Physics

Ray Optics and Mirrors

115 Questions

Ray optics covers the principles of light reflection and refraction through mirrors and lenses. The questions focus on focal length, magnification, and image formation. This physics topic is highly relevant for exams requiring science aptitude.

Concave mirror imagesConvex lens formulasFocal length calculationsMagnification ratiosImage distance

Ray Optics and Mirrors Questions

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

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

  1. $\displaystyle { 42 }^{ \circ }$
  2. $\displaystyle { 49 }^{ \circ }$
  3. $\displaystyle { 22 }^{ \circ }$
  4. $\displaystyle { 1 }^{ \circ }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

The critical angle of a medium B with respect another medium A is 45$^{\circ}$ and the critical angle of a medium C with respect the medium B is 30$^{\circ}$. The critical angle of medium C with respect to A is :

  1. Less than 30$^{\circ}$
  2. Greater than 30$^{\circ}$
  3. 30$^{\circ}$

  4. Cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ \mu _{B} sin 45^{\circ}=\mu _{A} $


$ \mu _{C} sin 30^{\circ}=\mu _{B} $

$ \mu _{C} sin 30 sin 45=\mu _{A} $

$ \mu _{C} \dfrac{1}{2\sqrt{2}}=\mu{A} $

$ \therefore critical \ angle = sin ^{-1} \left ( \dfrac{1}{2\sqrt{2}} \right )<30^{\circ} $

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

Magnification produced by a convex mirror is $\frac { 1 }{ 3 }$, then distance of the object from mirror is

  1. $\frac { f }{ 3 }$
  2. $\frac { 2f }{ 3 }$
  3. $1f$
  4. $2f$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Magnification m = -v/u = 1/3 for a convex mirror. Using the mirror formula 1/v + 1/u = 1/f, we substitute v = -u/3. This gives -3/u + 1/u = 1/f, leading to -2/u = 1/f, so u = -2f. The distance is 2f.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A convex lens of focal length 30 cm forms an image of height 2 cm for an object situated at infinity. If a concave lens of focal length 20 cm is placed coaxially at a distance of 26 cm in front of convex lens. then size of final image would be:

  1. $1.25cm$
  2. $2.5 cm$
  3. $2 cm$
  4. $0.75cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

An object is placed at a distance of 40 cm in front of a concave mirror of focal length 20 cm. Determine the ratio of the size of the image and the size of object

  1. 2:1

  2. 1:2

  3. 1:1

  4. 4:1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since focal length of concave mirror given is 20cm l. Object is at 40cm distance which means it is at centre of curvature. hence, image will be formed of same size and at centre focus only but real and inverted.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The focal length of a concave mirror is 50 cm where an object is to be placed so that its image is two times magnified, real and inverted :

  1. 75 cm

  2. 72 cm

  3. 63 cm

  4. 50 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

Focal length, $f=50\,cm$

Magnification, $m=2$

$ m=\dfrac{v}{u} = 2$

$ v=2u $

From mirror formula,

$ \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} $

$ \dfrac{1}{f}=\dfrac{1}{2u}+\dfrac{1}{u}$

$ u=\dfrac{3f}{2}=\dfrac{3\times 50}{2}=75\ cm $

Object Is placed at $75\,cm$ from the mirror.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A convex mirror has a focal length $f$.A real object is placed at a distance $f$ in front of it from the pole, produces an image at:

  1. $\infty$
  2. $f$
  3. $\dfrac{f}{2}$
  4. $2f$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a convex mirror, f is positive. With object distance u = -f, the mirror formula 1/v + 1/u = 1/f becomes 1/v - 1/f = 1/f. Thus, 1/v = 2/f, so v = f/2.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The distance between an object its doubly magnified by a concave mirror of focal length $f$ is

  1. $3 f/2$
  2. $2 f/3$
  3. $3\ f$

  4. Depend on whether the image is real or virtual
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

Focal length $=f$

We know that, the magnification is

$m=\dfrac{-v}{u}$

Now, magnification is double and image is real

  $ -2=\dfrac{-v}{u} $

 $ v=2u $

Now, using formula of mirror

  $ \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} $

 $ \dfrac{1}{f}=\dfrac{1}{2u}+\dfrac{1}{u} $

 $ \dfrac{1}{f}=\dfrac{3}{2u} $

 $ u=\dfrac{3f}{2} $

Hence, the distance is $\dfrac{3f}{2}$ 

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

An object $2.5\ cm$ high is placed at a distance of $10\ cm$ from a concave mirror of radius of curvature $30\ cm$. The size of the image is:

  1. $9.2\ cm$
  2. $10.5\ cm$
  3. $5.6\ cm$
  4. $7.5\ cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that,

Distance $u=-10\,cm$

Height $h=2.5\,cm$

Radius of curvature $R=30\,cm$

We know that,

  $ -f=\dfrac{R}{2} $

 $ -f=\dfrac{30}{2} $

 $ f=-15\,cm $

Now, using mirror’s formula

  $ \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} $

 $ -\dfrac{1}{15}=\dfrac{1}{v}-\dfrac{1}{10} $

 $ \dfrac{1}{v}=-\dfrac{1}{15}+\dfrac{1}{10} $

 $ v=30\,cm $

Now, the magnification is

  $ m=\dfrac{-v}{u} $

 $ m=\dfrac{30}{10} $

 $ m=3 $

We know that,

  $ m=\dfrac{h'}{h} $

 $ 3=\dfrac{h'}{2.5} $

 $ h'=7.5\,cm $

Hence, the size of the image is $7.5\ cm$

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The focal length $f$ of a mirror is given by $\cfrac{1}{f}=\cfrac{1}{u}+\cfrac{1}{v}$, where $u$ and $v$ represent object and image distances, respectively

  1. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}$
  2. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{v}+\cfrac{\Delta v}{u}$
  3. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}-\cfrac{\Delta (u+v)}{u+v}$
  4. $\cfrac{\Delta f}{f}=\cfrac{\Delta u}{u}+\cfrac{\Delta v}{v}+\cfrac{\Delta U}{u+v}+\cfrac{\Delta v}{u+v}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The focal length $f$ of a mirror is given by
$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$. . . . . . . .(1)
where $v=$ image distance
$u=$ object distance
from equation (1),
$f=\dfrac{uv}{u+v}=uv(u+v)^{-1}$
Taking log both sides we get
$logf=logu+log v+log(u+v)^{-1}$
$log f=logu+logv-log(u+v)$
Differentiating with respect to each variable
$\dfrac{\Delta f}{f}=\dfrac{\Delta u}{u}+\dfrac{\Delta v}{v}+\dfrac{\Delta (u+v)}{u+v}$
$\dfrac{\Delta f}{f}=\dfrac{\Delta u}{u}+\dfrac{\Delta v}{v}+\dfrac{\Delta u}{u+v}+\dfrac{\Delta v}{u+v}$
The correct option is D.