Tag: reflection of light at curved surfaces

Questions Related to reflection of light at curved surfaces

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

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

  1. 4/9

  2. 5/8

  3. 9/4

  4. 8/5

Show answer
Correct Option: B
Explanation:

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

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

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:

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

  1. 8/3

  2. 2.5

  3. 2

  4. 1.5

Show answer
Correct Option: D
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 $-\dfrac{v}{u}= -\dfrac{30}{20}= -1.5$

Mark the correct statement(s) w.r.t. a concave spherical mirror

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

  1. for real extended object, it can form a diminished virtual image

  2. for real extended object, it can form a magnified virtual image

  3. for virtual extended object, it can form a diminished real image

  4. for virtual extended object, it can form a magnified real image

Show answer
Correct Option: B,C
Explanation:
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$v= \dfrac{fu}{u-f}$

magnification is $\dfrac{-v}{u}=\dfrac{f}{f-u}$

if $f>|u|$ (u<0) then a magnified image is formed which is virtual 

if $u>2f$ (u>0) then a diminished image is formed which is real 

option $B,C$ are correct

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.

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

  1. 2 (virtual, inverted)

  2. 3 (real, inverted)

  3. 3, (virtual, enlarged)

  4. +1 (real, enlarged)

Show answer
Correct Option: A
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.

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 :

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

  1. $2$ (Virtual, Inverted)

  2. $3$ (Real, Inverted)

  3. $5$ (Real, Erect)

  4. $2$ (Virtual, Erect)

Show answer
Correct Option: D
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. 

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

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

  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

Show answer
Correct Option: C
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 $f<0.25m$

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

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

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

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

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

  4. 1

Show answer
Correct Option: A
Explanation:

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

The linear magnification for a mirror is the ratio of the size of the image to the size of the object, and is denoted by $'m'$. Then $m$ is equal to (symbols have their usual meanings)

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$.........(1)


multiplyng by u in eq.(1)

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

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

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

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

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

hence, $m=\dfrac{f}{u-f}$

If linear magnification for a spherical mirror is $\dfrac{3}{2}$, then we may write: (symbols have their usual meanings) 
physics reflection of light at curved surfaces problems on mirror and magnification formula derivation of formula for curved mirrors mirror formula and magnification

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

  2. $f=\dfrac{3u}{2}$

  3. $f=\dfrac{3u}{5}$

  4. None of these

Show answer
Correct Option: C
Explanation:

Mirror equation is: $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Multiplying both sides by $u$, we get:
$\dfrac{u}{f}=\dfrac{u}{v}+1$
Or  $\dfrac{u}{v}=\dfrac{u}{f}-1=\dfrac{u-f}{f}$
Or  $\dfrac{v}{u}=\dfrac{f}{u-f}$
Now, magnification, $m=\dfrac{v}{u}=\dfrac{3}{2}$
$\therefore$ $\dfrac{f}{u-f}=\dfrac{3}{2}$
Solving the above equation we get  $\dfrac{5}{2}f=\dfrac{3}{2}u$
 $\implies f=\dfrac{3u}{5}$

Magnification produced by a convex mirror is always:

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

  1. equal to 1

  2. less than 1

  3. more than 1

  4. zero

Show answer
Correct Option: B
Explanation:

A convex mirror always creates a virtual image which is diminished. So, magnification produced by convex mirror is always less than one.