Physics

Ray Optics and Mirrors

141 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 physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A short linear object of length $L$ lies on the axis of a spherical mirror of focal length $f$ at a distance $u$ from the mirror. Its image has an axial length $L$ equal to :

  1. $L{ \left[ \cfrac { f }{ \left( u-f \right) } \right] }^{ 1/2 }$
  2. $L{ \left[ \cfrac { u+f }{ \left( f \right) } \right] }^{ 1/2 }$
  3. $L{ \left[ \cfrac { u+f }{ \left( f \right) } \right] }^{ 2 }$
  4. $L{ \left[ \cfrac { f }{ \left( u-f \right) } \right] }^{ 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From mirror formula,       $\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } $


On differentiating, we get


      $\cfrac { -dv }{ { v }^{ 2 } } -\cfrac { du }{ { u }^{ 2 } } =0\\ \therefore dv=-du{ \left( \cfrac { v }{ u }  \right)  }^{ 2 }\\ as\quad \cfrac { v }{ u } =\cfrac { f }{ u-f } \\ \therefore dv=-du{ \left[ \cfrac { f }{ u-f }  \right]  }^{ 2 }\\ { L }^{ \prime  }=L{ \left[ \cfrac { f }{ u-f }  \right]  }^{ 2 }$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

If an object is placed at a distance of 20cm from the pole of a concave mirror, the magnification of its real image is 3. If the object is moved away from the mirror by 10cm, then the magnification is -1.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$M= \frac{f}{f-d _0}$ and real image has M negative

$-3= \frac{f}{f-20}$

$-3f+60=f$

$f=15 cm$

$M= \frac{15}{15-30}$

$M= -1$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A convex lens is given, for which the minimum distance between an object and its rel image is $40cm$. An object is placed at a distance of $15cm$ from this lens. The liner magnification of adjustment will be 

  1. $\dfrac{5}{3}$
  2. $-2$
  3. $2$
  4. $\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given object distance $u=15$ cm

Distance between object and real image produced $=40 $cm
Thus image distance $v=40-15=25$ cm
Also we know linear magnification,
$m=\dfrac{-v}{u}=\dfrac{-25}{-15}=\dfrac{5}{3}$ 

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

An object of length $6\ cm$ is placed on the principle axis of a concave mirror of focal length $f$ at a distance of $4\ f$. The length of the image will be

  1. $2\ cm$
  2. $12\ cm$
  3. $4\ cm$
  4. $1.2\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

The object distance $u=-6\,cm$

Now, magnification is

  $ m=\dfrac{I}{O} $

 $ m=\dfrac{f}{f-u} $

 $ \dfrac{I}{6}=\dfrac{-f}{-f-\left( -4f \right)} $

 $ I=-2\,cm $

Hence, the length of image is -$2\ cm$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

In the displacement method, a convex lens is placed in between an object and a screen. If one of the magnification is $3$ and the displacement of the lens between the two positions is $24$cm, then the focal length of the lens is:

  1. $10$ cm
  2. $9$ cm
  3. $6$ cm
  4. $16/3$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given magnification $M=\dfrac{v}{u}=3$


Thus $v=3u$, where v and u are the image and object distance respectively.

Also Distance between lenses$=v-u=24$
Thus $u=12 cm$, than $v=36 cm$

From lens formula we have,
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

$\dfrac{1}{f}=\dfrac{1}{36}+\dfrac{1}{12}$

$\dfrac{1}{f}=\dfrac{4}{36}$

$f=\dfrac{36}{4}$

$f=9 cm$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A concave mirror of focal length $20\ cm$ produces an image twice the height of the object. If the image is real, then the distance of the object from the mirror is:

  1. $20\ cm$
  2. $60\ cm$
  3. $10\ cm$
  4. $30\ cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a real image, magnification m = -v/u = -2. Thus v = 2u. Using 1/v + 1/u = 1/f with f = -20 cm, we have 1/(2u) + 1/u = 1/-20. This simplifies to 3/(2u) = -1/20, so 2u = -60, u = -30 cm. The distance is 30 cm.

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

In a concave mirror an object is placed at a distance x from the focus, and the image is formed at a distance y from the focus. The focal length of the mirror is

  1. $xy$
  2. $\sqrt{xy} $
  3. $\dfrac{x+y}{2} $
  4. $\sqrt{\dfrac{x}{y} }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a property of concave mirrors where the focal length f is the geometric mean of the distances of the object and image from the focus. Specifically, f^2 = x * y, so f = sqrt(xy).

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

The focal length of a concave mirror is f and the distance from the object to the principal focus is p. The ratio of the size of the real image to the size of the object is:

  1. $-\displaystyle \frac{f}{p}$
  2. $\displaystyle \left(\frac{f}{p}\right)^2$
  3. $\displaystyle \left(\frac{f}{p}\right)^{\frac{1}{2}}$
  4. $-\displaystyle \frac{p}{f}$
  5. $-fp$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance of object is $u= -(f+p)$
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ gives:
$\dfrac{1}{v}-\dfrac{1}{f+p}= -\dfrac{1}{f}$
or, $\dfrac{1}{v}= -\dfrac{1}{f}+\dfrac{1}{f+p}$
or, $\dfrac{1}{v}= -\dfrac{p}{(f+p)\times f}$
or, $v= -\dfrac{(f+p)\times f}{p}$      (-ve sign indicates image is real) 
   Magnification $=-\dfrac{v}{u}$ 
           $=-\dfrac{(f+p)\times f}{p\times (f+p)}$   
           $=-\dfrac{f}{p}$  (-ve sign indicates inverted)
    So, ratio of size of image to that of object is: $-\dfrac{f}{p}$
Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

An object is kept at 15 cm from a convex mirror of focal length 25 cm. What is the magnification?

  1. 4/9

  2. 5/8

  3. 9/4

  4. 8/5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Magnification for a mirror, $m = \dfrac{f}{f-u}$

As per sign convention: $u = -15\ cm$, $f = 25\ cm$
So, $m=\dfrac{25}{25+15}=\dfrac{5}{8}$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

The image of an object placed on the principal axis of a concave mirror of focal length 12 cm is formed at a point which is 10 cm more distance from the mirror than the object. The magnification of the image is:

  1. 8/3

  2. 2.5

  3. 2

  4. 1.5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the object distance be $u$ then image distance is $u+10$
$u= -u$ ; $v= -(u+10)$ ; $f= -12$
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{-1}{u+10}+\dfrac{-1}{u}=\dfrac{-1}{12}$
$\dfrac{2u+10}{u(u+10)}=\dfrac{1}{12}$
$u=20$cm
$v=-30$cm
Magnification is $-\dfrac{v}{u}= -\dfrac{30}{20}= -1.5$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A beam of light converges towards a point O, behind a convex mirror of focal length 20 cm. Find the magnification and nature of the image when point O is 30 cm behind the mirror.

  1. 2 (virtual, inverted)

  2. 3 (real, inverted)

  3. 3, (virtual, enlarged)

  4. +1 (real, enlarged)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$u=30$ ; $f=20$ 


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

$\dfrac{1}{v}+\dfrac{1}{30}=\dfrac{1}{20}$

$v= 60$

Image is virtual (v>0) 

Magnification is $-\dfrac{v}{u}= -\dfrac{60}{30}= -2$ (<0) hence it is inverted.

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A beam of light converges towards a point O, behind a convex mirror of focal length 20 cm. Find the magnification and nature of the image when point O is 10 cm behind the mirror :

  1. $2$ (Virtual, Inverted)
  2. $3$ (Real, Inverted)
  3. $5$ (Real, Erect)
  4. $2$ (Virtual, Erect)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$u=10$ ; $f=20$ 
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\dfrac{1}{v}+\dfrac{1}{10}=\dfrac{1}{20}$
$v= -20$
Image is real  , Magnification is $-\dfrac{v}{u}= -\dfrac{-20}{10}=2$ ( > 0) Hence, it is erect. 
Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placing

  1. a concave mirror of suitable focal length

  2. a convex mirror of suitable focal length

  3. a convex lens of focal length less than 0.25 m

  4. a concave lens of suitable focal length
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Image can be formed on the screed if it is real. Real image of reduced size can be formed can be formed by a concave mirror or a convex lens.


The object is beyond $2f$. 

So let $u=2f+x$

And using lens formula we have

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

or

$\dfrac{1}{2f+x}+\dfrac{1}{v}=\dfrac{1}{f}$

or

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{2f+x}$

Solving we get

$v=\dfrac{f(2f+x)}{f+x}$

We have $u+v=1$

or

$2f+x+\dfrac{f(2f+x)}{f+x}=1$

or

$\dfrac{(2f+x)^2}{f+x}<1$

$(2f+x)^2<(f+x)$

This is valid only when $f<0.25m$

Multiple choice physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

An object is placed at a distance $2 f$ from the pole of a convex mirror of focal length $f$. The linear magnification is:

  1. $\displaystyle \frac {1}{3}$
  2. $\displaystyle \frac {2}{3}$
  3. $\displaystyle \frac {3}{4}$

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

$\displaystyle \frac {1}{V} - \frac {1}{2f} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac {3}{2f}  \Rightarrow v = \frac{2}{3}f$
$\therefore m = \displaystyle \frac {u}{v} = \frac{2}{3} \frac{f}{2f} = \frac {1}{3}$