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Questions Related to physics

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

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

  1. $x=30$
  2. $x=31$
  3. $x=500$
  4. $x=1/30$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$ 
$\implies x = 31$

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

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 :

  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. $\dfrac {(u _2-u _1)}{(m _2)^{-1}-(m _1)^{-1}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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: $\dfrac{f}{u-f}$
$\dfrac{m _{1}}{m _{2}}=\dfrac{\frac{f}{u _{1}-f}}{\dfrac{f}{u _{2}-f}}$

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

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

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

  1. $\sqrt {D(D+4f)}$
  2. $\sqrt {D(D-4f)}$
  3. $\sqrt {2D(D-4f)}$
  4. $\sqrt {D^2+4f}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 
Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

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 :

  1. 3 diopter

  2. 5 diopter

  3. 2 diopter

  4. 9 diopter
Reveal answer Fill a bubble to check yourself
B Correct answer
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 
Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

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 :

  1. upright and magnified

  2. inverted and magnified

  3. inverted and diminished

  4. upright and diminished
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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 :

  1. 21 cm

  2. 30 cm

  3. 40 cm

  4. 67 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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 :

  1. 8 cm

  2. 10 cm

  3. 12 cm

  4. 6cm
Reveal answer Fill a bubble to check yourself
A Correct answer
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} $

$\Longrightarrow$ $OS = 8$

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

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

  1. 12 cm

  2. 24 cm or 18cm

  3. 18 cm or 6 cm

  4. 24 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $

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

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 ? 

  1. $50\ cm$
  2. $-36.67\ cm$
  3. $-26.67\ cm$
  4. $80\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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