Tag: reflection of light at curved surfaces

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Questions Related to reflection of light at curved surfaces

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

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

  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
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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
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

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

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 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)

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $m=\dfrac{f}{u-f}$

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 linear magnification for a spherical mirror is $\dfrac{3}{2}$, then we may write: (symbols have their usual meanings) 
  1. $f=\dfrac{u}{2}$
  2. $f=\dfrac{3u}{2}$
  3. $f=\dfrac{3u}{5}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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  $\dfrac{5}{2}f=\dfrac{3}{2}u$
 $\implies f=\dfrac{3u}{5}$