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Image formation by lenses - class-XII

Description: image formation by lenses
Number of Questions: 45
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Tags: optics physics ray optics and optical instruments
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If David is observing his image in plane mirror. The distance between the mirror and his image is 4 m. If he moves 1 m towards the mirror, then the distance between David and his image will be :

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. 3 m

  2. 5 m

  3. 6 m

  4. 8 m

Show answer
Correct Option: C
Explanation:
Plane mirror produces erect image and is of same size and also at the same distance from the mirror as the object.

if he moves 1 m, distance of object from mirror= 3 m

distance of image from mirror= 3 m

distance between image and object =6 m.

When the reflecting or refracting rays do not actually intersect but appear to intersect when produced backwards, 

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. a real image is formed

  2. a virtual image is formed

  3. either a real or a virtual image is formed

  4. neither a real nor a virtual image is formed

Show answer
Correct Option: B
Explanation:

A virtual image is formed when the refracting or reflecting rays do not actually intersect but appear to intersect when they are produced backwards.

The image which can be obtained on a screen is called

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. a virtual image

  2. a real image

  3. either a real or a virtual image

  4. neither a real nor a virtual image

Show answer
Correct Option: B
Explanation:

The image which can be obtained on a screen is called a real image.

A virtual image is

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. always erect

  2. either erect or inverted

  3. always inverted

  4. neither erect nor inverted

Show answer
Correct Option: A
Explanation:

A virtual image is always erect.

Comparing real and virtual images, we may say that

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. real images can not be obtained on a screen and virtual images can be

  2. virtual images can not be obtained on a screen and real images can be

  3. both real and virtual images can be obtained on a screen

  4. neither real nor virtual images can be obtained on a screen

Show answer
Correct Option: B
Explanation:

Real images can be obtained on a screen. Virtual images can not be obtained on a screen.

A real image is formed by

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. actual intersection of the refracting or reflecting rays

  2. imaginary intersection of the refracting or reflecting rays produced backwards

  3. either actual or imaginary intersection of the reflecting or refracting rays

  4. neither actual nor imaginary intersection of the reflecting or refracting rays

Show answer
Correct Option: A
Explanation:

When the refracting or reflecting rays actually intersect, a real image is formed.

A real image is

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. always erect

  2. either erect or inverted

  3. always inverted

  4. neither erect nor inverted

Show answer
Correct Option: C
Explanation:

A real image is always inverted.

A thin convex lens of focal length $30.00\ cm$ forms an image $2.00\ cm$ high, of an object at infinity. A thin concave lens of focal length $20.00\ cm$ is placed $26.00\ cm$ from the convex lens on the side of the image. The height of the image now is

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $1.00\ cm$

  2. $1.25\ cm$

  3. $2.00\ cm$

  4. $2.50\ cm$

Show answer
Correct Option: D

A meter stick lies along the optic axis of a convex lens of focal length 40 cm its nearer end 60 cm from the mirror surface. How long is the image of stick?

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $200 cm$

  2. $\dfrac {200}{3} cm$

  3. $8 cm$

  4. $\dfrac{210}{6} cm$

Show answer
Correct Option: C

When the distance between the object and the screen is more than 4f, we can obtain the image of the object on the screen for the two positions of the lens. It is called displacement method.In one case, the image is magnified. If $I _1$ and $I _2$ be the sizes of the two images, then the size of the object is

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $(I _1+I _2)/2$

  2. $I _1-I _2$

  3. $\sqrt{I _1\,I _2}$

  4. $\sqrt{I _1/I _2}$

Show answer
Correct Option: C

If $I _1$ and $I _2$ be the size of the images respectively for the two positions of lens in the displacement method, then the size of the object is given by

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $I _1/I _2$

  2. $I _1\times I _2$

  3. $\sqrt{I _1\times I _2}$

  4. $\sqrt{I _1/I _2}$

Show answer
Correct Option: C

The distance between a point source of light and a screen which is $60$ $cm$ is increased to 180 cm. The intensity on the screen as compared with the original intensity will be : 

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $\cfrac { 1 } { 9 }$ times

  2. $\cfrac { 1 } {3 }$ time

  3. $3$ times

  4. $9$ times

Show answer
Correct Option: A

A convex lens is placed between object and a screen. The size of object is $3 cm$ and an image of height $9 cm$ is obtained on the screen. When the lens is displaced to a new position, what will be the size of image on the screen?

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $2 cm$

  2. $6 cm$

  3. $4 cm$

  4. $1 cm$

Show answer
Correct Option: D
Explanation:

The given problem is an example of displacement method, which is generally used to measure the focal length of the lens. In this method, the two image sizes and the object size are related as:

$O = \sqrt{I _1 I _2}$
$\implies 3 = \sqrt {9 \times I _2}$
$I _2 = 1\ cm$

A point object is placed on the principle axis of a converging lens and its image $(I _{1})$ is formed on its principle axis. If the lens is rotated by an small angle $\theta$ about its optical centre such that its principle axis also rotates by the same amount then the image $(I _{2})$ of the same object is formed at point $P$. Choose the correct option.

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. Point $P$ lies on the new principle axis.

  2. Point $P$ lies on the old principle axis.

  3. Point $P$ is anywhere between the two principle axes

  4. None of these

Show answer
Correct Option: C

Optical axis of a thin equi-convex lens is the $X-$axis. The coordinate of a point object and its image are ($-20\ cm, 1\ cm$) and ($25\ cm,-2\ cm$) respectively:-

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. the lens is located at $x=5\ cm$

  2. the lens is located at $x=-5\ cm$

  3. the focal length of the lens is $10\ cm$

  4. the focal length of the lens is $15\ cm$

Show answer
Correct Option: B,C

An object and a screen are mounted on an optical bench and a converging lens is placed between them so that a sharp image is obtained on the screen. The linear magnification of the image is 25. The lens is now moved 30 cm towards the screen and a sharp image is again formed on the screen. Find the focal length of the lens.

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $1.2 cm$

  2. $14.3 cm$

  3. $14.6 cm$

  4. $14.9 cm$

Show answer
Correct Option: B

A convex lens forms an image of an object on a screen. The height of the image is 9 cm. The lens is now displaced until an image is again obtained on the screen. The height of this image is 4 cm. The distance between the object and the screen is 90 cm.

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. The distance between the two positions of the lens is 30 cm.

  2. The distance of the object from the lens in its first position is 36 cm.

  3. The height of the object is 6 cm.

  4. The focal length of the lens is 21.6 cm.

Show answer
Correct Option: D
Explanation:

$h^{2} _{object}=h _{image1} \times h _{image2}$


$h _{object}=\sqrt{36}=6$

magnification of image is $\dfrac{v}{u}=\dfrac{9}{6}$

                                            $v= \dfrac{3u}{2}$

in lens displacement method , $u+v=d$ ; $uv=df$

$u+\dfrac{3u}{2}=90$

$u=36$   => $v=54$

$uv=df$ 

$f=\dfrac{36\times 54}{90}=21.6$

option $D$ is correct 

In a converging lens of focal length f and the distance between real object and its real image is 4f. If the object moves $x _1$ distance towards lens its image moves $x _2$ distance away from the lens and when object moves $y _1$ distance away from the lens its image moves $y _2$ distance towards the lens, then choose the correct option:-

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $x _1>x _2 $ and $y _1>y _2$

  2. $ x _1 < x _2 $ and $ y _1 < y _2 $

  3. $ x _1 < x _2 $ and $y _1>y _2$

  4. $x _1>x _2 $ and $y _2>y _1$

Show answer
Correct Option: A

A double convex lens is made of glass which has refractive index $1.55$ for violet rays and $1.50$ for red rays. If the focal length for violet rays is $25$ cm, the focal length for red rays will be nearly

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. 37.5 cm

  2. 17.5 cm

  3. 27.5 cm

  4. 35 cm

Show answer
Correct Option: A

A points object at 15 cm from a concave mirror of radius of curvature 20 cm is made to oscillate along the pressure axis with amplitude 2 mm. The amplitude of its image will be 

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. 2 mm

  2. 4 mm

  3. 8 mm

  4. 16 mm

Show answer
Correct Option: D

The diameter of the sun is $1.4 \times 10 ^ { 9 } \mathrm { m }$ and its distance from the earth is $1.5 \times 10 ^ { 11 } \mathrm { m } .$ The radius of the image of the sun formed by a lens of focal length $20 \mathrm { cm }$ is

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $93 mm$

  2. $0.093mm$

  3. $9.3 mm$

  4. $0.93 mm$

Show answer
Correct Option: D

An object of length 2.0 cm is placed perpendicular to the principal axis of a convex lens of focal length 12 cm. Find the size of the image if the object is at a distance of 8.0 cm from the lens.

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $6 cm$

  2. $4 cm$

  3. $5 cm$

  4. $1 cm$

Show answer
Correct Option: A

A man wearing glasses of focal length  $+ 1 \mathrm { m }$  cannot clearly see beyond  $1\mathrm { m }$

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. if he is farsighted

  2. if he is nearsighted

  3. if his vision is normal

  4. in each of these case

Show answer
Correct Option: D
Explanation:
That is the person is near-sighted.
And the condition is known as myopia.
Ans. (b)

A thin lens produces an image of the same size as the object. Then from the optical centre of the lens, the distance of the object is

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. zero

  2. $4f$

  3. $2f$

  4. $f/2$

Show answer
Correct Option: C

Optical axis of a thin equi-convex lens  is the X-axis. The co-ordinate of a point object and its image are (-20 cm, 1 cm) and (25 cm, -2 cm) respectively:-

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. The lens is located at X = 5 cm

  2. The lens is located at X = -5 cm

  3. The focal length of the lens is 10 cm

  4. The focal length of the lens is 15 cm

Show answer
Correct Option: B,C

A point object $O$ is placed on the principle axis of a convex lens of focal length $20\ cm$ at a distance of $40\ cm$ to the left of it. The diameter of the lens is $10\ cm$. If the eye is placed $60\ cm$ to the right of the lens at a distance $h$ below the principle axis, then the maximum value of $h$ to see the image will be

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $2.5\ cm$

  2. $5\ cm$

  3. $0\ cm$

  4. $10\ cm$

Show answer
Correct Option: A

For two position of lens, the images are obtained on a fixed screen. If the size of the object is 2 cm and size of diminished image is 0.5 cm, the size of the other image will be

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. 1 cm

  2. 4 cm

  3. 8 cm

  4. 16 cm

Show answer
Correct Option: C
Explanation:

$height^{2} _{object}=height _{image1}\times height _{image2}$


$2 \times 2=0.5 \times h$

$h=8$

option $C$ is correct

A lens is placed between the source of light and a wall. It forms images of area $ { A } _{ 1 }$ and  ${ A } _{ 2 }$ on the wall for its two different positions. The area of the source of light is :

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $ \sqrt { { A } _{ 1 }{ A } _{ 2 } } $

  2. $ \dfrac { { A } _{ 1 }+{ A } _{ 2 } }{ 2 } $

  3. $ { \left( \dfrac { \sqrt { { A } _{ 1 } } +\sqrt { { A } _{ 2 } } }{ 2 } \right) }^{ 2 } $

  4. $None\ of\ these$

Show answer
Correct Option: A
Explanation:

$height^{2} _{object}=height _{image1} \times height _{image2}$


$r^{2} _{source}=r _{image1} \times r _{image2}$

$\pi r^{2} _{source}=\pi r _{image1} \times r _{image2}$

$A _{source}=\sqrt{\pi^{2} r^{2} _{image1}r^{2} _{image2}}$

$A _{source}=\sqrt{A _{1}A _{2}}$

option $A$ is correct 

A student focused the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle, screen and the lens as under position of candle $=12.0\ cm$position of convex lens $=50.0\ cm$position of the screen $=88.0\ cm$. Where will the image be formed, if he shifts the candle towards the lens at a position of $31.0\ cm$?

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. 19cm

  2. 48cm

  3. Infinity

  4. at center of curvature

Show answer
Correct Option: C

A magnifying glass is to be used at the fixed object distance of $1$ inch. If it is to produce an erect image $5$ items magnified, its focal length should be- 

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $0.2"$

  2. $0.8"$

  3. $1.25"$

  4. $5"$

Show answer
Correct Option: B

An object is placed at a distance of $30\ cm$ from a concave lens of focal length $15\ cm$. What is the height of the object if the height of the image is $3\ cm$?

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $3\ cm$

  2. $1\ cm$

  3. $6\ cm$

  4. $9\ cm$

Show answer
Correct Option: A

In a slide show program, the image on the screen has an area 900 times that of the slide. If the distance between the slide and the screen is $x$ times the distance between the slide and the projector lens, then

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $x=30$

  2. $x=31$

  3. $x=500$

  4. $x=1/30$

Show answer
Correct Option: B
Explanation:

Magnification of area = 900 times

So linear magnification = $\sqrt (\text{Area magnification})$ = 30 times

Let distance between slide and projector (u) be $a$

So, distance between projector and screen (v) = $m \times u = 30 a$

Distance between slide and screen = $x + 30x = 31a$

By question $ 31a = x \times  a$ 
$\implies x = 31$

The lateral magnification of the lens with an object located at two different positions $u _1$ and $u _2$ are $m _1$ and $m _2$, respectively. Then the focal length of the lens is :

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $f=\sqrt {m _1m _2}(u _2-u _1)$

  2. $\dfrac{m _2u _2 - m _1u _1}{m _2-m _1}$

  3. $\dfrac {(u _2-u _1)}{\sqrt {m _2m _1}}$

  4. $\dfrac {(u _2-u _1)}{(m _2)^{-1}-(m _1)^{-1}}$

Show answer
Correct Option: B
Explanation:

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
$u= -u$ ; $f= f$

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$v= \dfrac{fu}{u-f}$

Magnification is: $\dfrac{f}{u-f}$
$\dfrac{m _{1}}{m _{2}}=\dfrac{\frac{f}{u _{1}-f}}{\dfrac{f}{u _{2}-f}}$

$f=\dfrac{u _{2}m _{2}-u _{1}m _{1}}{m _{2}-m _{1}}$

A luminous object and a screen are at fixed distance D apart. A converging lens of focal length f is placed between the object and screen. A real image of the object in formed on the screen for two lens positions if they are separated by a distance d equal to

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $\sqrt {D(D+4f)}$

  2. $\sqrt {D(D-4f)}$

  3. $\sqrt {2D(D-4f)}$

  4. $\sqrt {D^2+4f}$

Show answer
Correct Option: B
Explanation:
$u+v=D$

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

$\dfrac{1}{D-u}+\dfrac{1}{u}=\dfrac{1}{f}$

$u^{2}-Du+Df=0$

$u _{1}= \dfrac{D+\sqrt{D(D-4f)}}{2}$ and $u _{2}=\dfrac{D-\sqrt{D(D-4f)}}{2}$

$u _{1}-u _{2}=\sqrt{D(D-4f)}$

option $B$ is correct 

The distance between an object and the screen is 100 cm. A lens produces an image on the screen when the lens is placed at either of the positions 40 cm apart. The power of the lens is nearly :

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. 3 diopter

  2. 5 diopter

  3. 2 diopter

  4. 9 diopter

Show answer
Correct Option: B
Explanation:
Given, 

$u _{1}+u _{2}=100$

$u _{1}-u _{2}=40$

=>  $u _{1}=70$ and $u _{2}=30$

for $u _{1}= -70$ $v _{1}$ will be $+30$

From lens formula, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{30}+\dfrac{1}{70}=\dfrac{1}{f}$

$\dfrac{1}{f}=\dfrac{1}{21}$

$power=\dfrac{1}{21}\times 100=5(approx)$

option $B$ is correct 

Heeltes of the other. If focal length material is $31 2$. then radius of curvature of of double convex lens is! the lens material is $3/2 \,h$ Radius of curvature of one surface of double con of the lens is $30 \,cm$ and refractive index of the small surface is

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $20 \,cm$

  2. $40 \,cm$

  3. $80 \,cm$

  4. $100 \,cm$

Show answer
Correct Option: A

A candle is placed at a distance of 20 cm from a converging lens of focal length 15 cm. The image obtained on the screen is :

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. upright and magnified

  2. inverted and magnified

  3. inverted and diminished

  4. upright and diminished

Show answer
Correct Option: B
Explanation:

When an object is placed between $F$ and $2F$ then image will formed between $F$ and $2F$ on opposite side of lens  and Image formed is real, Inverted and 

magnified. 

here $F= 15 cm$ then $2F = 30 cm$

object distance $u = 20 cm$ which lies between $F$ and $2F$

therefore image formed will real, inverted and magnified.

Thus Option B is correct.

A light source is placed 100 cm away from a screen. A converging lens placed at a certain position between the source and the screen focuses the image of the source on the screen. The lens is moved a distance of 40 cm and it is found that it again focuses the image of the source on the screen. The focal length of the lens is :

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. 21 cm

  2. 30 cm

  3. 40 cm

  4. 67 cm

Show answer
Correct Option: A
Explanation:

Answer is A.

The expression for focal length by displacement method is given as follows.
$f=\frac { { D }^{ 2 }-{ x }^{ 2 } }{ 4D } $
where,
D - the distance between the object and screen
x - the distance between the two positions of the lens.
Here, D = 100 cm and x = 40 cm.
So, $f=\frac { { D }^{ 2 }-{ x }^{ 2 } }{ 4D } =\frac { { 100 }^{ 2 }-{ 40 }^{ 2 } }{ 4\times 100 } =21\quad cm$.
Hence, the focal length of the lens is 21 cm.

A convex lens forms a real image 4 cm long on a screen. When the lens is shifted to a new position without disturbing the object or the screen, again real image is formed on the screen which is 16 cm long. The length of the object is :

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. 8 cm

  2. 10 cm

  3. 12 cm

  4. 6cm

Show answer
Correct Option: A
Explanation:

If the image sizes be $ {I} _{1} \  and \  {I} _{2} $,


By Displacement Method, object size ($OS$) is given by :
$ OS = \sqrt{{I} _{1} {I} _{2}} $

Thus, OS = $ \sqrt{64} $

$\Longrightarrow$ $OS = 8$

The distance between two point sources of light is 24 cm and a converging lens is kept in between two sources. The object distances of two sources from a converging lens of focal length of 9 cm, so that the image distances  of two sources are equal

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. 12 cm

  2. 24 cm or 18cm

  3. 18 cm or 6 cm

  4. 24 cm

Show answer
Correct Option: C
Explanation:

Here $u _1 + u _2  =-24$.......(1).

$\dfrac{1}{v _1}-\dfrac{1}{u _1}=\dfrac{1}{9}$
and for the virtual image 
 $\dfrac{1}{-v _1}- \dfrac{1}{u _2}= \dfrac{1}{9}$

$ -(\dfrac{1}{u _1}+ \dfrac{1}{u _2})= \dfrac{2}{9} \implies u _1 u _2=108$.........(2)
On solving (1) and (2) 
We get $u^2+24u+108=0 \implies u= -18, \ - 6 $

The image of a candle flame formed by a lens is obtained on a  screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is $80\ cm$, at what distance should the candle be placed from the lens ? 

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $50\ cm$

  2. $-36.67\ cm$

  3. $-26.67\ cm$

  4. $80\ cm$

Show answer
Correct Option: C
Explanation:

Given, magnification $=-\dfrac{v}{u}=3$ and $v=80\ cm$
So, object distance, $u=-\dfrac{v}{3}=-\dfrac{80}{3}=-26.67\ cm$

An object of  $5\mathrm { cm }$  is placed before a concave mirror at a distance of  $40\mathrm { cm } .$  If its focal length is  $20\mathrm { cm }$  then what is the magnification of the image.

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $40$

  2. $20$

  3. $5$

  4. $-1$

Show answer
Correct Option: D
Explanation:

$\begin{array}{l} \frac { 1 }{ v } =\frac { 1 }{ t } -\frac { 1 }{ u }  \ =\frac { { -1 } }{ { 20 } } -\frac { 1 }{ { -40 } } =\frac { { -1 } }{ { 40 } }  \ v=-40 \ m=-\frac { v }{ u } =-\frac { { -40 } }{ { -40 } } =-1 \ \therefore \, \, 1\times 1=1 \ Ans.\, \, (D) \end{array}$

In displacement method, the distance between object and screen is 96 cm. The ratio of lengths of two images formed by a converging lens placed between them is 4. Then :

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. ratio of the length of object to the length of shorter image is 2

  2. distance between the two positions of the lens is 32 cm

  3. focal length of the lens is 64/3 cm

  4. when the shorter image is formed on screen, distance of the lens from the screen is 32 cm

Show answer
Correct Option: A,B,C,D
Explanation:
Given -  Distance between object and screen $a=96cm$ ,

             Ratio of lengths of images $=4:1$ ,

Let length of larger image is $II'=4x$ ,

      length of smaller image is $II''=x$ ,

      length of object is $OO'$ .

we know that ,  $OO'=\sqrt{II'\times II''}$ ,

                         $OO'=\sqrt{4x\times x}=2x$ ,

(A) Hence ratio of length of object to the length of shorter image will be ,

          $\dfrac{OO'}{II''}=\dfrac{2x}{x}=2$

(B) We have ,

                   $\dfrac{II''}{OO'}=\dfrac{u}{d+u}$ ,

                    $\dfrac{1}{2}=\dfrac{u}{d+u}$ ,

or                $d=u$ ,

now , by    $u=\dfrac{a-d}{2}$ ,

or              $d=\dfrac{96-d}{2}$ ,

or              $d=32cm$

(C) By using , $f=\dfrac{a^{2}-d^{2}}{4a}$ ,

or                  $f=\dfrac{96^{2}-32^{2}}{4\times96}$ ,

or                  $f=64/3cm$ 

(D) When shorter image is on the screen , 

                   $u=\dfrac{a-d}{2}$ ,

or              $u=\dfrac{96-32}{2}$ ,

or              $u=32cm$

A lens forms a real image of an object on a screen placed at a distance of 100 cm from the screen. If the lens is moved by 20 cm towards the screen, another image of the object is formed on the screen. The focal length of the lens is:

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. 12 cm

  2. 24 cm

  3. 36 cm

  4. 48 cm

Show answer
Correct Option: B
Explanation:

From lens formula, $\displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$


$\displaystyle \frac{1}{100-u}+\frac{1}{u}=\frac{1}{f}$.....(1)

$\displaystyle \frac{1}{80-u}+\frac{1}{u+20}=\frac{1}{f}$........(2)

From (1) and (2),

$\displaystyle \frac{1}{100-u}+\frac{1}{u}=\frac{1}{80-u}+\frac{1}{u+20}$

$\displaystyle \frac{20}{\left ( u \right )\left ( u+20 \right )}=\frac{20}{\left ( 80-u \right )\left ( 100-u \right )}$

$\Rightarrow u^{2}+20 u=u^{2}-180 u+8000$

$\Rightarrow u=40$

$\displaystyle \frac{1}{60}+\frac{1}{40}=\frac{1}{f}$

$\Rightarrow f=24cm$

An object and a screen are mounted on an optical bench and a converging lens is placed between them so that a sharp image is received on the screen. The linear magnification of the image is 2.5. The lens is now moved 30 cm nearer to the screen and a sharp image is again formed on the screen. The focal length of the lens is:

image formation by lens image formation by lenses ray optics and optical instruments optics physics

  1. $14.0 cm$

  2. $14.3 cm$

  3. $14.6 cm$

  4. $14.9 cm$

Show answer
Correct Option: B
Explanation:

Given $\displaystyle \frac{v}{u} = 2.5$
$v= 2.5 u$
again $ v-u = 30$
$v=30+u$
$2.5 u = 30 +u$
$1.5 u=30$
$u=20$
Now, $\displaystyle \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{u+v}{uv}$
$f = \displaystyle \frac{uv}{u+v}= \frac{2.5 u^2}{3.5 u}$
   $\displaystyle =\frac{5}{7}u = \frac{5 \times 20}{7} = \frac{100}{7}=14.3cm$

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