Tag: image formation by lenses

Questions Related to image formation by lenses

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. $9\ cm$

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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. $x=1/30$

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

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

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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

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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. $100 \,cm$

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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

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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

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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

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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

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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 $