Tag: ray optics and optical instruments

Quizzes (21) Questions (129)

Questions Related to ray optics and optical instruments

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, $u=-\dfrac{v}{3}=-\dfrac{80}{3}=-26.67\ cm$

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

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.

  1. $40$
  2. $20$
  3. $5$
  4. $-1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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 :

  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
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
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              $u=32cm$
Multiple choice image formation by lens image formation by lenses ray optics and optical instruments optics physics

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:

  1. 12 cm

  2. 24 cm

  3. 36 cm

  4. 48 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
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),

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

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

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

$\Rightarrow u=40$

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

$\Rightarrow f=24cm$

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

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:

  1. $14.0 cm$
  2. $14.3 cm$
  3. $14.6 cm$
  4. $14.9 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $\displaystyle \frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{u+v}{uv}$
$f = \displaystyle \frac{uv}{u+v}= \frac{2.5 u^2}{3.5 u}$
   $\displaystyle =\frac{5}{7}u = \frac{5 \times 20}{7} = \frac{100}{7}=14.3cm$

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

Two thin lenses of focal length $f _1$ and $f _2$ are in contact and coaxial. The power of the combination is

  1. $\sqrt{\frac{f _1}{f _2}}$
  2. $\sqrt{\frac{f _2}{f _1}}$
  3. $\frac{f _1+f _2}{f _1f _2}$
  4. $\frac{f _1-f _2}{f _1f _2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the lenses placed coaxially, 

We have the relationship $P=P _1+P _2+P _3+......$ 
So,for given case P=$P _1+P _2$ or $P $= $\dfrac{1} {f _1} $+ $\dfrac{1} {f _2} $

So,C is the correct answer. 

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

A glass slab of thickness 3 cm and refractive index 3/2 is placed on ink mark on a picec of paper, For a person looking at the mark at a distance 2 cm above it, the distance of the mark will paper to be 

  1. 3 cm

  2. 4 cm

  3. 4.5 cm

  4. 5 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The apparent shift due to a glass slab is given by t(1 - 1/n). Here, 3(1 - 1/(3/2)) = 3(1 - 2/3) = 3(1/3) = 1 cm. The mark appears 1 cm closer to the surface. Since the person is 2 cm above the slab (total distance 5 cm from the mark), the apparent distance is 5 - 1 = 4 cm.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

Three lenses have a combined power of $2.7 D$. If the powers of two lenses are $2.5 D$ and $1.7 D$ respectively, find the focal length of the third lens.

  1. $-66.66 cm$
  2. $-6.666 cm$
  3. $-66.66 m$
  4. $-6.666 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total power, $P _{net} = P _1 + P _2 + P _3$


$P _3 = P _{net} - P _1 - P _2$

$P _3 = 2.7 - 2.5 - 1.7$

      $= - 1.5 = \dfrac{1}{f _3}$

${f _3} = - \dfrac{1}{1.5} = -0.666m$

       $= -66.66cm$