bigstar{} \underline{\sf{Using \: formula: - }}$

Lens formula:-

• $\underline{ \boxed{ \sf{ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} }}}$

Where,

• v is distance of image
• f is focal length
• u is distance of image

$\red\bigstar{} \underline{\sf{Substituting \: the \: values: - }}$

Remember that distance of object and distance of image would be negative.

$\implies \sf{ \dfrac{1}{f} = \: \dfrac{1}{ - 24} - \dfrac{1}{6} \: }$

$\implies \sf{ \dfrac{1}{f} = \: \dfrac{1}{8} }$

$\implies \boxed{\bf{f \: = \: 8 }}$

$\underline{ \therefore \: \sf{Focal \: length \: is \: 8 cm}}$