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Questions Related to 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.

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

  1. $40$

  2. $20$

  3. $5$

  4. $-1$

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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

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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

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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. $14.9 cm$

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

Critical angle of glass is $\theta _1$ and that of water is $\theta _2$. The critical angle for water and glass surface would be $(\mu _g=3/2, \mu _w=4/3)$

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. less than $\theta _2$

  2. between $\theta _1$ and $\theta _2$

  3. greater than $\theta _2$

  4. less than $\theta _1$

Show answer
Correct Option: C
Explanation:

At critical angle. $\mu _{dense} \times sin \theta _ {dense} = \mu _{rare} \times sin 90°$


For glass - air: $(3/2) \times Sin \theta _ {1} = 1 => \theta _{1} = 41.81 ^\circ$
For water - air: $ (4/3) \times  Sin \theta _ {2} = 1 => \theta _{2} = 48.59^\circ$

For water- glass, glass is denser: $ (3/2) \times  Sin \alpha = (4/3)$ => $ \alpha = 62.72^\circ$

A ray of light travelling in a transparent medium of refractive index $\mu$, falls on a surface separating the medium from air at an angle of incidence of $45^o$ . For which of the following value of $\mu$ the ray can undergo total internal reflection ?

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $\mu = 1.33$

  2. $\mu = 1.40$

  3. $\mu = 1.50$

  4. $\mu = 1.25$

Show answer
Correct Option: C
Explanation:

For total internal reflection, 
sin i > sin c 
where, i = angle of incidence, C = critical angle 


But, $sin \, C \, = \, \dfrac{1}{\mu} \,\, \therefore \, sin \, i \, = \, \dfrac{1}{\mu} \,\, or \,\, \mu \, = \, \dfrac{1}{sin \, i}$

$\therefore \, \mu \, > \, \dfrac{1}{sin \, 45} \,\, or \,\, \mu \, > \, \sqrt 2 \,\, (i \, = \, 45 \, Given)$

Hence, option (c) is correct.

A ray of light travelling in a transparent medium of refractive index $\mu $, falls on a surface separating the medium from air at an angle of incidence of $45^o$. For which of the following value of $\mu $ the ray can undergo total internal reflection?

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $\mu =1.33$

  2. $\mu =1.40$

  3. $\mu =1.50$

  4. $\mu =1.25$

Show answer
Correct Option: C
Explanation:

Given: The angle of incidence of the ray is $45^\circ$.


To find: The refractive index of the medium for Total Internal Reflection to occur.

For total internal reflection, the angle of incidence should be greater than the critical angle.
The critical angle for the ray can be given by:
$sin\ C=\dfrac{1}{\mu}$

$C=sin^{-1}\dfrac{1}{\mu}$

and $sin\ C>sin\ 45^\circ$

$\therefore sin 45>\dfrac{1}{\mu}$
$\mu>\dfrac{1}{sin\ 45}$

$\mu>1.41$

Option $(C)$ is correct.

Critical angle for light going from medium (i) to (ii) is $\theta $. The speed of light in medium (i) is v, then the speed of light in medium (ii) is

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $v\left( 1-\cos { \theta } \right) $

  2. $\dfrac { v }{ \sin { \theta } } $

  3. $\dfrac { v }{ \cos { \theta } } $

  4. $\dfrac { v }{ \left( 1-\sin { \theta } \right) } $

Show answer
Correct Option: B
Explanation:
Critical angle$=\theta$
Speed of light in medium(i)$=$v
$\sin\theta=\cfrac{1}{\mu}$
$\mu=\cfrac{1}{\sin\theta}$
$\cfrac{\mu _{2}}{\mu _{1}}=\cfrac{v _{2}}{v _{1}}$
where, $\mu _{2}=\mu$, $\mu _{1}=1$, $v _{1}=v$
$\Rightarrow \cfrac{\mu}{1}=\cfrac{v _{2}}{v}$
$\Rightarrow \cfrac{v _{2}}{v _{1}}=\cfrac{1}{\sin\theta}$
$\Rightarrow v _{2}=\cfrac{v}{\sin\theta}$

What will be the critical angle of water if $ _a\mu _w=\frac{4}{3}$

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $\displaystyle { 42 }^{ \circ }$

  2. $\displaystyle { 49 }^{ \circ }$

  3. $\displaystyle { 22 }^{ \circ }$

  4. $\displaystyle { 1 }^{ \circ }$

Show answer
Correct Option: B
Explanation:

$ _a\mu _w=\frac{4}{3}$

$sin(i _c)=\frac{1}{ _a\mu _w}=\frac{3}{4}$
So, $i _c=49^0$

If a solid transparent object has an refractive index of $2.90$ and a clear liquid has a refractive index of $1.45$ then, which of the following must be true for total internal reflection to occur at the interface between these two media?

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. Incident beam originates in the solid at an angle of incidence greater than $30^o$

  2. Incident beam originates in the liquid at an angle of incidence greater than $30^o$

  3. Incident beam originates in the solid at an angle of incidence greater than $60^o$

  4. Incident beam originates in the liquid at an angle of incidence greater than $60^o$

  5. Total internal reflection cannot occur

Show answer
Correct Option: A
Explanation:

There are two necessary conditions for total internal reflection

 (i) The light beam must go from denser to rarer medium.
(ii) Angle of incidence must be greater than critical angle.
here solid medium has greater refractive index therefore it is denser medium and beam must go from solid medium to liquid medium i.e. it must originates in solid medium.
   now angle of incidence is given by  
         $\sin C= \frac{{\mu} _{rarer}}{{\mu} _{denser}}=\frac{1.45}{2.90}=0.5000$
    or   $\sin C=\sin30$
    or  $C=30$
therefore angle of incidence must be greater than $30$  degree.