Tag: mirror formula and magnification

Questions Related to mirror formula and magnification

A glass hemisphere of radius R and of material having refractive index 1.5 is silvered on its flat face as shown in figure . a small object of height h is located at distance 2R from the surface of hemisphere as shown in the figure. the final image will form

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

  1. At a distance of R from silvered surface, on the right side

  2. on the object itself

  3. at hemisphere surface

  4. refractive index

Show answer
Correct Option: B

A small piece of wire bent into L shape such that the upright and horizontal portions are of equal length. It is placed with the horizontal portion along the axis of concave mirror of radius of curvature 20 cm. If the bend is 40 cm from the pole of the mirror, then the ratio of the length of the images of the upright and horizontal portions of the wire 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. 1 : 9

  2. 1 : 3

  3. 3 : 1

  4. 2 :1

Show answer
Correct Option: C
Explanation:
$u= -40$ ; $f= -10$

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

$\dfrac{1}{v}-\dfrac{1}{40}=\dfrac{-1}{10}$

$v=\dfrac{-40}{3}$

magnification=$\dfrac{-v}{u}=\dfrac{-1}{3}$

lateral magnification is $\dfrac{f^{2}}{(u-f)^{2}}$

                                 =  $\dfrac{100}{(40-10)^{2}}$

                                 =  $\dfrac{1}{9}$

ratio of up-right portion to lateral portion is $\dfrac{\dfrac{1}{3}}{\dfrac{1}{9}}=3:1$, hence option $C$ is correct 

A small piece of wire bent into an L shape, with upright and horizontal portions of equal lengths, is placed with the horizontal portion along the axis of the concave mirror whose radius of curvature is 10 cm. If the bend is 20 cm from the pole of the mirror, then the ratio of the lengths of the images of the upright and horizontal portions of the wire 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. 1:2

  2. 3:1

  3. 1:3

  4. 2:1

Show answer
Correct Option: B
Explanation:
Given, $u= -20$ ; $f= -5$

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

$\dfrac{1}{v}-\dfrac{1}{20}=\dfrac{-1}{5}$

$v=\dfrac{-20}{3}$

magnification=$\dfrac{-v}{u}=\dfrac{-1}{3}$

lateral magnification is $\dfrac{f^{2}}{(u-f)^{2}}$

                                 =  $\dfrac{25}{(20-5)^{2}}$

                                 =  $\dfrac{1}{9}$
Ratio of up-right portion to lateral portion is $\dfrac{\dfrac{1}{3}}{\dfrac{1}{9}}=3:1$

option $B$ is correct 

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:

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