Tag: optics

Quizzes (33) Questions (198)

Questions Related to optics

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 optics scattering of light and its applications some natural phenomena of light effects of light

In Raman effect, the wavelength of the incident radiation is $5890\overset{o}{A}$. The wavelengths of Stocks and anti-stocks lines are respectively.

  1. $5880\overset{o}{A}$ and $5900\overset{o}{A}$
  2. $5900\overset{o}{A}$ and $5880\overset{o}{A}$
  3. $5900\overset{o}{A}$ and $5910\overset{o}{A}$
  4. $5870\overset{o}{A}$ and $5880\overset{o}{A}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Incident wavelength=$5890 A^{0}$

  • Stocks  lines have wavelength more than the incident radiation
$\lambda _{s}>$ incident wavelength
  • Anti stocks lines have wavelength less than incident radiation
$\lambda _{as}<$ incident wavelength
So the correct option is B