# a 5 cm tall object is placed perpendicular to the principal axis of a convex lens

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## A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 18 cm at a distance of 12 cm from it. Use lens formula to determine the position, size and nature to the image formed.

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## A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 18 cm at a distance of 12 cm from it. Use lens formula to determine the position, size and nature to the image formed.

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Updated on : 2022-09-05

According to the question:Solution Verified by Toppr

Object distance, u=β12 cm

Focal length, f=18 cm (since, convex lens)

Let the Image distance be v.

By lens formula: v 1 β β u 1 β = f 1 β [4pt] β v 1 β β β12 cm 1 β = 18 cm 1 β [4pt] β v 1 β = β12 cm 1 β + 18 cm 1 β [4pt] β v 1 β = 36 cm β3+2 β [4pt] β v 1 β =β 36 cm 1 β [4pt] β΄v=β36 cm

Thus, image is obtained at 36 cm on the same side of the mirror as the object.

Now, Height of object, h 1 β =5 cm Magnification, m= h 1 β h 2 β β = u v β

Putting values of v and u:

Magnification m= 5 cm h 2 β β = β12 cm β36 cm β β 5 cm h 2 β β =3 [4pt]; βh 2 β =3Γ5 cm=15 cm

Thus, the height of the image is 15 cm and the positive sign means the image is virtual and erect.

Thus virtual, erect and enlarged image is formed.

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## A5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the position and nature of the image formed.A. Position of image =60 cm, image is virtual and erectB. Position of image = 60 cm, image is virtual and erectC. Position of image =60 cm, image is real and invertedD. Position of image =12 cm, image is real and inverted

A5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the position and nature of the image formed.A. Position of image =60 cm, image is virtual and erectB. Position of image = 60 cm, image is virtual and erectC. Position of image =60 cm, image is real and invertedD. Position of image =12 cm, image is real and inverted

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A5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the position and nature of the image formed.A. Position of image =60 cm, image is virtual and erectB. Position of image = 60 cm, image is virtual and erectC. Position of image =60 cm, image is real and invertedD. Position of image =12 cm, image is real and inverted

Question

A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the position and nature of the image formed.

A

Position of image = 60 cm, image is virtual and erect

B

Position of image = - 60 cm, image is virtual and erect

C

Position of image = 60 cm, image is real and inverted

D

Position of image = 12 cm, image is real and inverted

Open in App Solution

The correct options are

**C**Position of image = 60 cm, image is real and inverted

Given object size h = 5 cm

Object distance from lens

u = β 30 cm Focal length f

= 20 cm (as per sign conventions)

We have to find image distance

v

and the nature of the image formed.

Let h i

be the height of the image.

Using the lens formula

1 f = 1 v - 1 u

Putting the values we have,

1 20 = 1 v - 1 β 30 1 20 - 1 30 = 1 v 1 v = 3 β 2 60 or v = 60 c m .

We know, magnification,

M = h i h = v u = h i 5 = 60 β 30 β h i = β 10 c m

As per sign conventions, since the image is formed towards right side of lens, and height of the image is

β 10 c m

, the image is real and inverted.

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SIMILAR QUESTIONS

**Q.**

A convex lens forms an inverted image of size same as that of the object which is placed at a distance 60 cm in front of the lens. Find :

(a) the position of image, and

(b) the focal length of the lens

**Q.**

An object is placed at the following distances from a convex lens of focal length 15 cm : (a) 35 cm (b) 30 cm (c) 20 cm (d) 10 cm Which position of the object will produce : (i) a magnified real image ? (ii) a magnified virtual image ? (iii) a diminished real image ? (iv) an image of same size as the object ?

**Q.**

An object is placed infront of a converging lens of focal length

20 c m

. The distance of the object from the lens is

30 c m . Find:

1. the position of the image formed,

2. the magnification and

3. the nature of the image formed.

**Q.**Find the position and nature of the image formed by the concave mirror of focal length

20 cm , if the object O

is placed at a distance

30 cm from the pole.

**Q.**

An object is placed perpendicular to principal axis of a convex lens of focal length 20cm.the distance of object from lens is 30cm

Find the position,magn magnifica and Nature of image

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## A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the (i) position, (ii) nature, and (iii) size of the image formed.

Here h = +5 cm, f = +20 cm, u = - 30 cm (i) Using lens formula (1)/(v) - (1)/(u) = (1)/(f), we have (1)/(v) = (1)/(f) + (1)/(u) = (1)/((20)) + (1)/((-30))= (3-2)/(60) = (1)/(60) therefore v = + 60 cm (ii) +ve sign of v means that image is being formed on the other side of lens i.e., the image is a real image. (iii) As m = (h.)/(h) = (v)/(u) therefore" "h. = (v)/(u) . h = ((+60))/((-30)) xx (+5) = - 10 cm

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A 5 cm tall object is placed p...

A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the (i) position, (ii) nature, and (iii) size of the image formed.

Text Solution Open Answer in App Solution

Here h = +5 cm, f = +20 cm, u = - 30 cm

(i) Using lens formula

1 v β 1 u = 1 f 1v-1u=1f , we have 1 v = 1 f + 1 u = 1 (20) + 1 (β30) = 3β2 60 = 1 60

1v=1f+1u=1(20)+1(-30)=3-260=160

β΄ β΄ v = + 60 cm

(ii) +ve sign of v means that image is being formed on the other side of lens i.e., the image is a real image.

(iii) As m= h. h = v u m=h.h=vu β΄ h.= v u .h= (+60) (β30) Γ(+5)=β10

β΄ h.=vu.h=(+60)(-30)Γ(+5)=-10

cm

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